Which statement correctly defines the sample space of a random experiment?
Any single outcome that the experimenter expects to occur
A collection of outcomes whose probabilities sum to less than one
The most likely outcome among all possibilities
The set of all possible outcomes of the experiment
Correct answer: The set of all possible outcomes of the experiment
The correct choice is that the sample space is the set of all possible outcomes of the experiment. By definition, the sample space (often denoted S or Omega) lists every outcome the experiment could produce, and any event is a subset of it. It is not a single outcome, a partial collection, nor the most likely outcome, since it must contain every possibility.
Under the axioms of probability, what value must the probability of the entire sample space equal?
0
0.5
1
Infinity
Correct answer: 1
The correct answer is 1. The probability axioms (Kolmogorov's axioms) require that P(S) = 1 for the sample space S, meaning some outcome in S is certain to occur. A value of 0 would mean the experiment yields nothing, 0.5 has no basis, and probabilities can never exceed 1 or be infinite.
For any event A in a sample space, which inequality must always hold according to the axioms of probability?
-1 <= P(A) <= 1
0 <= P(A) <= 1
0 < P(A) < 1
P(A) >= 1
Correct answer: 0 <= P(A) <= 1
The correct statement is 0 <= P(A) <= 1. A core axiom of probability is nonnegativity combined with the bound that no event can be more certain than the whole sample space, so every probability lies between 0 and 1 inclusive. Negative probabilities are disallowed, the strict inequalities wrongly exclude impossible and certain events, and probabilities cannot exceed 1.
Two events A and B are mutually exclusive. What is the probability that both A and B occur on a single trial?
P(A)P(B)
P(A) + P(B)
1
0
Correct answer: 0
The correct value is 0. Mutually exclusive (disjoint) events cannot happen at the same time, so their intersection is empty and P(A and B) = 0. The product P(A)P(B) would apply only to independent events, the sum gives the probability of their union, and 1 has no justification here.
For events A and B that are NOT necessarily mutually exclusive, which expression correctly gives P(A union B)?
P(A) + P(B)
P(A)P(B)
P(A) + P(B) - P(A and B)
P(A) + P(B) + P(A and B)
Correct answer: P(A) + P(B) - P(A and B)
The correct expression is P(A) + P(B) - P(A and B). The general addition rule subtracts the overlap once because outcomes in both events are otherwise counted twice. Simply adding P(A) and P(B) double-counts the intersection, the product applies to independence not unions, and adding the intersection inflates the result further.
An insurer finds that 30% of policyholders file an auto claim, 20% file a home claim, and 8% file both in a year. What is the probability a randomly chosen policyholder files at least one of these claims?
0.42
0.50
0.58
0.34
Correct answer: 0.42
The correct probability is 0.42. Using the addition rule, P(auto or home) = 0.30 + 0.20 - 0.08 = 0.42, where subtracting the 0.08 overlap prevents double-counting those who file both. Adding without subtracting gives 0.50, and the other values come from incorrect arithmetic.
Events A and B are independent with P(A) = 0.6 and P(B) = 0.5. What is P(A and B)?
1.1
0.5
0.3
0.6
Correct answer: 0.3
The correct answer is 0.3. For independent events the multiplication rule gives P(A and B) = P(A)P(B) = 0.6 x 0.5 = 0.3. Adding the probabilities produces 1.1 which is invalid, while 0.5 and 0.6 simply restate the inputs.
Which condition is both necessary and sufficient for two events A and B (each with positive probability) to be independent?
P(A and B) = P(A)P(B)
P(A and B) = 0
P(A union B) = P(A) + P(B)
P(A) = P(B)
Correct answer: P(A and B) = P(A)P(B)
The defining condition is P(A and B) = P(A)P(B). Independence means the occurrence of one event does not change the probability of the other, which is captured exactly by the joint probability factoring into the product. An intersection of 0 describes mutually exclusive events, the additive relation describes disjoint events, and equal marginal probabilities are unrelated to independence.
A fair coin is flipped three times, with flips independent. What is the probability of obtaining heads on all three flips?
21
83
81
61
Correct answer: 81
The correct probability is 1/8. Because the flips are independent, the multiplication rule gives 212121 = 1/8. The value 83 confuses this with exactly one specific arrangement count, 21 ignores the multiple flips, and 61 has no basis.
The conditional probability P(A given B) is defined for P(B) > 0 by which formula?
P(A and B) / P(A)
P(A) / P(B)
P(A and B) x P(B)
P(A and B) / P(B)
Correct answer: P(A and B) / P(B)
The correct definition is P(A and B) / P(B). Conditional probability rescales the joint probability of A and B by the probability of the conditioning event B, reflecting that we now restrict attention to outcomes where B occurred. Dividing by P(A) reverses the roles, the ratio of marginals ignores the joint behavior, and multiplying does not yield a conditional probability.
In a group, 40% of people own a dog and, among dog owners, 25% also own a cat. What is the probability a randomly chosen person owns both a dog and a cat?
0.65
0.10
0.625
0.15
Correct answer: 0.10
The correct probability is 0.10. Using P(dog and cat) = P(dog) x P(cat given dog) = 0.40 x 0.25 = 0.10. Adding the percentages gives 0.65, dividing gives 0.625, and 0.15 results from incorrect subtraction; only the multiplication rule with the conditional probability is valid here.
A box contains 7 red and 3 blue marbles. Two marbles are drawn without replacement. What is the probability that both drawn marbles are red?
10049
157
4521
103
Correct answer: 157
The correct probability is 7/15. Without replacement, P(both red) = 10796 = 9042 = 157, where the second factor reflects that one red marble is already removed. The value 10049 wrongly assumes replacement, 4521 is unsimplified and incorrect, and 103 ignores the joint draw.
How many distinct 3-letter arrangements (order matters, no repeats) can be formed from the 5 letters A, B, C, D, E?
10
15
60
125
Correct answer: 60
The correct count is 60. Because order matters and letters are not repeated, this is a permutation: 5 x 4 x 3 = 60. The value 10 is the number of combinations (order ignored), 15 has no basis, and 125 comes from allowing repeats (5 cubed).
A committee of 3 people is to be chosen from a group of 8 people, where order does not matter. How many different committees are possible?
56
24
336
512
Correct answer: 56
The correct count is 56. Since order does not matter, this is a combination: 8 choose 3 = (8 x 7 x 6)/(3 x 2 x 1) = 56. The value 336 is the permutation count where order matters, 24 and 512 come from incorrect setups.
An urn has 4 defective and 6 working components. Three are selected at random without replacement. Which expression gives the probability that exactly one is defective?
The correct expression is (4 choose 1)(6 choose 2) / (10 choose 3). The numerator counts ways to pick exactly one of the 4 defectives and two of the 6 working units, while the denominator counts all ways to choose 3 of the 10 total components. The other denominators use wrong totals, and omitting the (6 choose 2) factor undercounts the favorable selections.
In a Venn diagram, region A contains 18 elements total, region B contains 12 elements total, and 5 elements lie in the overlap of A and B. How many elements lie in A but not in B?
18
13
23
7
Correct answer: 13
The correct count is 13. The elements in A but not B equal the total in A minus those in the overlap: 18 - 5 = 13. The value 18 ignores the shared region, 7 mistakenly uses B's count, and 23 incorrectly adds rather than subtracts.
A survey shows 70% of customers like product X and 55% like product Y, while 90% like at least one of the two. Using a Venn-diagram approach, what percentage like both products?
35%
25%
45%
15%
Correct answer: 35%
The correct percentage is 35%. By the inclusion-exclusion (addition) rule, P(both) = P(X) + P(Y) - P(at least one) = 0.70 + 0.55 - 0.90 = 0.35. The other values come from adding instead of subtracting or from arithmetic errors in applying the overlap relationship.
A factory has machines M1, M2, and M3 producing 50%, 30%, and 20% of output, with defect rates of 2%, 4%, and 5% respectively. Using the law of total probability, what is the overall probability that a randomly selected item is defective?
0.110
0.032
0.037
0.030
Correct answer: 0.032
The correct probability is 0.032. The law of total probability weights each machine's defect rate by its share of output: (0.50)(0.02) + (0.30)(0.04) + (0.20)(0.05) = 0.010 + 0.012 + 0.010 = 0.032. Simply averaging the rates or summing them gives the other values, which ignore the production weights.
A test for a rare disease has 99% sensitivity and 95% specificity, and the disease affects 1% of the population. Using Bayes' Theorem, which expression gives the probability a person actually has the disease given a positive test result?
The correct expression is (0.99)(0.01) / [(0.99)(0.01) + (0.05)(0.99)]. Bayes' Theorem divides the probability of a true positive (sensitivity times prevalence) by the total probability of a positive result, which adds true positives and false positives, where false positives equal (1 - specificity) times the healthy proportion, namely (0.05)(0.99). The other options misuse specificity in place of the false-positive rate or attach the wrong sensitivity factor to the numerator.
For a continuous random variable X with probability density function f(x), which property must the function satisfy over its entire support?
F(x) must integrate to 1 over the support and be nonnegative everywhere
F(x) must be less than or equal to 1 at every point
F(x) must equal the probability that X takes a specific value
F(x) must be a decreasing function of x
Correct answer: F(x) must integrate to 1 over the support and be nonnegative everywhere
The defining requirements are that f(x) integrate to 1 over the support and be nonnegative everywhere. A valid probability density function must never be negative and must enclose a total area of 1, since total probability is certain. A density value can exceed 1, the probability that a continuous variable equals an exact point is 0 rather than f(x), and a density need not be decreasing.
The cumulative distribution function F(x) of a random variable gives which quantity?
The probability density at the point x
The probability that X exceeds x
The probability that X is less than or equal to x
The expected value of X up to the point x
Correct answer: The probability that X is less than or equal to x
The cumulative distribution function gives the probability that X is less than or equal to x, written F(x) = P(X <= x). It accumulates probability from the lower tail up to the point x. The probability of exceeding x is the complement 1 - F(x), the density is the derivative of F rather than F itself, and F is not an expected value.
For a continuous random variable, how is the probability density function obtained from the cumulative distribution function F(x)?
By integrating F(x)
By evaluating 1 - F(x)
By squaring F(x)
By taking the derivative of F(x) with respect to x
Correct answer: By taking the derivative of F(x) with respect to x
The density is obtained by taking the derivative of F(x) with respect to x, since f(x) = F'(x) for a continuous variable. The cumulative distribution function accumulates the density, so differentiation reverses that accumulation. Integrating F gives no meaningful density, the complement 1 - F is the survival function, and squaring has no basis.
A discrete random variable X takes values 0, 1, and 2 with probabilities 0.5, 0.3, and 0.2 respectively. What is the expected value of X?
1.0
0.5
1.5
0.7
Correct answer: 0.7
The expected value is 0.7. Computing the probability-weighted sum gives (0)(0.5) + (1)(0.3) + (2)(0.2) = 0 + 0.3 + 0.4 = 0.7. The value 1.0 is the simple average of the outcomes ignoring probabilities, while 0.5 and 1.5 come from incorrect weighting.
For a random variable X, which identity correctly expresses the variance in terms of expected values?
E(X2)−[E(X)]2
[E(X)]2−E(X2)
E(X2) - E(X)
E[(X)2]+[E(X)]2
Correct answer: E(X2)−[E(X)]2
The correct identity is E(X2)−[E(X)]2. Variance is the expected squared deviation from the mean, which algebraically simplifies to the second moment minus the square of the first moment. Reversing the terms can give a negative result, subtracting E(X) instead of its square is dimensionally wrong, and adding the terms does not yield variance.
A loss distribution has a mean of 200 and a standard deviation of 50. What is the coefficient of variation?
4.0
0.25
250
0.50
Correct answer: 0.25
The coefficient of variation is 0.25. It is defined as the standard deviation divided by the mean, so 50 divided by 200 equals 0.25. The value 4.0 inverts the ratio, 250 adds the two figures, and 0.50 results from an arithmetic error; only the ratio of standard deviation to mean is correct.
The moment generating function of a random variable X is M(t). How is the expected value E(X) obtained from it?
Take the first derivative of M(t) and evaluate at t = 0
Evaluate M(t) at t = 1
Take the second derivative of M(t) and evaluate at t = 0
Integrate M(t) from 0 to 1
Correct answer: Take the first derivative of M(t) and evaluate at t = 0
The expected value is found by taking the first derivative of M(t) and evaluating at t = 0, giving M'(0) = E(X). The moment generating function produces moments through successive derivatives at the origin. The second derivative at 0 gives the second moment E(X2), evaluating at t = 1 has no general meaning, and integration does not recover a moment.
For a continuous distribution, the value m that satisfies F(m) = 0.5, where F is the cumulative distribution function, is called what?
The mode
The median
The mean
The variance
Correct answer: The median
That value is the median. The median splits a distribution so that half the probability lies below it, which is exactly the condition F(m) = 0.5. The mode is the most likely value where the density peaks, the mean is the probability-weighted average, and the variance measures spread, none of which is defined by F equaling 0.5.
For a continuous density f(x), the mode of the distribution is best described as which of the following?
The value where the cumulative distribution equals one half
The average of all possible values weighted by probability
The point where the density crosses zero
The value of x where the density f(x) attains its maximum
Correct answer: The value of x where the density f(x) attains its maximum
The mode is the value of x where the density f(x) attains its maximum, representing the most likely region of outcomes. The point where the cumulative distribution equals one half is the median, the probability-weighted average is the mean, and a density crossing zero does not define the mode.
For a random variable X, the value c such that P(X <= c) = 0.90 is called which of the following?
The 90th percentile
The 10th percentile
The mean
The interquartile range
Correct answer: The 90th percentile
That value is the 90th percentile. A percentile is the point below which a stated proportion of the distribution lies, so c with P(X <= c) = 0.90 marks the 90th percentile. The 10th percentile would correspond to a cumulative probability of 0.10, the mean is an average rather than a quantile, and the interquartile range measures spread.
In a binomial distribution with n trials and success probability p, what is the expected number of successes?
Np(1-p)
P/n
Np
N/p
Correct answer: Np
The expected number of successes is np. The binomial mean multiplies the number of trials by the per-trial success probability, reflecting the average count over n independent Bernoulli trials. The expression np(1-p) is the variance, while p/n and n/p invert or misplace the factors.
A multiple-choice quiz has 8 independent questions, each answered correctly with probability 0.25 by random guessing. What is the probability of getting exactly 2 correct, expressed using the binomial formula?
(8 choose 2)(0.25)6(0.75)^2
(8 choose 2)(0.25)2(0.75)^6
(0.25)2(0.75)^6
(8 choose 2)(0.25)2(0.75)^8
Correct answer: (8 choose 2)(0.25)2(0.75)^6
The correct expression is (8 choose 2)(0.25)2(0.75)^6. The binomial probability multiplies the number of ways to place 2 successes among 8 trials by the success probability squared and the failure probability raised to the remaining 6 trials. Swapping the exponents misassigns successes and failures, omitting the combination undercounts arrangements, and using exponent 8 exceeds the number of trials.
For a Poisson distribution, which statement about its mean and variance is true?
The variance equals the square of the mean
The mean and variance are both equal to the rate parameter lambda
The mean equals lambda and the variance equals lambda squared
The mean is lambda and the variance is lambda divided by 2
Correct answer: The mean and variance are both equal to the rate parameter lambda
The correct statement is that the mean and variance are both equal to the rate parameter lambda. A distinctive feature of the Poisson distribution is the equality of its mean and variance. The other options incorrectly square lambda or scale it, which would break this defining equidispersion property.
Customer arrivals at a help desk follow a Poisson distribution with a mean of 3 per hour. What is the probability that exactly 0 customers arrive in a given hour?
3 e−3
1 - e−3
e−31
e−3
Correct answer: e−3
The probability is e−3. The Poisson probability of k events is e−λλk / k!, so for k = 0 and lambda = 3 it reduces to e−3 since λ0 and 0! both equal 1. The value 3 e−3 is the probability of exactly one arrival, 1 - e−3 is the probability of at least one, and e−31 uses the wrong rate.
A geometric random variable counts the number of independent trials until the first success, with success probability p on each trial. What is its expected value?
P
1/p2
1/p
(1-p)/p
Correct answer: 1/p
The expected value is 1/p. For the geometric distribution counting trials until the first success, the mean number of trials is the reciprocal of the success probability. The value p is the per-trial probability itself, 1/p2 confuses it with a second-moment quantity, and (1-p)/p is the mean when counting failures before the first success rather than total trials.
A salesperson makes calls until the first sale, with each independent call resulting in a sale with probability 0.2. What is the probability that the first sale occurs on exactly the fourth call?
(0.2)3(0.8)
(0.8)4(0.2)
(0.2)4
(0.8)3(0.2)
Correct answer: (0.8)3(0.2)
The probability is (0.8)3(0.2). The geometric model requires the first three calls to fail, each with probability 0.8, followed by a success on the fourth with probability 0.2. Swapping the roles of success and failure misassigns the probabilities, using a fourth power of 0.8 adds an extra failure, and (0.2)4 ignores the required failures.
The hypergeometric distribution is the appropriate model in which of the following sampling situations?
Counting successes in a fixed number of independent trials with replacement
Sampling a fixed number of items without replacement from a finite population of two types
Counting the number of rare events over a fixed time interval
Counting trials until the first success with replacement
Correct answer: Sampling a fixed number of items without replacement from a finite population of two types
The hypergeometric distribution models sampling a fixed number of items without replacement from a finite population of two types, such as defective and non-defective. Because draws are made without replacement, the trials are dependent. Independent trials with replacement describe the binomial, rare events over time describe the Poisson, and trials until the first success describe the geometric.
A box contains 5 red and 7 green balls. Four balls are drawn at random without replacement. Which expression gives the probability that exactly 2 are red?
The correct expression is (5 choose 2)(7 choose 2) / (12 choose 4). The hypergeometric formula chooses 2 of the 5 red balls and 2 of the 7 green balls to fill the four-ball sample, divided by the number of ways to choose any 4 of the 12 balls. The other options use the wrong green count, the wrong total in the denominator, or omit the green selection factor.
A negative binomial random variable models which of the following quantities?
The number of successes in a fixed number of trials
The number of events in a continuous interval
The number of independent trials needed to achieve a fixed number of successes
The proportion of successes in a large sample
Correct answer: The number of independent trials needed to achieve a fixed number of successes
The negative binomial models the number of independent trials needed to achieve a fixed number of successes, generalizing the geometric distribution to more than one required success. Counting successes in a fixed number of trials is the binomial, events in a continuous interval is the Poisson, and a proportion is not a count distribution.
The standard normal distribution has which mean and standard deviation?
Mean 0 and standard deviation 1
Mean 1 and standard deviation 0
Mean 0 and standard deviation 0
Mean 1 and standard deviation 1
Correct answer: Mean 0 and standard deviation 1
The standard normal distribution has mean 0 and standard deviation 1. Standardizing any normal variable shifts it to be centered at 0 and scaled to unit spread, which is why its tables are universally tabulated. The other options assign nonzero means or zero spread, neither of which describes the standard normal.
A normally distributed claim size has mean 500 and standard deviation 80. What is the standardized z-score corresponding to a claim of 660?
1.6
0.5
2.0
-2.0
Correct answer: 2.0
The z-score is 2.0. Standardizing subtracts the mean and divides by the standard deviation, so (660 - 500) divided by 80 equals 160 divided by 80, which is 2.0. The value 1.6 forgets to divide by the standard deviation properly, 0.5 inverts the ratio, and -2.0 has the wrong sign since 660 lies above the mean.
For a normally distributed variable, approximately what percentage of the distribution lies within one standard deviation of the mean?
50%
95%
99.7%
68%
Correct answer: 68%
Approximately 68% lies within one standard deviation of the mean. Under the empirical rule for the normal distribution, about 68% of probability falls within one standard deviation, about 95% within two, and about 99.7% within three. The value 50% understates the central concentration, while 95% and 99.7% correspond to two and three standard deviations.
For an exponential distribution with a mean of 5, what is the variance?
5
2.5
25
0.2
Correct answer: 25
The variance is 25. For the exponential distribution the variance equals the square of the mean, so with a mean of 5 the variance is 5 squared, which is 25. The value 5 confuses the variance with the mean, 2.5 halves the mean, and 0.2 is the rate parameter rather than the variance.
The exponential distribution is characterized by which special property that makes it model the time until an event without aging?
The symmetry property
The memoryless property
The additivity property
The bounded-support property
Correct answer: The memoryless property
The exponential distribution is characterized by the memoryless property, meaning the probability of waiting an additional amount of time does not depend on how much time has already elapsed. This is why it models lifetimes without aging. Symmetry describes the normal distribution, additivity is not a defining trait here, and the exponential has unbounded support.
A component's lifetime follows an exponential distribution with a mean of 1000 hours. Given that it has already survived 1000 hours, what is the probability it survives at least another 500 hours?
The same as a brand-new component surviving 500 hours
Higher than for a brand-new component because it has proven reliable
Lower than for a brand-new component because it is aging
Zero because it has exceeded its mean lifetime
Correct answer: The same as a brand-new component surviving 500 hours
The probability is the same as a brand-new component surviving 500 hours. The memoryless property of the exponential distribution means the past survival time provides no information about future survival, so a used component behaves like a new one. It is neither more nor less likely to survive based on age, and exceeding the mean does not force failure.
The gamma distribution can be viewed as a generalization of which simpler distribution when its shape parameter equals 1?
The normal distribution
The binomial distribution
The exponential distribution
The uniform distribution
Correct answer: The exponential distribution
The gamma distribution reduces to the exponential distribution when its shape parameter equals 1. A gamma variable can be seen as a sum of independent exponential variables, so a single one recovers the exponential. The normal, binomial, and uniform distributions are not special cases of the gamma at shape parameter 1.
The beta distribution is most commonly used to model a random variable defined over which interval?
All real numbers
The nonnegative real numbers
The integers from 0 to n
The interval from 0 to 1
Correct answer: The interval from 0 to 1
The beta distribution is defined over the interval from 0 to 1, making it a natural model for proportions and probabilities. Its bounded support distinguishes it from distributions on all real numbers like the normal, on the nonnegative reals like the exponential, or on a discrete integer range like the binomial.
A continuous uniform random variable is defined on the interval from 2 to 10. What is its expected value?
6
4
8
5
Correct answer: 6
The expected value is 6. For a continuous uniform distribution the mean is the midpoint of the interval, equal to (2 + 10) divided by 2, which is 6. The values 4 and 8 are interior points but not the center, and 5 is the half-width of the interval rather than its midpoint.
In actuarial loss modeling, what is the key distinction between a loss random variable and a payment random variable?
The loss is always smaller than the payment
The payment is the amount before applying any deductible
They are always equal regardless of policy terms
The loss is the full amount of damage, while the payment is what the insurer actually pays after policy provisions
Correct answer: The loss is the full amount of damage, while the payment is what the insurer actually pays after policy provisions
The key distinction is that the loss is the full amount of damage, while the payment is what the insurer actually pays after policy provisions such as deductibles, limits, and coinsurance are applied. The payment is generally smaller than or equal to the loss, not larger, and it reflects deductibles rather than ignoring them, so the two are not always equal.
An insurance policy has an ordinary deductible of 500. If a covered loss X exceeds the deductible, the payment to the insured equals which of the following?
X
X - 500
500
500 - X
Correct answer: X - 500
The payment equals X - 500. Under an ordinary deductible, the insurer pays the loss amount reduced by the deductible whenever the loss exceeds it, so the insured absorbs the first 500. Paying the full X ignores the deductible, paying a flat 500 confuses the deductible with the payment, and 500 - X is negative for large losses and has no meaning.
A policy pays losses subject to 80% coinsurance after the deductible is satisfied. If the loss above the deductible is 2000, how much does the insurer pay for that portion?
2000
1600
400
2500
Correct answer: 1600
The insurer pays 1600. Under 80% coinsurance, the insurer covers 80% of the eligible loss, so 0.80 times 2000 equals 1600, with the insured responsible for the remaining 20%. Paying the full 2000 ignores the coinsurance, 400 applies only 20%, and 2500 incorrectly grosses up the amount.
An insurance policy has a benefit limit (policy limit) of 10,000. If a covered loss of 13,000 occurs with no deductible, how much does the insurer pay?
13,000
3,000
10,000
11,500
Correct answer: 10,000
The insurer pays 10,000. A benefit limit caps the maximum payment, so any loss above the limit results in a payment equal to the limit itself, leaving the insured to absorb the excess. Paying 13,000 ignores the cap, 3,000 is the uncovered excess rather than the payment, and 11,500 has no basis.
For two discrete random variables X and Y, the joint cumulative distribution function F(x, y) gives which quantity?
The probability that X equals x and Y equals y exactly
The probability that X is at most x and Y is at most y simultaneously
The product of the marginal densities of X and Y
The probability that X is at most x for any value of Y
Correct answer: The probability that X is at most x and Y is at most y simultaneously
The joint cumulative distribution function F(x, y) is the probability that X is at most x and Y is at most y simultaneously, accumulating joint probability over the lower-left region. The exact joint probability at a single point is the joint mass function, not the CDF; a product of marginals describes independence; and accumulating in only one variable while ignoring the other describes a marginal CDF.
A joint cumulative distribution function F(x, y) for two random variables is evaluated as both x and y grow without bound, approaching positive infinity in each variable. What value does F(x, y) approach?
0
Infinity
The larger of the two marginal limits
1
Correct answer: 1
As both x and y approach positive infinity, the joint cumulative distribution function approaches 1, because it accumulates the entire probability mass of the pair. It approaches 0 only when either variable goes to negative infinity, it cannot exceed 1 to become infinite, and a probability is not the larger of two marginal limits.
For independent continuous random variables X and Y, the joint cumulative distribution function F(x, y) can be expressed in terms of their individual cumulative distribution functions in what way?
As the product of the marginal CDF of X and the marginal CDF of Y
As the sum of the marginal CDF of X and the marginal CDF of Y
As the difference of the two marginal CDFs
As the larger of the two marginal CDFs
Correct answer: As the product of the marginal CDF of X and the marginal CDF of Y
For independent variables, the joint cumulative distribution function equals the product of the marginal CDF of X and the marginal CDF of Y, mirroring how independence factors joint probabilities into products. Adding, subtracting, or taking the maximum of the marginal CDFs does not reproduce the joint distribution function of independent variables.
Four independent observations are drawn from a continuous distribution with cumulative distribution function F. An actuary wants the distribution of the largest of the four values. The cumulative distribution function of this maximum at a point t is given by which expression?
Four times F(t)
One minus F(t) raised to the fourth power
F(t) raised to the fourth power
F(t) divided by four
Correct answer: F(t) raised to the fourth power
The cumulative distribution function of the maximum of four independent observations at t is F(t) raised to the fourth power, because the maximum is at most t exactly when all four observations are at most t and independence multiplies their probabilities. Multiplying F(t) by four overcounts, the survival-based one minus F(t) to the fourth power describes the minimum, and dividing has no basis.
Five independent and identically distributed continuous observations are ordered from smallest to largest. The actuary refers to the value that lands in the third position of this ordering. What is this value called?
The sample range
The third order statistic
The sample variance
The fifth order statistic
Correct answer: The third order statistic
The value in the third position of the ordered sample is called the third order statistic, since order statistics are the observations arranged from smallest to largest and indexed by their rank. The sample range is the difference between the largest and smallest, the sample variance measures spread rather than a single ranked value, and the fifth order statistic is the maximum.
For random variables X and Y and constants a and b, the covariance Cov(aX, bY) equals which expression?
A plus b times Cov(X, Y)
Ab times Cov(X, Y)
Cov(X, Y) divided by ab
A squared times b squared times Cov(X, Y)
Correct answer: Ab times Cov(X, Y)
The covariance Cov(aX, bY) equals ab times Cov(X, Y), because covariance is bilinear and constant factors pull straight out of each argument. Adding the constants ignores the multiplicative scaling, dividing reverses the operation, and squaring the constants applies the rule for variance rather than covariance.
A random variable X has variance 7. Using properties of covariance, what is the value of Cov(X, X)?
0
49
1
7
Correct answer: 7
The covariance of a variable with itself, Cov(X, X), equals its variance, so here it is 7. Covariance reduces to variance when both arguments are the same variable; it is not 0 unless the variance is 0, it is not forced to 1, and 49 squares the variance rather than reporting it.
Random variables X, Y, and Z satisfy Cov(X, Y) = 3 and Cov(X, Z) = 5. Using the linearity of covariance, what is Cov(X, Y + Z)?
15
2
8
3.5
Correct answer: 8
By the linearity of covariance in each argument, Cov(X, Y + Z) equals Cov(X, Y) plus Cov(X, Z) = 3 + 5 = 8. Multiplying the two covariances gives 15, subtracting them gives 2, and averaging them gives 3.5, none of which follow from the additive property.
An insurer models X = number of small claims and Y = number of large claims with a joint probability function. Knowing exactly how many small claims occurred changes the probabilities for the number of large claims. What does this dependence tell you about the conditional distribution of Y?
The conditional distribution of Y given X equals the marginal distribution of Y
The conditional distribution of Y given X depends on the observed value of X
The conditional distribution of Y is undefined
The conditional distribution of Y must be uniform
Correct answer: The conditional distribution of Y given X depends on the observed value of X
Because knowing X changes the probabilities for Y, the conditional distribution of Y given X depends on the observed value of X, which is precisely the meaning of statistical dependence. If it equaled the marginal of Y the variables would be independent, the conditional distribution is well defined wherever the conditioning event has positive probability, and there is no reason it must be uniform.
Discrete random variables X and Y have joint probabilities p(0,0) = 0.1, p(0,1) = 0.2, p(0,2) = 0.1, p(1,0) = 0.2, p(1,1) = 0.3, and p(1,2) = 0.1. Given that X = 0, what is the conditional probability that Y = 1?
0.50
0.20
0.40
0.30
Correct answer: 0.50
Given X = 0, the conditional probability that Y = 1 is 0.50. The marginal P(X = 0) = 0.1 + 0.2 + 0.1 = 0.4, and dividing the joint p(0,1) = 0.2 by it gives 0.2 / 0.4 = 0.50. The value 0.20 is the joint probability before conditioning, 0.40 is the conditioning marginal, and 0.30 is a probability tied to a different cell.
Discrete random variables X and Y have a joint probability function p(x, y) = (x + y) / 30 for x in {1, 2} and y in {1, 2, 3, 4}. What is the probability that X = 1 and Y = 3, written as a joint probability value?
303
301
307
304
Correct answer: 304
The joint probability that X = 1 and Y = 3 is 304, found by substituting x = 1 and y = 3 into (x + y)/30 to get (1 + 3)/30 = 4/30. Using 303 plugs in only y, 301 plugs in only x, and 307 mistakenly adds the two values to 7 rather than to 4.
Continuous random variables X and Y have joint density f(x, y) = k(x + y) on the unit square 0 < x < 1 and 0 < y < 1. What value of k makes this a valid joint density?
21
2
1
41
Correct answer: 1
The constant k must be 1. Integrating x + y over the unit square gives the integral of x plus the integral of y, each equal to 21, for a total of 1, so k times 1 = 1 forces k = 1. The value 21 and 41 understate the required normalizer, and 2 doubles it.
A discrete joint table for X and Y has marginal column totals for Y of 0.2, 0.5, and 0.3. The marginal distribution of Y is used to compute which quantity directly?
The conditional distribution of X given Y
The expected value of Y, E[Y]
The covariance of X and Y
The joint probability p(x, y) for a specific pair
Correct answer: The expected value of Y, E[Y]
The marginal distribution of Y, given by its probabilities across all Y values, is used directly to compute the expected value of Y, E[Y], by summing each value of Y weighted by its marginal probability. The conditional distribution of X requires joint probabilities and a Y-marginal denominator, the covariance needs the joint behavior of both variables, and a single joint probability cannot be recovered from a marginal alone.
Continuous random variables X and Y have joint density f(x, y) = 4xy on 0 < x < 1 and 0 < y < 1. What is the marginal density of Y at a point y in (0, 1)?
4y
4xy
Y
2y
Correct answer: 2y
The marginal density of Y at y is 2y. Integrating f(x, y) = 4xy over x from 0 to 1 gives 4y times the integral of x, which is 4y times 21 = 2y. The value 4y omits the factor of 21 from integrating x, 4xy is the joint density before integrating, and y drops the constant entirely.
An actuary aggregates 400 independent and identically distributed claim payments, each with mean 250 and standard deviation 100. Using the Central Limit Theorem, what is the approximate mean and standard deviation of the total payment?
Mean 100,000 and standard deviation 2,000
Mean 250 and standard deviation 100
Mean 100,000 and standard deviation 40,000
Mean 1,000 and standard deviation 5
Correct answer: Mean 100,000 and standard deviation 2,000
The total has approximate mean 100,000 and standard deviation 2,000. Means add, so 400 times 250 = 100,000, and for independent terms variances add, giving 400 times 100 squared = 4,000,000, whose square root is 2,000. Reporting 250 and 100 gives a single claim's values, using 40,000 reports a variance-scaled figure, and the last choice misapplies the count.
Why does the Central Limit Theorem make the normal distribution especially convenient for approximating the total of many independent insurance claims, even when individual claim amounts follow a non-normal distribution?
Because the normal distribution is the only distribution with a finite mean
Because the standardized sum becomes approximately normal as the number of claims grows
Because individual claims become normal once they are added
Because the total of claims always has zero variance
Correct answer: Because the standardized sum becomes approximately normal as the number of claims grows
The Central Limit Theorem is convenient because the standardized sum of many independent claims becomes approximately normal as the number of claims grows, letting actuaries use normal probabilities for aggregate totals. The normal distribution is not the only one with a finite mean, individual claims do not themselves become normal, and a sum of variable claims does not have zero variance.
Two random variables X and Y are known to be independent. What is the value of their correlation coefficient?
1
-1
0
Undefined
Correct answer: 0
If X and Y are independent, their correlation coefficient is 0, because independence makes the covariance 0 and correlation is covariance rescaled by the standard deviations. A correlation of 1 or -1 would indicate a perfect linear relationship, and the correlation is well defined as long as both variables have positive, finite variances.
For independent random variables X and Y with Var(X) = 8 and Var(Y) = 6, what is the variance of the sum X + Y?
2
48
10
14
Correct answer: 14
The variance of X + Y is 14. For independent variables the covariance term is 0, so the variance of the sum is simply Var(X) + Var(Y) = 8 + 6 = 14. The value 2 subtracts the variances, 48 multiplies them, and 10 does not match any valid combination of these variances.
In a sample of independent and identically distributed observations, the sample range is defined using two particular order statistics. Which two?
The largest order statistic minus the smallest order statistic
The median order statistic minus the smallest order statistic
The largest order statistic plus the smallest order statistic
The second-largest order statistic minus the second-smallest order statistic
Correct answer: The largest order statistic minus the smallest order statistic
The sample range is the largest order statistic minus the smallest order statistic, capturing the full spread between the maximum and minimum observed values. Using the median, adding rather than subtracting, or using interior order statistics does not give the range between the extreme values.
De Morgan's laws relate the complement of a union to operations on individual complements. Which expression is correct?
The complement of (A union B) equals (A complement) intersect (B complement)
The complement of (A union B) equals (A complement) union (B complement)
The complement of (A union B) equals A intersect B
The complement of (A union B) equals (A complement) union B
Correct answer: The complement of (A union B) equals (A complement) intersect (B complement)
The correct statement is that the complement of (A union B) equals the intersection of the complements, (A complement) intersect (B complement). De Morgan's law says 'not (A or B)' is the same as 'not A and not B,' since an outcome avoids the union only when it lies outside both sets.
For three events A, B, and C, the inclusion-exclusion principle gives P(A union B union C) as which expression?
P(A)+P(B)+P(C) - P(A and B and C)
P(A)+P(B)+P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C)
P(A)+P(B)+P(C) + P(A and B) + P(A and C) + P(B and C) - P(A and B and C)
P(A)+P(B)+P(C) - P(A and B) - P(A and C) - P(B and C)
Correct answer: P(A)+P(B)+P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C)
The answer adds the single probabilities, subtracts all three pairwise intersections, then adds back the triple intersection: P(A)+P(B)+P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C). This corrects for elements counted multiple times across the overlapping sets.
If the odds in favor of an event are 3 to 2, what is the probability that the event occurs?
0.4
0.667
0.6
1.5
Correct answer: 0.6
The probability is 0.6. Odds in favor of 3 to 2 mean 3 favorable outcomes out of 3+2 = 5 total equally weighted parts, so the probability equals 53 = 0.6.
The probability that an event occurs is 0.8. What are the odds against the event?
4 to 1
1 to 5
4 to 5
1 to 4
Correct answer: 1 to 4
The odds against are 1 to 4. With probability 0.8 of occurring and 0.2 of not occurring, the odds against equal the ratio of failure to success, 0.2 to 0.8, which simplifies to 1 to 4.
A point is chosen uniformly at random on a line segment of length 10. What is the probability that the point lies within distance 2 of the left endpoint?
0.2
0.4
0.1
0.5
Correct answer: 0.2
The probability is 0.2. For a uniform choice on a segment, probability equals favorable length over total length; the points within 2 of the left endpoint form a sub-segment of length 2, so the probability is 102 = 0.2.
A dart lands uniformly at random inside a square of side 4 that contains an inscribed circle of radius 2. What is the probability the dart lands inside the circle?
2π
4π
8π
16π
Correct answer: 4π
The probability is 4π. For uniform placement, probability equals the circle's area over the square's area: the circle area is pi times 2 squared = 4pi, and the square area is 16, giving 164π = 4π.
How many distinct outcomes are possible when a fair six-sided die is rolled four times in sequence?
24
360
1296
4096
Correct answer: 1296
There are 1296 outcomes. By the multiplication (counting) principle, each of the four rolls has 6 possibilities independently, so the total is 6 raised to the 4th power, which equals 1296.
A password consists of 2 distinct letters from the 26-letter alphabet followed by 2 distinct digits from 0-9, with order mattering. How many such passwords are possible?
67600
65000
325 times 45
58500
Correct answer: 58500
There are 58500 passwords. The two distinct ordered letters give 26 times 25 = 650 arrangements, the two distinct ordered digits give 10 times 9 = 90 arrangements, and by the multiplication principle 650 times 90 = 58500.
In how many ways can 5 distinct books be arranged in a row on a shelf?
120
25
60
3125
Correct answer: 120
There are 120 arrangements. The number of orderings of 5 distinct objects is 5 factorial = 5 times 4 times 3 times 2 times 1 = 120.
From 9 distinct points on a circle, how many distinct triangles can be formed using these points as vertices?
504
84
27
729
Correct answer: 84
There are 84 triangles. A triangle is an unordered choice of 3 points from 9, which is the combination (39) = (9 times 8 times 7)/(3 times 2 times 1) = 84.
A class of 20 students splits into a group of 12 and a group of 8 for a project. In how many ways can the group of 12 be chosen?
240
96
125970
2432902008176640000
Correct answer: 125970
There are 125970 ways. Choosing the group of 12 from 20 is the combination (1220), which equals (820) = 125970; selecting the 12 automatically determines the remaining 8.
How many distinct ways can 4 identical red flags and 3 identical blue flags be arranged in a row of 7 positions?
5040
12
210
35
Correct answer: 35
There are 35 arrangements. With repeated identical items, the count is 7 factorial divided by (4 factorial times 3 factorial) = 5040/(24 times 6) = 35, equivalent to choosing which 4 of the 7 positions hold red flags.
Two fair six-sided dice are rolled. What is the probability that the sum of the two dice equals 8?
365
61
91
366
Correct answer: 365
The probability is 5/36. Among the 36 equally likely ordered outcomes, the pairs summing to 8 are (2,6),(3,5),(4,4),(5,3),(6,2) -- exactly 5 outcomes -- giving 5/36.
A fair coin is flipped 5 times. What is the probability of getting exactly 3 heads?
53
165
21
1610
Correct answer: 165
The probability is 5/16. The number of ways to place 3 heads among 5 flips is (35) = 10, and each specific sequence has probability (21) to the 5th = 321, so 3210 = 5/16.
A box has 5 red and 5 blue balls. Three balls are drawn without replacement. What is the probability that all three are the same color?
121
81
61
41
Correct answer: 61
The probability is 1/6. The ways to draw 3 of one color are (35) = 10 for red plus 10 for blue = 20, and the total ways to draw 3 from 10 is (310) = 120, giving 12020 = 1/6.
A drawer contains 6 black socks and 4 white socks. Two socks are pulled out at random without replacement. What is the probability they form a matching pair (same color)?
158
21
31
157
Correct answer: 157
The probability is 7/15. Matching pairs are (26) = 15 black plus (24) = 6 white = 21, out of (210) = 45 total pairs, so 4521 = 7/15.
If P(A) = 0.5, P(B) = 0.6, and A and B are independent, what is the probability that exactly one of the two events occurs?
0.5
0.3
0.8
0.2
Correct answer: 0.5
The probability is 0.5. Exactly one occurs means (A and not B) or (B and not A): 0.5 times 0.4 = 0.20 plus 0.5 times 0.6 = 0.30, summing to 0.50.
Events A and B are independent. Which statement about A complement and B complement must be true?
A complement and B complement are mutually exclusive
A complement and B complement are also independent
A complement and B complement are dependent
P(A complement and B complement) = 0
Correct answer: A complement and B complement are also independent
The correct statement is that A complement and B complement are also independent. If A and B are independent, then their complements are independent as well, since P(A' and B') = P(A')P(B') follows directly from the independence of A and B.
For mutually exclusive events A and B with P(A) = 0.3 and P(B) = 0.45, what is P(A union B)?
0.135
0.615
0.15
0.75
Correct answer: 0.75
The probability is 0.75. For mutually exclusive events the intersection probability is 0, so P(A union B) = P(A) + P(B) = 0.30 + 0.45 = 0.75.
Can two events that are mutually exclusive (each with positive probability) also be independent?
No, because mutual exclusivity forces P(A and B) = 0 while independence requires P(A)P(B) > 0
No, because independent events must be equally likely
Correct answer: No, because mutual exclusivity forces P(A and B) = 0 while independence requires P(A)P(B) > 0
No: mutual exclusivity forces P(A and B) = 0, while independence with positive probabilities requires P(A and B) = P(A)P(B) > 0. These two conditions cannot both hold, so positively probable mutually exclusive events are necessarily dependent.
In a deck of 52 cards, one card is drawn. Given that the card is a face card (Jack, Queen, or King), what is the probability it is a King?
131
524
31
41
Correct answer: 31
The probability is 1/3. There are 12 face cards and 4 of them are Kings, so the conditional probability is 124 = 1/3.
A family has two children, and at least one of them is a girl (assume each child is independently a boy or girl with probability 21). What is the probability that both children are girls?
21
41
32
31
Correct answer: 31
The probability is 1/3. The equally likely outcomes given at least one girl are GG, GB, BG (BB excluded), and only GG has both girls, giving 1/3.
Three cards are drawn without replacement from a standard 52-card deck. What is the probability that all three are hearts?
521351125011
(5213)3
521351135013
133122111
Correct answer: 521351125011
The probability is 521351125011. Using the multiplication rule for dependent draws, each successive heart reduces both the hearts remaining and the total cards remaining, producing this product of conditional probabilities.
A bag has 3 gold and 7 silver coins. Coins are drawn one at a time without replacement until the first gold coin appears. What is the probability the first gold coin appears on the second draw?
103
307
10021
21
Correct answer: 307
The probability is 7/30. The first draw must be silver (107) and the second gold (93), and 10793 = 9021 = 7/30.
An event A satisfies P(A) = 0. Which statement about its complement is guaranteed by the probability axioms?
P(A complement) = 0
P(A complement) is undefined
P(A complement) = 1
P(A complement) = 0.5
Correct answer: P(A complement) = 1
The guaranteed result is P(A complement) = 1. The complement rule states P(A) + P(A complement) = 1, so if P(A) = 0 then P(A complement) = 1.
If A is a subset of B (event A implies event B), which inequality must hold?
P(A) >= P(B)
P(A) = P(B)
P(A) + P(B) = 1
P(A) <= P(B)
Correct answer: P(A) <= P(B)
The required inequality is P(A) <= P(B). Monotonicity of probability states that if A is contained in B, then every outcome in A is also in B, so P(A) cannot exceed P(B).
For any two events, which relationship is always true (Boole's inequality for two events)?
P(A union B) <= P(A) + P(B)
P(A union B) >= P(A) + P(B)
P(A union B) = P(A) + P(B)
P(A union B) <= P(A) P(B)
Correct answer: P(A union B) <= P(A) + P(B)
The always-true relationship is P(A union B) <= P(A) + P(B). Because P(A union B) = P(A) + P(B) - P(A and B) and the intersection probability is nonnegative, the union probability is at most the sum of the individual probabilities.
A jar contains 2 red, 3 green, and 5 blue beads. One bead is drawn at random. What is the probability it is NOT blue?
51
103
21
52
Correct answer: 21
The probability is 1/2. There are 10 beads, 5 of which are blue, so 5 are not blue, giving 105 = 21 by the complement of drawing blue.
Two fair dice are rolled. What is the probability that at least one die shows a 6?
61
31
3612
3611
Correct answer: 3611
The probability is 11/36. Using the complement, the probability that neither die is a 6 is 6565 = 3625, so at least one 6 has probability 1 - 3625 = 11/36.
Three friends each independently choose one of 4 restaurants at random. What is the probability that all three choose the same restaurant?
161
41
641
43
Correct answer: 161
The probability is 1/16. The first friend's choice is free; the second matches with probability 41 and the third with probability 41, so 4141 = 1/16.
In a group of 4 people, assuming birthdays are equally likely among 365 days and independent, which expression gives the probability that all 4 have different birthdays?
(4 times 3 times 2 times 1)/3654
(365 times 364 times 363 times 362)/3654
3654/(365 times 364 times 363 times 362)
(3651)4
Correct answer: (365 times 364 times 363 times 362)/3654
The expression is (365 times 364 times 363 times 362)/3654. The numerator counts ordered selections of 4 distinct birthdays and the denominator counts all equally likely birthday assignments, so their ratio is the probability of no shared birthday.
A multiple-choice quiz has 3 questions, each with 4 options and one correct answer. A student guesses every answer independently and at random. What is the probability of getting all 3 correct?
43
121
641
41
Correct answer: 641
The probability is 1/64. Each guess is correct with probability 41, and since the guesses are independent, the probability of all three correct is (41) cubed = 1/64.
A retailer finds 20% of buyers purchase a warranty, and warranty purchases are independent across buyers. Among 3 randomly selected buyers, what is the probability none purchases a warranty?
0.8
0.2
0.6
0.512
Correct answer: 0.512
The probability is 0.512. Each buyer declines the warranty with probability 0.8, and by independence the probability all three decline is 0.8 cubed = 0.512.
A test for a condition is administered to a population where 2% have the condition. The test has a 95% true-positive rate and a 90% true-negative rate. Which quantity is the denominator in Bayes' formula for P(condition given positive)?
(0.02)(0.95) + (0.98)(0.10)
(0.02)(0.95)
(0.02)(0.95) + (0.98)(0.90)
(0.98)(0.10)
Correct answer: (0.02)(0.95) + (0.98)(0.10)
The denominator is (0.02)(0.95) + (0.98)(0.10). By the law of total probability, the overall probability of a positive test sums the true positives among the diseased and the false positives (1 - 0.90 = 0.10) among the healthy.
An insurer's portfolio is 40% standard and 60% preferred risks, with claim probabilities 0.10 and 0.03. A randomly selected policy files a claim. What is the probability it is a standard risk?
0.04/0.10
0.04/0.058
0.058/0.04
0.04/0.018
Correct answer: 0.04/0.058
The probability is 0.04/0.058. The standard-risk claim probability is (0.40)(0.10) = 0.04, and the total claim probability is 0.04 + (0.60)(0.03) = 0.04 + 0.018 = 0.058, so Bayes' Theorem gives 0.04/0.058.
Events A, B, and C partition the sample space with P(A) = 0.5 and P(B) = 0.3. An event E has P(E given A) = 0.2, P(E given B) = 0.4, and P(E given C) = 0.6. What is P(E)?
0.40
0.30
0.34
0.22
Correct answer: 0.34
The probability is 0.34. With P(C) = 1 - 0.5 - 0.3 = 0.2, the law of total probability gives P(E) = (0.5)(0.2) + (0.3)(0.4) + (0.2)(0.6) = 0.10 + 0.12 + 0.12 = 0.34.
A survey of 200 people finds 120 own a smartphone, 90 own a tablet, and 60 own both. How many own a smartphone but not a tablet?
120
30
90
60
Correct answer: 60
The number is 60. Smartphone owners who do not own a tablet equal total smartphone owners minus those owning both: 120 - 60 = 60.
A point is chosen uniformly at random on the interval from 0 to 10. What is the probability that the chosen point lies between 3 and 7?
0.4
0.7
0.3
0.04
Correct answer: 0.4
The probability is 0.4. For a uniform choice on a line segment, the probability of landing in a subinterval equals the subinterval length divided by the total length: (7-3)/(10-0) = 104 = 0.4. Geometric probability uses length ratios for one-dimensional sample spaces.
A claims adjuster must assign 10 distinct files into two teams of 5 each, where the two teams are labeled Team Day and Team Night. How many ways can the assignment be made?
252
3628800
100000
120
Correct answer: 252
There are 252 ways. Because the teams are labeled, choosing which 5 files go to Team Day determines the rest, giving the combination (510) = 10!/(5!5!) = 252. No additional division by 2 is needed since the labels make the two groups distinguishable.
A system has three components that operate independently, each functioning with probability 0.9. What is the probability that at least one component functions?
0.999
0.729
0.9
0.271
Correct answer: 0.999
The probability is 0.999. The complement of at least one functioning is none functioning, which by independence is (0.1)(0.1)(0.1) = 0.001, so the answer is 1 - 0.001 = 0.999. The complement rule turns an at-least-one question into a single product.
An actuary partitions a portfolio into events E1, E2, and E3 that are mutually exclusive and exhaustive. Which property must this partition satisfy?
The events have no common outcomes and together cover the entire sample space, so their probabilities sum to 1
The events are independent and each has probability 31
The events overlap pairwise but their union is the sample space
The events have no common outcomes but need not cover the sample space
Correct answer: The events have no common outcomes and together cover the entire sample space, so their probabilities sum to 1
A mutually exclusive and exhaustive partition has no shared outcomes and its union is the whole sample space, forcing P(E1)+P(E2)+P(E3) = 1. Mutually exclusive means disjoint events; exhaustive means they collectively exhaust every possible outcome.
A drawer holds 5 red socks and 3 blue socks. Two socks are drawn without replacement. What is the probability that the two socks are of different colors?
2815
145
2813
21
Correct answer: 2815
The probability is 15/28. A mixed pair occurs as red-then-blue or blue-then-red: 8573 + 8375 = 5615 + 5615 = 5630 = 15/28. Summing the two ordered sequences captures both ways to obtain one sock of each color.
A continuous random variable X has density f(x) = c x2 on the interval from 0 to 3 and 0 elsewhere. What is the value of c?
271
31
3
91
Correct answer: 91
The value of c is 1/9. The density must integrate to 1, so the integral of c x2 from 0 to 3 equals c times (327) = 9c, and setting 9c = 1 gives c = 1/9.
A discrete random variable X has probability mass function P(X = k) = c/2k for k = 1, 2, 3, and so on. What is the value of c?
41
2
1
21
Correct answer: 1
The value of c is 1. The probabilities must sum to 1, and the geometric series of 1/2k for k starting at 1 sums to 1, so c times 1 = 1, giving c = 1.
A continuous random variable X has density f(x) = e−x for x greater than or equal to 0. What is the probability that X is greater than 2?
1 - e−2
2 e−2
e−1
e−2
Correct answer: e−2
The probability is e−2. Integrating the density e−x from 2 to infinity gives e−2, which is the survival probability beyond x = 2.
A random variable X has the property that E(X) = 6 and Var(X) = 4. What is E(X2)?
36
10
40
32
Correct answer: 40
The answer is 40. Using Var(X)=E(X2)−(E(X))2, we get E(X2)=Var(X)+(E(X))2=4+36=40.
For a random variable X with E(X) = 5, a transformed variable Y = -2X + 10 is defined. What is E(Y)?
20
-10
5
0
Correct answer: 0
The answer is 0. By linearity of expectation, E(Y) = -2 E(X) + 10 = -2(5) + 10 = 0.
A random variable X has standard deviation 3. A new variable Y = 4X + 2 is defined. What is the standard deviation of Y?
14
144
3
12
Correct answer: 12
The standard deviation of Y is 12. The constant 2 does not affect spread, and multiplying by 4 scales the standard deviation by the absolute value of 4, giving 4 times 3 = 12.
The third central moment of a distribution is positive. What does this indicate about the shape of the distribution?
It is perfectly symmetric
It is negatively (left) skewed
It is positively (right) skewed
It has heavy tails on both sides
Correct answer: It is positively (right) skewed
A positive third central moment indicates the distribution is positively (right) skewed, meaning it has a longer tail extending to the right of the mean.
The kurtosis of a distribution measures which characteristic relative to the normal distribution?
The location of the center
The total area under the curve
The heaviness of the tails and peakedness
The degree of asymmetry
Correct answer: The heaviness of the tails and peakedness
Kurtosis measures the heaviness of the tails and peakedness of a distribution relative to the normal distribution; higher kurtosis indicates more extreme outliers.
A discrete random variable X takes the value 10 with probability 0.4 and the value 20 with probability 0.6. What is the variance of X?
24
100
16
4.9
Correct answer: 24
The variance is 24. The mean is 0.4(10) + 0.6(20) = 16, and E(X2) is 0.4(100) + 0.6(400) = 280, so Var(X) = 280 - 256 = 24.
The moment generating function of a random variable X is M(t) = e3t. Which type of random variable does this describe?
An exponential variable with mean 3
A Poisson variable with mean 3
A constant equal to 3
A normal variable with mean 3
Correct answer: A constant equal to 3
This describes a constant equal to 3. The moment generating function of a degenerate random variable that always equals a constant c is ect, so here c = 3.
The moment generating function of X is M(t). How is the second moment E(X2) obtained?
M(t) evaluated at t = 2
The first derivative of M(t) evaluated at t = 0
The integral of M(t) from 0 to 2
The second derivative of M(t) evaluated at t = 0
Correct answer: The second derivative of M(t) evaluated at t = 0
The second moment E(X2) is the second derivative of M(t) evaluated at t = 0. In general the nth moment equals the nth derivative of the moment generating function at zero.
A continuous random variable X has density f(x) = (41) e−x/4 for x greater than or equal to 0. What is E(X)?
2
16
41
4
Correct answer: 4
The expected value is 4. This is an exponential density with rate 41, so its mean equals the reciprocal of the rate, which is 4.
A loss random variable X is exponential with mean 100. An insurer applies a deductible of 50. By the memoryless property, what is the expected payment per loss that exceeds the deductible?
75
50
100
150
Correct answer: 100
The expected payment per loss exceeding the deductible is 100. By the memoryless property of the exponential distribution, the excess over the deductible has the same exponential distribution with mean 100.
A binomial random variable has 10 trials and success probability 0.5. What is its mode (the most likely number of successes)?
5
10
0
2.5
Correct answer: 5
The mode is 5. For a symmetric binomial with n = 10 and p = 0.5, the most probable outcome is at n times p = 5.
A Bernoulli random variable equals 1 with probability 0.3 and 0 otherwise. What is its variance?
0.09
0.21
0.7
0.3
Correct answer: 0.21
The variance is 0.21. For a Bernoulli variable the variance equals p times (1 - p) = 0.3 times 0.7 = 0.21.
A random variable follows a discrete uniform distribution on the integers 1 through 6 (a fair die). What is its expected value?
3
4
6
3.5
Correct answer: 3.5
The expected value is 3.5. For a discrete uniform on integers 1 through n, the mean is (n + 1)/2, which for n = 6 equals 3.5.
A claim count X follows a Poisson distribution with mean 9. Using the normal approximation, what is the approximate standard deviation used to standardize X?
9
4.5
3
81
Correct answer: 3
The approximate standard deviation is 3. A Poisson variable has variance equal to its mean of 9, so the standard deviation is 9, which is 3.
For a standard normal random variable Z, what is P(Z greater than 0)?
0.6826
0
1.0
0.5
Correct answer: 0.5
The probability is 0.5. The standard normal distribution is symmetric about 0, so exactly half of its area lies above 0.
A normally distributed loss has mean 1000 and standard deviation 100. What loss value corresponds to a z-score of -1.5?
850
950
1150
985
Correct answer: 850
The loss value is 850. Converting back from the z-score gives x = mean + z times standard deviation = 1000 + (-1.5)(100) = 850.
A continuous random variable X is uniform on the interval from 4 to 10. What is the probability that X is greater than 7?
21
103
107
31
Correct answer: 21
The probability is 1/2. The interval above 7 has length 3 out of the total length 6, so the probability is 63 = 1/2.
For an exponential distribution with mean theta, what is the value of its median?
Theta times ln 2
2 theta
Theta
Theta divided by 2
Correct answer: Theta times ln 2
The median equals theta times ln 2. Setting the cumulative distribution 1 - e−x/θ equal to 0.5 and solving gives x = theta times the natural log of 2.
A random variable X has density f(x)=4x3 on the interval from 0 to 1. What is P(X less than 0.5)?
21
161
81
41
Correct answer: 161
The probability is 1/16. The cumulative distribution is x4, so F(0.5) = (0.5)4 = 1/16.
An insurance payment per loss is capped at a policy limit of 8000 with no deductible. If losses are uniform on the interval from 0 to 20000, what is the probability that the insurer pays the full 8000?
0.4
0.2
0.6
0.8
Correct answer: 0.6
The probability is 0.6. The insurer pays the full limit whenever the loss is at least 8000, which for a uniform variable on 0 to 20000 has probability (20000 - 8000)/20000 = 0.6.
A random variable X has E(X) = 8 and E(X2) = 80. By Chebyshev's inequality, at most what fraction of the distribution lies more than 2 standard deviations from the mean?
1
81
21
41
Correct answer: 41
At most 41 lies beyond 2 standard deviations. Chebyshev's inequality bounds this probability by 1 divided by k squared, which for k = 2 gives 41, regardless of the specific moments.
A continuous random variable X has cumulative distribution function F(x)=1−(10+x10)2 for x greater than or equal to 0. What is P(X greater than 10)?
21
101
41
43
Correct answer: 41
The probability is 1/4. The survival function is (10+x10)2, and at x = 10 this equals (2010)2=(21)2=41.
A discrete random variable counts the number of heads in 3 tosses of a fair coin. What is P(X = 2)?
41
81
83
21
Correct answer: 83
The probability is 3/8. There are 3 ways to get exactly 2 heads out of 3 tosses, each with probability 81, giving 3/8.
A random variable X is lognormally distributed, meaning the natural log of X follows which distribution?
A normal distribution
An exponential distribution
A uniform distribution
A Poisson distribution
Correct answer: A normal distribution
For a lognormal random variable, the natural log of X follows a normal distribution; this is the defining property used to model positively skewed quantities like claim sizes.
A random variable X has moment generating function M(t) = (1−2t)−3 for t less than 1/2. What is E(X)?
6
21
2
3
Correct answer: 6
The expected value is 6. This is a gamma moment generating function with shape 3 and scale 2, so the mean is shape times scale = 3 times 2 = 6.
A claim severity X has E(X) = 500 and the policy reimburses losses with a fixed coinsurance factor of 0.8 (no deductible or limit). What is the expected reimbursement?
400
500
100
625
Correct answer: 400
The expected reimbursement is 400. Multiplying every loss by a constant factor scales the expectation, so E(0.8 X) = 0.8 times 500 = 400.
A continuous random variable X has density that is symmetric about x = 5. What is its median?
0
Cannot be determined
5
10
Correct answer: 5
The median is 5. For any distribution symmetric about a point, that point of symmetry is the median because half the probability lies on each side.
A random variable X follows a negative binomial distribution counting failures before the second success with success probability 0.5. What is its expected number of failures?
4
1
2
0.5
Correct answer: 2
The expected number of failures is 2. For this parameterization the mean equals r times (1 - p) divided by p = 2 times 0.5 divided by 0.5 = 2.
A continuous random variable X has density f(x) = 2(1 - x) on the interval from 0 to 1. What is E(X)?
21
32
41
31
Correct answer: 31
The expected value is 1/3. Integrating x times 2(1 - x) from 0 to 1 gives 2 times (21 - 31) = 2 times 61 = 1/3.
A Poisson random variable models defects with mean 2.5 per item. By a property of the Poisson distribution, what is its variance?
5
2.5
6.25
1.58
Correct answer: 2.5
The variance is 2.5. A defining property of the Poisson distribution is that its variance equals its mean, here 2.5.
A random variable X is uniform on the interval from a to b. Which expression gives its variance?
(b−a)2 divided by 12
(a + b) divided by 2
(b - a) divided by 2
(b−a)2 divided by 4
Correct answer: (b−a)2 divided by 12
The variance of a continuous uniform distribution on a to b is (b - a) squared divided by 12, a standard result derived from its constant density.
A random variable X has variance 9. What is the variance of the standardized variable Z = (X - E(X)) divided by 3?
3
1
0
9
Correct answer: 1
The variance is 1. Standardizing by dividing the centered variable by its standard deviation (9, which is 3) always produces a variable with variance 1.
A reinsurance arrangement pays the insurer only the part of a loss above a retention of 100,000 (an excess-of-loss layer). For a loss random variable X, which expression represents this reinsured payment?
0.5 times X
The minimum of X and 100000
X minus 100000 for all X
The maximum of (X - 100000) and 0
Correct answer: The maximum of (X - 100000) and 0
The reinsured payment equals the maximum of (X - 100000) and 0. This excess-of-loss layer pays the amount by which the loss exceeds the retention and pays nothing when the loss is below it.
A continuous random variable X has density f(x)=4x3 on the interval from 0 to 1 and 0 elsewhere. What is the expected value of X?
0.75
0.5
1.0
0.8
Correct answer: 0.8
The expected value is 0.8. E(X) is found by integrating x times f(x) over the support: the integral of x⋅4x3 from 0 to 1 equals the integral of 4x4, which is 54=0.8. The other values come from misapplying the density or confusing the result with the mode at x = 1.
A continuous random variable X has density f(x)=4x3 on the interval from 0 to 1. What is E(X2)?
0.5
0.64
0.667
0.8
Correct answer: 0.667
E(X2) equals 0.667. It is computed as the integral of x2⋅4x3 from 0 to 1, which is the integral of 4x5, equal to 64=32=0.667. The value 0.64 incorrectly squares E(X), and 0.8 is E(X) itself rather than the second moment.
A random variable X has a probability density function f(x) = c on the interval from 3 to 8 and 0 elsewhere. What is the value of the constant c?
0.125
0.4
5
0.2
Correct answer: 0.2
The constant c equals 0.2. For a valid density the total area must equal 1, so c times the interval length (8 - 3 = 5) equals 1, giving c = 51 = 0.2. The value 0.125 would correspond to a width of 8 rather than 5, and 5 is the width itself, not the density height.
The lifetime of a machine part has a hazard rate (force of failure) that is constant over time. Which distribution describes this lifetime?
Normal distribution
Uniform distribution
Exponential distribution
Beta distribution
Correct answer: Exponential distribution
A constant hazard rate corresponds to the exponential distribution. Its memoryless property produces a hazard function h(x) = lambda that does not depend on x, meaning the instantaneous failure rate is the same at every age. The normal, uniform, and beta distributions all have hazard rates that change with x.
A random variable X has moment generating function M(t) = (1−4t)−2 for t less than 1/4. What is E(X)?
4
2
16
8
Correct answer: 8
E(X) equals 8. The first moment is M'(0); differentiating gives M'(t) = 8(1−4t)−3, and evaluating at t = 0 yields 8. This MGF is that of a gamma distribution with shape 2 and scale 4, whose mean alpha times theta equals 2 times 4 = 8. The value 4 confuses the scale with the mean.
For an exponential random variable with mean 10, what is the probability that X exceeds 20?
e−20
e−0.5
1 - e−2
e−2
Correct answer: e−2
The probability is e−2. For an exponential with mean 10 the rate is lambda = 101, and the survival function P(X > x) = e−x/10. Evaluating at x = 20 gives e−1020 = e−2. The expression 1 - e−2 is the CDF (probability of being at or below 20), not the survival probability.
A binomial random variable has 100 trials and success probability 0.25. What is its standard deviation?
18.75
5
4.33
25
Correct answer: 4.33
The standard deviation is 4.33. The variance of a binomial is n p (1 - p) = 100 times 0.25 times 0.75 = 18.75, and the standard deviation is 18.75, which is about 4.33. The value 18.75 is the variance, and 25 is the mean n p.
A random variable X is uniform on the interval from 0 to 10. A new variable is defined as Y = 2X + 3. What is the variance of Y?
33.33
16.67
8.33
100
Correct answer: 33.33
The variance of Y is 33.33. The variance of a uniform on (0, 10) is (10−0)2 / 12 = 12100 = 8.33. Multiplying by 2 scales the variance by 22 = 4, and adding the constant 3 has no effect, so Var(Y) = 4 times 8.33 = 33.33. The value 8.33 is Var(X) before scaling.
A random variable X has variance 9. What is the variance of the transformed variable Y = -4X + 2?
36
146
144
-144
Correct answer: 144
The variance of Y is 144. Variance scales by the square of the multiplicative constant, so Var(-4X + 2) = (−4)2 times Var(X) = 16 times 9 = 144. Variance is never negative, ruling out -144, and the additive constant 2 does not change the variance.
A discrete random variable X takes the value 0 with probability 0.6 and the value 5 with probability 0.4. What is the variance of X?
4
2
10
6
Correct answer: 6
The variance is 6. First E(X) = 0 times 0.6 + 5 times 0.4 = 2, and E(X2) = 0 times 0.6 + 25 times 0.4 = 10. The variance is E(X2)−[E(X)]2=10−4=6. The value 2 is the mean, and 10 is the second moment, not the variance.
The number of claims filed in a day follows a Poisson distribution with mean 6. What is the variance of the number of claims?
3
2.45
36
6
Correct answer: 6
The variance is 6. A defining feature of the Poisson distribution is that its mean and variance are both equal to lambda, so a mean of 6 implies a variance of 6. The value 2.45 is the standard deviation (6), and 36 incorrectly squares the mean.
A continuous random variable X has cumulative distribution function F(x) = 1 - (4/x)2 for x greater than or equal to 4. What is the probability that X is greater than 8?
0.0625
0.75
0.25
0.5
Correct answer: 0.25
The probability is 0.25. The survival function P(X > x) = 1 - F(x) = (4/x)2, and at x = 8 this equals (84)2 = (0.5)2 = 0.25. The value 0.75 is F(8), the probability of being at or below 8, which is the complement.
A geometric random variable X counts the number of failures before the first success, with success probability 0.2 per trial. What is the expected number of failures?
0.2
0.8
5
4
Correct answer: 4
The expected number of failures is 4. For the failures-before-first-success form of the geometric distribution, E(X) = (1 - p)/p = 0.8/0.2 = 4. The value 5 corresponds to the alternative form counting the total trials (1/p), which includes the successful trial.
For a random variable X with E(X) = 5 and standard deviation 2, what is E(X2)?
29
25
21
4
Correct answer: 29
E(X2) equals 29. Using the identity E(X2)=Var(X)+[E(X)]2, and noting Var(X)=22=4, gives E(X2)=4+25=29. The value 25 is the squared mean alone, and 21 incorrectly subtracts rather than adds the variance.
An insurance policy has an ordinary deductible of 300, and losses are uniformly distributed on the interval from 0 to 1000. What is the probability that a loss results in no payment to the policyholder?
0.03
0.3
0.3333
0.7
Correct answer: 0.3
The probability of no payment is 0.3. A payment of zero occurs when the loss falls at or below the deductible of 300. For a uniform loss on (0, 1000), P(X less than or equal to 300) = 1000300 = 0.3. The value 0.7 is the probability that a payment is made.
A random variable X has density f(x) = 0.5 e−0.5x for x greater than 0. What is the median of X?
2
0.693
0.5
1.386
Correct answer: 1.386
The median is about 1.386. This is an exponential distribution with rate 0.5 (mean 2). The median m satisfies F(m) = 1 - e−0.5m = 0.5, giving e−0.5m = 0.5 and m = ln(2)/0.5 = 0.693/0.5 = 1.386. The value 2 is the mean, which exceeds the median for this right-skewed distribution.
A negative binomial random variable counts the number of failures before achieving 3 successes, with success probability 0.5 per trial. What is the expected number of failures?
6
0.5
3
1.5
Correct answer: 3
The expected number of failures is 3. For this form of the negative binomial, E(X) = r(1 - p)/p = 3 times 0.5/0.5 = 3. The value 6 would result from using the total-trials form r/p = 3/0.5, which counts the successful trials as well.
A continuous random variable X has density f(x) = (81)x on the interval from 0 to 4. What is P(X less than 2)?
0.25
0.125
0.75
0.5
Correct answer: 0.25
The probability is 0.25. Integrating the density from 0 to 2 gives the integral of (81)x, which is (161)x2 evaluated from 0 to 2, equal to 164 = 0.25. The value 0.5 would result from treating the density as uniform rather than increasing linearly.
The moment generating function of a random variable is M(t) = e5(et−1). Which distribution does X follow?
Poisson with mean 5
Normal with mean 5
Exponential with mean 5
Binomial with mean 5
Correct answer: Poisson with mean 5
X follows a Poisson distribution with mean 5. The MGF of a Poisson with parameter lambda is eλ(et−1), and matching gives lambda = 5. The binomial MGF has the form (1−p+pet)n, and the exponential and normal MGFs do not involve et inside an exponent of this form.
An exponential random variable has a mean of 4. What is the probability that X is less than its own mean?
0.5
1 - e−4
1 - e−1
e−1
Correct answer: 1 - e−1
The probability is 1 - e−1, about 0.632. For an exponential with mean 4 the CDF is F(x) = 1 - e−x/4, and at x = 4 (the mean) this equals 1 - e−1. Because the exponential is right-skewed, this probability exceeds 0.5, so the median is below the mean. The expression e−1 is the survival probability P(X > 4).
A random variable X has E(X) = 2 and E(X2) = 13. A new variable is defined as Y = 3X. What is the variance of Y?
27
81
9
117
Correct answer: 81
The variance of Y is 81. First Var(X)=E(X2)−[E(X)]2=13−4=9. Scaling by 3 multiplies the variance by 32=9, so Var(Y) = 9 times 9 = 81. The value 9 is Var(X) before scaling, and 27 incorrectly scales by 3 instead of 9.
A binomial random variable has 12 trials and success probability 0.5. Which value is both the mean and the value at which the probability mass function is maximized?
3
0.5
6
12
Correct answer: 6
The value is 6. The mean of a binomial is n p = 12 times 0.5 = 6, and for a symmetric binomial (p = 0.5) the mode coincides with the mean at the center of the distribution. The value 12 is the number of trials, and 0.5 is the success probability.
A loss random variable X is exponential with mean 500. A policy applies an ordinary deductible of 200. What is the expected payment per loss, accounting for losses below the deductible producing zero payment?
200 times e−0.4
500
300
500 times e−0.4
Correct answer: 500 times e−0.4
The expected payment per loss is 500 times e−0.4, about 335. For an exponential loss with mean theta and ordinary deductible d, the expected payment per loss equals theta times e−d/θ due to the memoryless property; here theta = 500 and d/theta = 500200 = 0.4. The value 300 is simply mean minus deductible, which ignores losses falling below the deductible.
A random variable X is normal with mean 0 and standard deviation 1. What is P(X greater than 1.96), approximately?
0.05
0.5
0.025
0.975
Correct answer: 0.025
The probability is approximately 0.025. For the standard normal distribution, P(Z less than 1.96) is about 0.975, so the upper-tail probability P(Z greater than 1.96) = 1 - 0.975 = 0.025. The value 0.05 is the two-tailed probability beyond plus or minus 1.96, not the single upper tail.
A continuous random variable X has density f(x)=3x2 on the interval from 0 to 1. What is the variance of X?
0.5625
0.75
0.0375
0.6
Correct answer: 0.0375
The variance is 0.0375. E(X)= integral of x⋅3x2=43=0.75, and E(X2)= integral of x2⋅3x2=53=0.6. The variance is 0.6−(0.75)2=0.6−0.5625=0.0375. The value 0.5625 is the squared mean, and 0.6 is the second moment.
A discrete random variable X has probability mass function P(X=k)=(21)k for k = 1, 2, 3, and so on. What is the expected value of X?
2
1
0.5
4
Correct answer: 2
The expected value is 2. This is a geometric distribution (counting total trials to first success) with success probability p = 0.5, whose mean is 1/p = 1/0.5 = 2. The value 0.5 is the success probability itself, not the expected number of trials.
An insurance policy has an ordinary deductible of 100 and a maximum covered loss such that the policy limit (maximum payment) is 400. What is the largest ground-up loss at which the payment is still increasing with the loss?
100
500
300
400
Correct answer: 500
The largest such loss is 500. The payment equals loss minus deductible until it reaches the policy limit of 400; that occurs when the ground-up loss equals deductible plus limit = 100 + 400 = 500. Beyond 500 the payment stays capped at 400. The value 400 is the maximum payment, not the loss at which capping begins.
A random variable X has density f(x) = c(1 - x2) on the interval from 0 to 1. What value of c makes f a valid density?
1.5
0.667
2
1
Correct answer: 1.5
The constant c equals 1.5. The integral of (1 - x2) from 0 to 1 is x - x3/3 evaluated at the endpoints, which is 1 - 31 = 2/3. Setting c times 32 = 1 gives c = 23 = 1.5. The value 0.667 is the integral itself, the reciprocal of the correct constant.
A Poisson random variable has mean 3. What is the probability of exactly 2 events?
4.5 e−3
9 e−3
3 e−3
e−3/2
Correct answer: 4.5 e−3
The probability is 4.5 e−3. Using the Poisson mass function e−λλk / k! with lambda = 3 and k = 2 gives e−3 times 29 = 4.5 e−3. The expression 9 e−3 omits division by 2! = 2, and 3 e−3 is the probability of exactly one event.
A gamma distribution with shape alpha and scale theta has mean 12 and variance 36. What is the value of theta?
4
12
3
6
Correct answer: 3
The scale theta equals 3. The mean is alpha theta = 12 and the variance is alpha θ2 = 36. Dividing the variance by the mean gives theta = 1236 = 3. The value 4 would be the shape parameter alpha = 312, and 6 is the standard deviation (36).
A continuous random variable X is uniform on the interval from 0 to 20. What is the conditional probability that X exceeds 15 given that X exceeds 10?
0.25
0.75
0.5
0.333
Correct answer: 0.5
The conditional probability is 0.5. By definition it equals P(X greater than 15) divided by P(X greater than 10) = (205) / (2010) = 0.25/0.5 = 0.5. The value 0.25 is the unconditional P(X greater than 15), which ignores the conditioning event.
A random variable X has third central moment equal to 0 and is described as symmetric. For such a distribution, which measure of skewness applies?
Positive skewness
Negative skewness
Undefined skewness
Zero skewness
Correct answer: Zero skewness
The distribution has zero skewness. Skewness is based on the standardized third central moment, and a symmetric distribution has a third central moment of zero, producing a skewness of zero. Positive skewness indicates a longer right tail and negative skewness a longer left tail, neither of which occurs for a symmetric distribution.
A loss is exponential with mean 1000. A policy has an ordinary deductible of 500. Given that a payment is made, what is the expected payment amount?
1000
1500
500
606
Correct answer: 1000
The expected payment given a payment is made is 1000. By the memoryless property of the exponential distribution, the conditional excess over the deductible has the same exponential distribution with mean 1000, regardless of the deductible amount. The value 500 incorrectly subtracts the deductible from the mean.
A binomial random variable has parameters n = 10 and p = 0.1. Which expression gives the probability of exactly 1 success?
0.1 times 0.99
10 times 0.1 times 0.99
0.910
10 times 0.19 times 0.9
Correct answer: 10 times 0.1 times 0.99
The correct expression is 10 times 0.1 times 0.99. The binomial mass function for exactly 1 success uses the combination of 10 choose 1 = 10, times p1 = 0.1, times (1−p)9 = 0.99. The expression 0.910 is the probability of zero successes, and dropping the combination factor of 10 omits the count of ways to place the single success.
A random variable X has cumulative distribution function F(x) = x2/16 on the interval from 0 to 4. What is the density function f(x) on that interval?
2x
X/16
x2/16
X/8
Correct answer: X/8
The density is x/8. The density function is the derivative of the cumulative distribution function, and differentiating x2/16 with respect to x gives 2x/16 = x/8. The expression x2/16 is the CDF itself, not its derivative, and x/16 omits the factor of 2 from the power rule.
For a right-skewed (positively skewed) continuous distribution with a long tail extending toward larger values, what is the typical ordering of the mean, median, and mode?
Mode is less than the median, which is less than the mean
Mean is less than the median, which is less than the mode
The mean, median, and mode are all equal
The median is greater than both the mean and the mode
Correct answer: Mode is less than the median, which is less than the mean
The correct ordering is mode less than median less than mean. In a right-skewed distribution the long upper tail pulls the mean toward the larger values more strongly than it affects the median, while the mode sits at the peak on the left. This produces the characteristic mode < median < mean relationship, the opposite of the left-skewed case.
A random variable X has mean 50 and standard deviation 5. Using Chebyshev's inequality, what is the smallest guaranteed lower bound on the probability that X lies within 15 units of its mean?
91
98
32
31
Correct answer: 98
The bound is 8/9. Chebyshev's inequality states P(|X - mu| < k*sigma) is at least 1 - 1/k2. Here 15 units equals k = 515 = 3 standard deviations, so the bound is 1 - 91 = 8/9. This bound holds for any distribution with finite variance, regardless of its shape.
For two discrete random variables X and Y, the expected value of their product, E[XY], is computed in what way?
By multiplying E[X] and E[Y] regardless of dependence
By summing x times y times the joint probability p(x, y) over all pairs
By summing x times y over all pairs without any probability weighting
By adding E[X] and E[Y]
Correct answer: By summing x times y times the joint probability p(x, y) over all pairs
E[XY] is found by summing x times y times the joint probability p(x, y) over all pairs, because the expectation of a function of two variables weights each outcome by its joint probability. Multiplying the marginal means only works under independence, omitting the probability weights ignores the distribution, and adding the means computes E[X+Y] instead.
Given Y = y, the conditional expectation of X is E[X given Y = y] = 3y + 1, and E[Y] = 2. Using the law of total expectation, what is E[X]?
6
7
3
4
Correct answer: 7
E[X] is 7. By the law of total expectation, E[X] = E[3Y + 1] = 3 times E[Y] + 1 = 3 times 2 + 1 = 7. The value 6 drops the added constant, 3 reports only the slope, and 4 omits the contribution of Y.
Two independent Poisson random variables have means 3 and 5. Using moment generating functions for the sum, what is the distribution of their sum?
Poisson with mean 15
Normal with mean 8
Poisson with mean 8
Poisson with mean 4
Correct answer: Poisson with mean 8
The sum is Poisson with mean 8. The product of two Poisson moment generating functions yields another Poisson MGF whose mean is the sum 3 + 5 = 8. The value 15 multiplies the means, a normal distribution misidentifies the family, and a mean of 4 averages rather than adds.
For independent continuous random variables X and Y, the joint density f(x, y) factors in what way?
As the product of the marginal density of X and the marginal density of Y
As the sum of the two marginal densities
As the marginal density of X divided by that of Y
As the larger of the two marginal densities
Correct answer: As the product of the marginal density of X and the marginal density of Y
For independent variables, the joint density factors as the product of the marginal density of X and the marginal density of Y. Summing, dividing, or taking the maximum of the marginal densities does not characterize independence and does not integrate to a valid joint density.
The conditional density of Y given X = x is obtained from the joint density f(x, y) by which operation?
Dividing the joint density f(x, y) by the marginal density of X at x
Multiplying the joint density f(x, y) by the marginal density of X at x
Dividing the joint density f(x, y) by the marginal density of Y at y
Subtracting the marginal density of X from the joint density
Correct answer: Dividing the joint density f(x, y) by the marginal density of X at x
The conditional density of Y given X = x equals the joint density f(x, y) divided by the marginal density of X at x, paralleling the definition of conditional probability. Multiplying instead of dividing, dividing by the wrong marginal, or subtracting densities all fail to produce a valid conditional density.
For two random variables, Var(X + Y) is expanded using covariance in which way?
Var(X) + Var(Y) - 2 times Cov(X, Y)
Var(X) + Var(Y) + 2 times Cov(X, Y)
Var(X) + Var(Y) only
Var(X) times Var(Y) + Cov(X, Y)
Correct answer: Var(X) + Var(Y) + 2 times Cov(X, Y)
Var(X + Y) equals Var(X) + Var(Y) + 2 times Cov(X, Y), where the covariance term accounts for how the variables move together. Subtracting the covariance applies to a difference, omitting it assumes independence, and multiplying the variances is not a valid variance rule.
Random variables X and Y have joint density f(x, y) = 2 on the triangle where 0 < x < y < 1, and 0 elsewhere. What is the marginal density of Y at a point y in (0, 1)?
2(1 - y)
2
Y
2y
Correct answer: 2y
The marginal density of Y at y is 2y. Integrating the constant density 2 over x from 0 to y gives 2 times y = 2y, since x ranges up to y on this triangle. The value 2 forgets to multiply by the interval length, y drops the constant, and 2(1 - y) uses the wrong limits for x.
Random variables X and Y have joint density f(x, y) = 2 on the triangle 0 < x < y < 1. What is the marginal density of X at a point x in (0, 1)?
2x
1 - x
2(1 - x)
2
Correct answer: 2(1 - x)
The marginal density of X at x is 2(1 - x). Integrating the density 2 over y from x to 1 gives 2 times (1 - x), since y ranges from x up to 1 for a fixed x. The value 2x integrates over the wrong interval, 1 - x drops the constant, and 2 ignores the variable limits.
An actuary wants to know whether two continuous random variables are independent using only the shape of the support of their joint density. Which condition is required for independence?
The support must be a rectangle aligned with the axes (and the density must factor on it)
The support must be a triangle below the line y equals x
The support must be a circle centered at the origin
The support may be any shape as long as the density is positive
Correct answer: The support must be a rectangle aligned with the axes (and the density must factor on it)
For independence the support must be a rectangle aligned with the axes, with the density factoring into a product of functions of x and y on it. A triangular or circular support links the range of one variable to the other, which forces dependence, and an arbitrary support does not guarantee independence even with a positive density.
Three independent and identically distributed exponential random variables each have mean 10. What is the expected value of their minimum?
10
30
About 3.33
5
Correct answer: About 3.33
The expected value of the minimum is about 3.33. The minimum of three independent exponentials with mean 10 (rate 0.1) is itself exponential with rate 3 times 0.1 = 0.3, so its mean is 1 divided by 0.3, about 3.33. The value 10 is a single variable's mean, 30 multiplies the means, and 5 has no basis.
For a sample of n independent and identically distributed continuous observations, the density of the minimum (first order statistic) at a point t is proportional to which expression?
N times the density only
N times the cumulative distribution function to the power n, times the density
The density divided by n
N times the survival function to the power n minus 1, times the density
Correct answer: N times the survival function to the power n minus 1, times the density
The density of the minimum at t is n times the survival function (one minus F) raised to the power n minus 1, times the density f(t), reflecting that all n observations must exceed t except the one equal to t. Using the CDF instead of the survival function describes the maximum, dividing by n is incorrect, and using n times the density alone omits the survival factor.
A bivariate distribution has E[X] = 4, E[Y] = 3, E[X squared] = 20, E[Y squared] = 13, and E[XY] = 15. What is Cov(X, Y)?
12
15
3
27
Correct answer: 3
Cov(X, Y) is 3. Covariance equals E[XY] minus the product of the means: 15 minus (4 times 3) = 15 - 12 = 3. The value 15 is E[XY] before subtracting, 12 is the product of the means, and 27 adds rather than subtracts.
Using the same data with E[X] = 4 and E[X squared] = 20, what is Var(X)?
20
4
16
36
Correct answer: 4
Var(X) is 4. Variance equals E[X squared] minus the square of the mean: 20 minus 4 squared = 20 - 16 = 4. The value 20 is the second moment, 16 is the squared mean, and 36 adds the two terms instead of subtracting.
A bivariate normal pair (X, Y) has correlation coefficient 0. What does this imply about X and Y in the specific case of the bivariate normal distribution?
X and Y are independent
X and Y are dependent but uncorrelated
X must equal Y
X and Y have equal variances
Correct answer: X and Y are independent
For a bivariate normal pair, a correlation of 0 implies that X and Y are independent, a special property of the normal where zero correlation upgrades to full independence. In general zero correlation does not force independence, the variables need not be equal, and the variances need not match.
Given Y, the conditional variance of X is Var(X given Y = y) = 4y, and E[Y] = 5. The conditional mean E[X given Y = y] = 2y has variance 16. Using the law of total variance, what is Var(X)?
4
20
16
36
Correct answer: 36
Var(X) is 36. The law of total variance gives the mean of the conditional variance, E[4Y] = 4 times 5 = 20, plus the variance of the conditional mean, which is 16, so 20 + 16 = 36. The value 20 keeps only the first piece, 16 keeps only the second, and 4 reports a coefficient rather than a variance.
For discrete X and Y, knowing that X and Y are independent, how does E[XY] simplify?
It equals E[X] plus E[Y]
It equals E[X] times E[Y]
It equals E[X] divided by E[Y]
It equals zero
Correct answer: It equals E[X] times E[Y]
Under independence, E[XY] equals E[X] times E[Y], because independence lets the expectation of the product factor into the product of expectations. Adding or dividing the means is not the product rule, and the product of means is not generally zero.
Continuous random variables X and Y have joint density f(x, y) = x + y on 0 < x < 1 and 0 < y < 1. What is the marginal density of X at a point x in (0, 1)?
X + 1
X + 21
X
21
Correct answer: X + 21
The marginal density of X at x is x + 1/2. Integrating x + y over y from 0 to 1 gives x times 1 plus the integral of y, which is x + 1/2. The value x + 1 overstates the integral of y, x drops it entirely, and 21 omits the x term.
For random variables X and Y, when is the covariance Cov(X, Y) guaranteed to be exactly zero?
Whenever both means are positive
Whenever X and Y have the same mean
Whenever X and Y have the same variance
When X and Y are independent
Correct answer: When X and Y are independent
The covariance is guaranteed to be zero when X and Y are independent, since independence makes E[XY] equal the product of the means. Sharing a common mean, a common variance, or having positive means does not by itself force the covariance to zero.
An actuary applies a linear transformation U = aX + b to a single random variable. How does the variance of U relate to the variance of X?
Var(U) equals a squared times Var(X)
Var(U) equals a times Var(X) plus b
Var(U) equals a times Var(X)
Var(U) equals Var(X) plus b squared
Correct answer: Var(U) equals a squared times Var(X)
Var(U) equals a squared times Var(X), because multiplying by a constant scales variance by that constant squared and adding b shifts the variable without changing spread. Including b, using a to the first power, or adding b squared all misstate how variance behaves under a linear transformation.
Discrete X and Y have a joint table with p(0,0) = 0.2, p(0,1) = 0.1, p(1,0) = 0.3, and p(1,1) = 0.4. What is the marginal probability P(X = 1)?
0.5
0.3
0.4
0.7
Correct answer: 0.7
The marginal P(X = 1) is 0.7, found by summing the joint probabilities over Y for X = 1: p(1,0) + p(1,1) = 0.3 + 0.4 = 0.7. The values 0.3 and 0.4 are individual joint cells, and 0.5 sums the wrong row.
Using the same joint table with p(0,0) = 0.2, p(0,1) = 0.1, p(1,0) = 0.3, p(1,1) = 0.4, what is the conditional probability P(Y = 1 given X = 1)?
0.4
About 0.571
0.7
0.3
Correct answer: About 0.571
The conditional probability is about 0.571. Dividing the joint p(1,1) = 0.4 by the marginal P(X = 1) = 0.7 gives 0.4 / 0.7, approximately 0.571. The value 0.4 is the joint probability before conditioning, 0.7 is the conditioning marginal, and 0.3 is a different joint cell.
An insurer sums losses from 100 independent policies, each with mean 80 and variance 400. What is the variance of the total loss?
8,000
400
40,000
4,000,000
Correct answer: 40,000
The variance of the total is 40,000. For independent terms variances add, so the total variance is 100 times 400 = 40,000. The value 400 is a single policy's variance, 8,000 scales the mean instead, and 4,000,000 squares the count incorrectly.
Two random variables X and Y are uncorrelated, meaning Cov(X, Y) = 0. What can be said about Var(X - Y)?
It equals Var(X) + Var(Y)
It equals Var(X) - Var(Y)
It equals Var(X) + Var(Y) - 2 Cov(X, Y) with a nonzero covariance term
It equals zero
Correct answer: It equals Var(X) + Var(Y)
With zero covariance, Var(X - Y) equals Var(X) + Var(Y), because the covariance term in Var(X - Y) = Var(X) + Var(Y) - 2 Cov(X, Y) vanishes, leaving the sum of variances. Subtracting the variances is never the rule, the covariance term is zero here, and the variance is not zero unless both variances are zero.
A continuous joint density is given by f(x, y) = c on the unit square 0 < x < 1, 0 < y < 1, making X and Y jointly uniform. What is the constant c?
21
2
1
41
Correct answer: 1
The constant c is 1. The area of the unit square is 1, and for a valid uniform density the constant times the area must equal 1, so c = 1. The values 2, 21, and 41 would not integrate to 1 over the unit square.
For jointly uniform X and Y on the unit square, with f(x, y) = 1, are X and Y independent?
It cannot be determined from the density
No, because uniform variables are always dependent
No, because the density is constant
Yes, because the density factors into a product of marginals on a rectangular support
Correct answer: Yes, because the density factors into a product of marginals on a rectangular support
Yes, X and Y are independent because the constant density factors as 1 times 1, the marginals are each uniform on (0,1), and the support is a rectangle. A constant density does not cause dependence, uniform variables are not inherently dependent, and independence is fully determined by the factoring density here.
An actuary needs the conditional expectation E[X given Y = y] for discrete variables. How is it computed?
By summing each x times the joint probability p(x, y)
By summing each x times the conditional probability of X given Y = y
By summing each x times the marginal probability of X
By averaging all possible x values equally
Correct answer: By summing each x times the conditional probability of X given Y = y
The conditional expectation E[X given Y = y] is the sum of each x times the conditional probability of X given Y = y, weighting outcomes by the conditional distribution. Using joint probabilities omits the conditioning denominator, using the marginal of X ignores the conditioning, and equal averaging discards the probabilities.
Discrete X takes values 0 and 1 with conditional probabilities given Y = 2 of 0.25 and 0.75 respectively. What is E[X given Y = 2]?
0.50
0.25
0.75
1.00
Correct answer: 0.75
E[X given Y = 2] is 0.75. The conditional expectation is 0 times 0.25 plus 1 times 0.75 = 0.75. The value 0.25 is the probability that X = 0, 0.50 is an unweighted midpoint, and 1.00 ignores that X can be 0.
For two independent random variables X and Y, what is E[(X + Y) squared] in terms of their moments, given E[X] = 1, E[Y] = 2, Var(X) = 3, and Var(Y) = 4?
16
9
7
25
Correct answer: 16
E[(X + Y) squared] is 16. It equals Var(X + Y) plus the square of E[X + Y]; the variance is 3 + 4 = 7 (independence drops covariance) and the mean is 1 + 2 = 3, so 7 + 3 squared = 7 + 9 = 16. The value 9 is only the squared mean, 7 is only the variance, and 25 squares the sum of all four numbers incorrectly.
Continuous X and Y have joint density f(x, y) = 8xy on 0 < x < y < 1. To find the marginal density of Y, over what range is x integrated?
From y to 1
From 0 to 1
From 0 to y
From y to infinity
Correct answer: From 0 to y
To find the marginal density of Y, x is integrated from 0 to y, because on the region 0 < x < y < 1 the variable x runs from 0 up to the current value of y. Integrating from 0 to 1 ignores the constraint x < y, from y to 1 reverses the inequality, and an infinite upper limit is outside the support.
An actuary models aggregate losses S = X1 + X2 + X3 from three independent policies with means 50, 60, and 70. What is E[S]?
540
60
210,000
180
Correct answer: 180
E[S] is 180. The expectation of a sum is the sum of the expectations, 50 + 60 + 70 = 180, regardless of dependence. The value 60 is a single mean, 210,000 multiplies the means, and 540 triples the total incorrectly.
Two random variables X and Y have correlation 1. What does a correlation of exactly 1 indicate about their relationship?
No relationship at all
A perfect positive linear relationship
A perfect negative linear relationship
A weak positive relationship
Correct answer: A perfect positive linear relationship
A correlation of exactly 1 indicates a perfect positive linear relationship, meaning the points lie exactly on an upward-sloping line. No relationship would be near 0, a perfect negative relationship would be -1, and a weak positive relationship would be a small positive value.
For discrete random variables, the marginal probability function of X is recovered from the joint function p(x, y) by which operation?
Summing p(x, y) over all values of x
Summing p(x, y) over all values of y
Multiplying p(x, y) across all values of y
Taking the maximum of p(x, y) over y
Correct answer: Summing p(x, y) over all values of y
The marginal of X is recovered by summing the joint p(x, y) over all values of y, collapsing out the second variable. Summing over x gives the marginal of Y instead, multiplying across y is not how probabilities aggregate, and taking a maximum discards the rest of the mass.
An actuary computes E[X given Y] and finds it is a constant that does not depend on Y. What does this most directly suggest about the covariance of X and Y?
The covariance equals the variance of X
The covariance of X and Y is positive
The covariance of X and Y is undefined
The covariance of X and Y is zero
Correct answer: The covariance of X and Y is zero
If E[X given Y] is constant in Y, the covariance of X and Y is zero, because a conditional mean that does not move with Y removes the linear association that covariance measures. A constant conditional mean does not produce a positive covariance, the covariance remains well defined, and it does not equal the variance of X.
Continuous random variables X and Y have joint density f(x, y) = 8xy on the triangular region 0 < x < y < 1. What is the marginal density f_X(x) of X for 0 < x < 1?
4x(1 - x2)
8x
4x3
2x(1 - x)
Correct answer: 4x(1 - x2)
The marginal density is f_X(x) = 4x(1 - x2). To get the marginal of X you integrate the joint density over y, and because the support is 0 < x < y < 1, y runs from x to 1: f_X(x) = integral from y = x to 1 of 8xy dy = 8x * [y2/2] from x to 1 = 4x(1 - x2). Forgetting the variable lower limit or using the wrong bounds produces the other expressions.
An actuary models the number of claims N as Poisson and, given N, total severity S has conditional variance Var(S | N) = 100N and conditional mean E[S | N] = 50N. With E[N] = 4 and Var(N) = 4, what is Var(S) using the law of total variance?
10400
400
10000
2400
Correct answer: 10400
The variance of S is 10400. By the law of total variance Var(S) = E[Var(S | N)] + Var(E[S | N]) = E[100N] + Var(50N) = 100 * E[N] + 502 * Var(N) = 100 * 4 + 2500 * 4 = 400 + 10000 = 10400. Using only one of the two components gives 400 or 10000, so both terms must be summed.
Two random variables X and Y are independent and each is exponential with mean 1. What is the conditional density of Y given X = x, denoted f(y | x), for y > 0?
e−y, the same as the unconditional density of Y
e−(x+y)
X e−xy
e−y / e−x
Correct answer: e−y, the same as the unconditional density of Y
The conditional density is e−y, identical to the unconditional density of Y. Because X and Y are independent, conditioning on X = x provides no information about Y, so f(y | x) = f_Y(y) = e−y for y > 0. The form e−(x+y) is the joint density, not the conditional, and the other expressions incorrectly mix in x even though independence removes any dependence on x.
Two events A and B satisfy P(A) = 0.55, P(B) = 0.40, and P(A and B) = 0.22. Are A and B independent?
Yes, because A and B are mutually exclusive
No, because P(A and B) is less than P(A)
No, because P(A) plus P(B) exceeds P(A and B)
Yes, because P(A)P(B) = 0.22 equals P(A and B)
Correct answer: Yes, because P(A)P(B) = 0.22 equals P(A and B)
Yes, A and B are independent. Independence holds exactly when P(A and B) = P(A)P(B). Here P(A)P(B) = 0.55 times 0.40 = 0.22, which matches the given P(A and B) = 0.22, so the events are independent.
A standard 52-card deck is dealt one card. Let A be 'the card is a diamond' and B be 'the card is a face card (J, Q, K).' What is P(A given B)?
523
131
41
133
Correct answer: 41
The answer is 1/4. Among the 12 face cards, exactly 3 are diamonds (jack, queen, king of diamonds), so P(A given B) = 123 = 1/4. Equivalently, the suit is independent of being a face card.
In a town, 35% of households subscribe to streaming service P and 50% subscribe to service Q. If 15% subscribe to both, what proportion subscribe to exactly one of the two services?
0.70
0.65
0.85
0.55
Correct answer: 0.55
The answer is 0.55. Exactly one equals P(P) + P(Q) minus twice the overlap: 0.35 + 0.50 minus 2 times 0.15 = 0.85 minus 0.30 = 0.55.
A drug screen is given to a workforce where 6% of workers use the substance. The test correctly flags users 88% of the time and incorrectly flags non-users 4% of the time. What is the probability a randomly chosen worker tests positive?
0.0528
0.1200
0.0904
0.0376
Correct answer: 0.0904
The answer is 0.0904. By the law of total probability, P(positive) = 0.06 times 0.88 + 0.94 times 0.04 = 0.0528 + 0.0376 = 0.0904.
A relay network sends a signal through two independent links in series; the signal succeeds only if both links work. Each link works with probability 0.95. What is the probability the signal fails to get through?
0.9025
0.0025
0.0500
0.0975
Correct answer: 0.0975
The answer is 0.0975. The signal succeeds only if both work: 0.95 times 0.95 = 0.9025. The failure probability is the complement, 1 minus 0.9025 = 0.0975.
A code is formed by choosing 3 distinct letters from {A, B, C, D, E, F} where order does not matter, then those 3 letters are always written in alphabetical order. How many distinct codes are possible?
120
15
216
20
Correct answer: 20
The answer is 20. Because order does not matter (the letters are forced into alphabetical order), this is a combination: 6 choose 3 = 6!/(3!3!) = 20.
A random variable X has probability mass function P(X = k) = k/10 for k = 1, 2, 3, 4. What is E(X)?
2
2.5
3
3.5
Correct answer: 3
The answer is 3. E(X) = sum of k times P(X=k) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0.
A continuous random variable X has density f(x) = (563)(x2) for x on the interval from 2 to 4 and 0 elsewhere. This is a valid density. What is P(X < 3)?
5627
569
5619
21
Correct answer: 5619
The answer is 19/56. Integrating f from 2 to 3 gives (563) times (x3/3) evaluated from 2 to 3, i.e. (561)(33 - 23) = (561)(27 - 8) = 19/56.
A random variable X has moment generating function M(t)=(0.6et+0.4)8. What is the variance of X?
1.92
2.4
4.8
1.44
Correct answer: 1.92
The answer is 1.92. This MGF is that of a binomial with n = 8 and p = 0.6, so Var(X) = np(1-p) = 8 times 0.6 times 0.4 = 1.92.
Phone calls to a hotline arrive as a Poisson process at 5 per hour. What is the probability that exactly 2 calls arrive in a 30-minute period?
e−2.5 times 2.52
e−5 times 2.52 / 2
e−2.5 times 2.52 / 2
e−5 times 52 / 2
Correct answer: e−2.5 times 2.52 / 2
The answer is e−2.5(2.52)/2. Over 30 minutes the mean is 5 times 0.5 = 2.5, so P(X=2) = e−2.5(2.52)/2! = e−2.5(6.25)/2.
An exponential random variable X has rate parameter 0.5 (mean 2). What is the 90th percentile of X?
-0.5 ln(0.1)
-2 ln(0.9)
2 ln(0.1)
-2 ln(0.1)
Correct answer: -2 ln(0.1)
The answer is -2 ln(0.1). Set F(x) = 1 - e−0.5x = 0.90, so e−0.5x = 0.10, giving x = -ln(0.1)/0.5 = -2 ln(0.1), about 4.61.
A normal random variable X has mean 60 and variance 64. What is P(X > 76) in terms of the standard normal cumulative function (Phi)?
1 - Phi(1)
1 - Phi(16)
Phi(2)
1 - Phi(2)
Correct answer: 1 - Phi(2)
The answer is 1 - Phi(2). The standard deviation is 64, which is 8, so the z-score is (76 - 60)/8 = 2, and P(X > 76) = 1 - Phi(2).
A continuous random variable X has cumulative distribution function F(x)=1−(x3)4 for x at least 3. What is the median of X?
3 times 24
3 times 221
3 times 241
6
Correct answer: 3 times 241
The answer is 3 times 241. Set F(m) = 1 - (3/m)4 = 0.5, so (3/m)4 = 0.5, giving 3/m = 0.541 and m = 3 / 0.541 = 3 times 241, about 3.57.
Items on an assembly line are defective independently with probability 0.05. What is the probability that the first defective item appears on the 4th inspection (using the geometric distribution for trials until first success)?
0.053 times 0.95
0.953
0.954 times 0.05
0.953 times 0.05
Correct answer: 0.953 times 0.05
The answer is 0.953 times 0.05. For the geometric distribution counting trials until the first success, P(X = 4) = (1−p)3 times p = 0.953 times 0.05.
A random variable X has E(X) = 7 and E(X2) = 65. A new variable is defined as W = 0.5X - 4. What is Var(W)?
4
16
2
8
Correct answer: 4
The answer is 4. First Var(X) = E(X2) - E(X)2 = 65 - 49 = 16. For W = 0.5X - 4, Var(W) = 0.52 times Var(X) = 0.25 times 16 = 4.
A discrete random variable X has P(X = -1) = 0.3, P(X = 0) = 0.4, and P(X = 2) = 0.3. What is E(X2)?
0.9
2.0
0.3
1.5
Correct answer: 1.5
The answer is 1.5. E(X2) = (−1)2(0.3) + 02(0.4) + 22(0.3) = 0.3 + 0 + 1.2 = 1.5.
A continuous uniform random variable is defined on the interval from 10 to 30. What is its median?
15
10
20
25
Correct answer: 20
The answer is 20. For a continuous uniform distribution on (a, b), the median equals the midpoint (a + b)/2 = (10 + 30)/2 = 20, by symmetry.
Discrete random variables X and Y have joint probabilities p(0,0) = 0.15, p(0,1) = 0.25, p(1,0) = 0.35, and p(1,1) = 0.25. What is the marginal probability P(Y = 1)?
0.60
0.25
0.40
0.50
Correct answer: 0.50
The answer is 0.50. The marginal P(Y = 1) sums over X: p(0,1) + p(1,1) = 0.25 + 0.25 = 0.50.
Discrete X and Y have the joint table p(0,0) = 0.15, p(0,1) = 0.25, p(1,0) = 0.35, p(1,1) = 0.25. What is the conditional probability P(X = 1 given Y = 0)?
0.70
0.35
0.50
0.30
Correct answer: 0.70
The answer is 0.70. P(X = 1 given Y = 0) = p(1,0) divided by P(Y = 0). Here P(Y = 0) = 0.15 + 0.35 = 0.50, so the result is 0.35/0.50 = 0.70.
Continuous X and Y have joint density f(x, y) = c(x + 2y) on 0 < x < 1 and 0 < y < 1, and 0 elsewhere. What value of c makes f a valid density?
32
1
23
31
Correct answer: 32
The answer is 2/3. Integrating x + 2y over the unit square gives the integral of x (which is 21) plus the integral of 2y (which is 1), totaling 3/2. Setting c times 23 = 1 gives c = 2/3.
Continuous X and Y have joint density f(x, y) = 3x on the region 0 < y < x < 1, and 0 elsewhere. What is the marginal density of X, f_X(x), for x in (0, 1)?
3x2
x3
(23)x2
3x
Correct answer: 3x2
The answer is 3x2. The marginal of X integrates the joint over y from 0 to x: the integral of 3x dy from 0 to x equals 3x⋅x=3x2.
Random variables X and Y have Var(X) = 12, Var(Y) = 27, and correlation coefficient 0.5. What is Cov(X, Y)?
18
6
4.5
9
Correct answer: 9
The answer is 9. Covariance equals correlation times the product of the standard deviations: 0.5 times 12 times 27 = 0.5 times 324 = 0.5 times 18 = 9.
For random variables X and Y, Var(X) = 5, Var(Y) = 8, and Cov(X, Y) = -2. What is Var(3X + 2Y)?
An actuary models aggregate claims S = X1 + X2 + X3 from three independent lines, each with variance 50. What is the standard deviation of S?
50
150
150
3 times 50
Correct answer: 150
The answer is 150. For independent variables, variances add: Var(S) = 50 + 50 + 50 = 150, so the standard deviation is 150, about 12.25.
Given Y = y, the conditional mean of X is E[X given Y = y] = 4y, and Y has mean 3 and variance 2. The conditional variance is Var(X given Y = y) = 6. Using the law of total variance, what is Var(X)?
32
48
38
6
Correct answer: 38
The answer is 38. By the law of total variance, Var(X) = E[Var(X|Y)] + Var(E[X|Y]) = 6 + Var(4Y) = 6 + 16Var(Y) = 6 + 16(2) = 6 + 32 = 38.
A claim count N has mean 5, and given N the total severity S satisfies E[S given N] = 300N. Using the double expectation theorem, what is E[S]?
1505
305
300
1500
Correct answer: 1500
The answer is 1500. By double expectation, E[S] = E[E[S given N]] = E[300N] = 300 times E[N] = 300 times 5 = 1500.
Continuous random variables X and Y are independent, with X uniform on (0, 1) and Y uniform on (0, 2). What is their joint density f(x, y) on that rectangle?
1
41
21
2
Correct answer: 21
The answer is 1/2. For independent variables the joint density factors: f_X(x) = 1 on (0,1) and f_Y(y) = 21 on (0,2), so the product is 1 times 21 = 1/2.
Discrete X and Y have joint probabilities p(1,1) = 0.2, p(1,2) = 0.1, p(2,1) = 0.1, p(2,2) = 0.6. What is E[XY]?
4.0
2.5
3.0
3.3
Correct answer: 3.0
The answer is 3.0. E[XY] sums xy times p(x,y): (1)(1)(0.2) + (1)(2)(0.1) + (2)(1)(0.1) + (2)(2)(0.6) = 0.2 + 0.2 + 0.2 + 2.4 = 3.0.
For independent random variables X and Y with E[X] = 3, E[Y] = 4, Var(X) = 2, and Var(Y) = 5, what is E[XY]?
12
17
19
7
Correct answer: 12
The answer is 12. For independent variables, E[XY] = E[X] times E[Y] = 3 times 4 = 12; the variances are not needed because independence makes the product expectation factor.
Random variables X and Y satisfy Var(X) = 4, Var(Y) = 9, and they are uncorrelated. What is Var(X - Y)?
1
13
36
5
Correct answer: 13
The answer is 13. Because Cov(X, Y) = 0, Var(X - Y) = Var(X) + Var(Y) = 4 + 9 = 13; the covariance cross-term vanishes for uncorrelated variables.
Continuous X and Y have joint density f(x, y) = 6(1 - y) on 0 < x < y < 1, and 0 elsewhere. What is the marginal density of Y, f_Y(y), for y in (0, 1)?
6y(1 - y)
6y
3(1−y)2
6(1 - y)
Correct answer: 6y(1 - y)
The answer is 6y(1 - y). The marginal of Y integrates the joint over x from 0 to y: 6(1 - y) times the length y gives 6y(1 - y).
For random variables X, Y, and Z, Cov(X, Z) = 4 and Cov(Y, Z) = -1. Using the bilinearity of covariance, what is Cov(X + Y, Z)?
3
5
-4
4
Correct answer: 3
The answer is 3. By bilinearity, Cov(X + Y, Z) = Cov(X, Z) + Cov(Y, Z) = 4 + (-1) = 3.
An actuary forms T = X - 2Y where Var(X) = 10, Var(Y) = 4, and Cov(X, Y) = 3. What is Var(T)?
A joint table has P(X = 1, Y = 1) = 0.3, with marginals P(X = 1) = 0.6 and P(Y = 1) = 0.5. Does the independence factorization condition hold at this cell?
No, because 0.3 is greater than 0.5
No, because the marginals sum to more than 1
Yes, because P(X=1) times P(Y=1) equals the joint probability 0.30
It cannot be assessed from probabilities at all
Correct answer: Yes, because P(X=1) times P(Y=1) equals the joint probability 0.30
The answer is yes: P(X=1) times P(Y=1) = 0.6 times 0.5 = 0.30, which equals the joint probability 0.30, so the factorization condition for independence holds at this cell.
Continuous X and Y have joint density f(x, y) = 4xy on 0 < x < 1 and 0 < y < 1. What is E[X]?
21
94
31
32
Correct answer: 32
The answer is 2/3. The marginal of X is f_X(x) = 2x on (0,1), so E[X] = integral of x⋅2x from 0 to 1 = integral of 2x2=32.
Random variables X and Y have Cov(X, Y) = 12, standard deviation of X equal to 4, and standard deviation of Y equal to 6. What is the correlation coefficient?
0.75
0.5
2.0
0.25
Correct answer: 0.5
The answer is 0.5. The correlation equals Cov divided by the product of standard deviations: 12 / (4 times 6) = 2412 = 0.5.
An actuary computes the conditional expectation E[Y given X = x] = 2x + 1 for a pair where E[X] = 4. Using the law of total expectation, what is E[Y]?
9
5
8
4
Correct answer: 9
The answer is 9. By the law of total expectation, E[Y] = E[E[Y given X]] = E[2X + 1] = 2E[X] + 1 = 2(4) + 1 = 9.
Aggregate losses are modeled as S = N times c where N (claim count) has mean 8 and variance 9, and c = 500 is a fixed payment per claim. What is Var(S)?
4000000
4500
2000000
2250000
Correct answer: 2250000
The answer is 2,250,000. For a constant multiplier, Var(S) = Var(500N) = 5002 times Var(N) = 250000 times 9 = 2,250,000.
Continuous random variables X and Y have joint density f(x, y) = c on the triangular region where 0 < y < x < 2. What value of c makes f a valid joint density?
21
41
1
2
Correct answer: 21
The answer is 1/2. A valid joint density must integrate to 1 over its region. The triangle defined by 0 < y < x < 2 is half of the 2-by-2 square, so its area is (21)(2)(2) = 2. For a constant density f = c, the total probability is c times the region's area, giving c times 2 = 1, so c = 1/2.
For two random variables X and Y, the correlation coefficient is defined so that it always lies within which interval, regardless of the distributions involved?
Between -1 and 1, inclusive
Between 0 and 1, inclusive
Between -infinity and infinity
Between 0 and infinity
Correct answer: Between -1 and 1, inclusive
The answer is between -1 and 1, inclusive. The correlation coefficient equals Cov(X, Y) divided by the product of the standard deviations, and the Cauchy-Schwarz inequality guarantees this standardized measure is bounded in magnitude by 1. Thus it always falls in the interval from -1 to 1.
An actuary models a claim count N with mean 6 and, given N = n, the conditional total severity S has mean 200n. Using the double expectation theorem, what is the unconditional expected total severity E[S]?
1200
200
206
1206
Correct answer: 1200
The answer is 1200. By the double expectation theorem, E[S] = E[E[S | N]] = E[200N] = 200 times E[N] = 200 times 6 = 1200. The conditional mean is a linear function of N, so taking its expectation simply substitutes the mean of N.
Random variables X and Y satisfy Var(X) = 9, Var(Y) = 25, and Cov(X, Y) = -6. What is the variance of the difference X - Y?
46
22
34
40
Correct answer: 46
The answer is 46. The variance of a difference is Var(X - Y) = Var(X) + Var(Y) - 2 Cov(X, Y). Substituting gives 9 + 25 - 2(-6) = 34 + 12 = 46. Because the covariance is negative, subtracting it increases the variance of the difference.
Continuous random variables X and Y have joint density f(x, y) = 2 e−xe−2y for x > 0 and y > 0, and zero otherwise. Are X and Y independent?
Yes, because the joint density factors into a function of x times a function of y over a rectangular region
No, because the density contains both x and y in the same expression
No, because the constant 2 cannot be split between the two variables
Cannot be determined without computing the covariance
Correct answer: Yes, because the joint density factors into a function of x times a function of y over a rectangular region
The answer is yes, because the joint density factors into a function of x times a function of y over a rectangular region. Here f(x, y) = (e−x) times (2 e−2y) on the product region x > 0, y > 0, which is exactly the factorization criterion for independence. The constant simply belongs to one of the factors.
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An insurer finds that 30% of policyholders file an auto claim, 20% file a home claim, and 8% file both in a year. What is the probability a randomly chosen policyholder files at least one of these claims?
Pick an answer to see the explanation
Click Start Test above to launch a full-length, 30-question Actuarial Exam P practice test weighted exactly like the real exam, or drill a single topic — General Probability, Univariate Random Variables, or Multivariate Random Variables. Every question includes a clear explanation so you learn the reasoning, not just the answer.
Actuarial Exam P (Probability) is a preliminary actuarial exam that develops knowledge of the fundamental probability tools for quantitatively assessing risk, with an emphasis on applying those tools to problems encountered in actuarial science.
It is administered by the Society of Actuaries (SOA) as a three-hour, computer-based test at Prometric centers, and it is recognized by the Casualty Actuarial Society (CAS) as its Exam 1.[1] A single pass counts toward both societies’ pathways.
These practice questions follow the published Exam P syllabus and topic weightings, mirroring the content and pacing of the real exam so you can build readiness across every topic.[2] To build readiness across all three areas, pair these with our free study guide, flashcards.
Fees, schedules, and policies change — always verify the current details at SOA.org before registering.
Actuarial Exam P at a Glance
Actuarial Exam P at a glance
Detail
Actuarial Exam P
Questions
30 multiple-choice questions (5 options, A-E)
Question type
Multiple choice (computer-based)
Time limit
3 hours (180 minutes)
Result
Scaled score 0-10; a 6 or higher is a pass
Administered by
Society of Actuaries (SOA) at Prometric centers; recognized by the CAS
Schedule
Offered in multiple computer-based testing windows per year
Cost
$275 exam fee (verify at SOA.org)
Topics
General Probability; Univariate Random Variables; Multivariate Random Variables
What Is on Exam P?
Exam P covers three topics totaling 30 multiple-choice questions: Univariate Random Variables (44-50%), General Probability (23-30%), and Multivariate Random Variables (23-30%).[2]
These topics and weight ranges come from the official SOA Exam P syllabus, with Univariate Random Variables carrying the most weight. Our full practice test mirrors these proportions:
Exam P weighting by topic
Univariate Random Variables47% · 44-50%
General Probability27% · 23-30%
Multivariate Random Variables27% · 23-30%
Practice Questions by Topic
Use Start Test for a full weighted Exam P simulation, or open the hub and pick a single topic to drill your weak area. After each full exam, your results show a per-topic breakdown so you know exactly where to focus — most candidates need the most reps on Univariate Random Variables.
Who Is Eligible to Take Exam P?
Exam P is open to anyone pursuing an actuarial credential — there is no degree prerequisite to register through the SOA.[1]
The syllabus assumes you are comfortable with calculus, including series, differentiation, and integration, and that you are familiar with the concepts in the SOA’s “Risk and Insurance” study note.
Most successful candidates are college students or recent graduates in mathematics, statistics, actuarial science, or a related quantitative field. Because Exam P is recognized by both the SOA and the CAS, a single pass advances you toward either society’s credentials.
How Do You Register for Exam P?
You register for Exam P online through the SOA, pay the $275 exam fee, and then schedule your computer-based appointment at a Prometric test center within the testing window.[3]
Exam P is offered in multiple testing windows throughout the year, so you can choose a date that fits your study timeline. Verify the current fee and window dates at SOA.org before registering, as both can change.
After you register, you select your Prometric center and appointment time. Because the computer-based exam runs over several days, statistical scaling keeps all forms comparable in content and passing criteria.
Bring a valid, government-issued photo ID whose name matches your registration, and review the SOA exam rules so nothing on test day comes as a surprise.
How Is Exam P Scored?
Exam P is reported on a scaled score from 0 to 10, and a scaled score of 6 or higher is a pass.[4]
The SOA does not publish a fixed raw cut score. Instead, your raw number-correct is measured against the pass mark for your form and converted to the 0-to-10 scale, which keeps the standard constant across testing windows.
For the computer-based test, unofficial pass/fail results are emailed within the hour to the address you used to schedule your appointment, with official results following after the window. A few unscored pilot questions are mixed into each form and do not affect your result.
How Hard Is Exam P?
Exam P is demanding because it pairs calculus-based probability with strict time pressure — roughly six minutes per question across 30 problems — and recent administrations have shown pass rates near 50 percent.[4] The practical challenge is applying distribution results quickly and accurately, not any single exotic topic.
Univariate Random Variables is the heaviest area, covering the common discrete and continuous distributions, expected value, variance, and insurance applications such as deductibles, coinsurance, and benefit limits.
General Probability rewards fluency with the axioms, combinatorics, conditional probability, and Bayes’ theorem, while Multivariate Random Variables tests joint, marginal, and conditional distributions, covariance, and transformations under time pressure.
0-10
Scaled score range
6+ is a pass
30
Questions total
in 3 hours
~50%
Recent pass rate
SOA administrations
The takeaway: drill until you’re consistently passing full-length, blueprint-weighted practice within the three-hour clock — especially Univariate Random Variables — before you book your exam date.
What to Expect on Exam Day
Arrive at your Prometric test center early to check in — bring a valid, unexpired government-issued photo ID whose name matches your Exam P registration.[3] You’ll store phones and personal items in a locker; no outside notes are allowed.
A normal-distribution table is provided on screen under an Exhibit button, so you do not need to bring one. You then work through 30 multiple-choice questions in the three-hour session, answering every item since unanswered questions are scored as incorrect.
Unofficial pass/fail results arrive by email within the hour, and the SOA releases official results after the window closes. Having simulated the full timing with practice tests makes the three-hour clock feel routine.
How to Use This Exam P Practice Test
Recreate exam conditions. Take the full 30-question test timed, with no notes.[2]
Diagnose, then drill. Use a full Exam P simulation to find weak topics, then drill them.
Prioritize Univariate Random Variables. It carries the most weight on the exam.
Learn the why. Read every explanation — understanding the calculus beats memorizing.
Answer everything. Unanswered questions are scored incorrect, so never leave one blank.
Why Exam P Matters
Passing Exam P is the first concrete milestone on the actuarial career path — it proves you can apply probability and calculus to real risk problems, and because it is recognized by both the SOA and the CAS, a single pass opens doors on either credentialing track.[1] Employers treat a passed preliminary exam as a strong signal when hiring entry-level actuaries and interns. These free Exam P practice tests are the most efficient way to get there.
Conclusion
Performing well on Exam P comes down to calculus-based probability fluency and the timing to sustain it across 30 questions in three hours. Use this free Actuarial Exam P practice test to find your weak topics, drill them to mastery, and pair it with our free study guide, flashcards to walk in confident on test day.
Actuarial Exam P Practice Test FAQ
Exam P (Probability) is a preliminary actuarial exam administered by the Society of Actuaries (SOA) that develops knowledge of the fundamental probability tools for quantitatively assessing risk. It is intended for students and early-career professionals pursuing an actuarial credential, and it is recognized by the Casualty Actuarial Society (CAS) as its Exam 1, so a single pass counts toward both societies' pathways.
Exam P is a three-hour exam that consists of 30 multiple-choice questions, administered as a computer-based test (CBT). Each problem has five answer choices labeled A through E, and only one is correct. A few unscored pilot questions are mixed in, and all unanswered questions are scored as incorrect, so you should answer every question.
Exam P is reported on a scaled score from 0 to 10, and a scaled score of 6 or higher is a pass. The SOA does not publish a fixed raw cut score; your raw number-correct is measured against the pass mark for your form and converted to the 0-to-10 scale. In practice the pass mark has worked out to roughly 70 to 75 percent correct, but the official standard is the scaled 6.
Exam P is challenging because it pairs calculus-based probability with strict time pressure — about six minutes per question across 30 problems — and recent SOA administrations have shown pass rates near 50 percent. The difficulty is less about any single exotic topic and more about applying univariate and multivariate distribution results quickly and accurately under exam conditions, which is exactly what timed, blueprint-weighted practice builds.
You register for Exam P online through the SOA, and the exam fee is $275 (verify the current amount at SOA.org, since fees change). After registering you schedule your computer-based appointment at a Prometric test center within the testing window. Exam P is offered in multiple windows throughout the year, so you can choose a date that fits your study timeline.
Exam P covers three topics with official weight ranges: General Probability (23 to 30 percent), Univariate Random Variables (44 to 50 percent), and Multivariate Random Variables (23 to 30 percent). Univariate Random Variables is the largest area and includes the common discrete and continuous distributions, expected value, variance, and insurance applications such as deductibles and benefit limits.
For the computer-based test, unofficial pass/fail results are sent within the hour to the email address you used to schedule your appointment. Official results are released by the SOA after the testing window. Because Exam P is delivered over several days, statistical scaling is used to keep all forms comparable in content and passing criteria.
The most effective preparation is repeated, full-length, blueprint-weighted practice under timed conditions, with extra reps on Univariate Random Variables since it carries the most weight. Read every rationale to learn the calculus-and-probability reasoning, not just the answer, and reinforce weak topics between sessions with a study guide, flashcards, and a cheat sheet.
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