Your FREE Exam P Flashcards 2026 – 200+ Cards

$P(B\mid A)=\dfrac{P(A\mid B)\,P(B)}{P(A)}$ — reverses a conditional probability to update a cause given evidence.

The set $S$ (or $\Omega$) of all possible outcomes of a random experiment. An event is any subset of $S$.

Any subset of the sample space. Its probability is a number in $[0,1]$.

$P(A)\ge 0$; $P(S)=1$; for disjoint events $P\!\left(\bigcup A_i\right)=\sum P(A_i)$.

$P(A^{c})=1-P(A)$.

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

Events that cannot both occur: $P(A\cap B)=0$, so $P(A\cup B)=P(A)+P(B)$.

$P(A\cap B)=P(A)\,P(B)$, equivalently $P(A\mid B)=P(A)$. Independence is about influence, not overlap.

$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ for $P(B)>0$.

$P(A\cap B)=P(A\mid B)\,P(B)=P(B\mid A)\,P(A)$.

For a partition $B_1,\dots,B_n$: $P(A)=\sum_i P(A\mid B_i)\,P(B_i)$.

Arrangements where order matters: $\,_nP_r=\dfrac{n!}{(n-r)!}$.

Selections where order does not matter: $\binom{n}{r}=\dfrac{n!}{r!\,(n-r)!}$.

If stage 1 has $n_1$ outcomes and stage 2 has $n_2$, the process has $n_1\times n_2$ outcomes.

$(A\cup B)^c=A^c\cap B^c$ and $(A\cap B)^c=A^c\cup B^c$.

$P(A\cup B\cup C)=\sum P(\cdot)-\sum P(\cdot\cap\cdot)+P(A\cap B\cap C)$.

If $P(A)=p$, the odds against are $(1-p):p$.

When $N$ outcomes are equally likely, $P(A)=\dfrac{\#\,\text{outcomes in }A}{N}$.

A collection of mutually exclusive, exhaustive events whose union is $S$.

$A$ and $B$ are independent given $C$ if $P(A\cap B\mid C)=P(A\mid C)\,P(B\mid C)$.

Pairwise independence (every pair independent) does not imply mutual independence of all events together.

$P(A\cup B)=P(A)+P(B)$ when $A$ and $B$ are mutually exclusive.

$P\!\left(\bigcup A_i\right)\le \sum P(A_i)$.

Successive draws are dependent; counts follow the hypergeometric distribution.

Successive draws are independent and identically distributed; counts follow the binomial distribution.

A diagram of sequential events whose branch probabilities multiply along a path.

$P(A)=\sum_i P(A\mid B_i)P(B_i)$ weights each conditional by how likely its cause is.

In Bayes, $P(B)$ is the prior and $P(B\mid A)$ — updated by evidence $A$ — is the posterior.

$P(\varnothing)=0$.

For any event $A$, $0\le P(A)\le 1$.

For events of positive probability, mutually exclusive implies dependent (not independent).

$P(A^c\mid B)=1-P(A\mid B)$.

$A\,\triangle\,B=(A\setminus B)\cup(B\setminus A)$ — outcomes in exactly one of $A,B$.

$P(B\mid A)=\dfrac{P(A\mid B)P(B)}{P(A\mid B)P(B)+P(A\mid B^c)P(B^c)}$.

If $A\subseteq B$ then $P(A)\le P(B)$.

$n!=n(n-1)\cdots 2\cdot 1$, with $0!=1$ by convention.

$\sum_{r=0}^{n}\binom{n}{r}=2^n$ — the number of subsets of an $n$-element set.

$\binom{n}{r}=\binom{n}{n-r}$.

$P(\text{at least one})=1-P(\text{none})$ — use the complement.

$P(A\cap B\cap C)=P(A)\,P(B\mid A)\,P(C\mid A\cap B)$.

Disjoint events whose union is the whole sample space; their probabilities sum to 1.

If $A,B$ are independent, so are $A,B^c$ and $A^c,B^c$.

A single outcome (element) of the sample space.

Choosing $r$ from $n$ with order and repetition allowed gives $n^r$ outcomes.

Over many trials, $P(A)$ is approximated by the fraction of trials in which $A$ occurs.

The $P(A)$ in Bayes' theorem is computed by the law of total probability.

$\dfrac{n!}{n_1!\,n_2!\cdots n_k!}$ counts arrangements of $n$ items in $k$ groups.

Counts rare events at rate $\lambda$: $P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}$; mean and variance both $\lambda$.

A function assigning a number to each outcome. Discrete (countable values) or continuous (interval of values).

For a discrete variable, $p(x)=P(X=x)$; values are $\ge 0$ and $\sum_x p(x)=1$.

For a continuous variable, $f(x)\ge 0$ with $\int_{-\infty}^{\infty} f(x)\,dx=1$; probability is area under $f$.

$F(x)=P(X\le x)$; nondecreasing from 0 to 1, and $f(x)=F'(x)$ for continuous $X$.

$P(a\le X\le b)=\int_a^b f(x)\,dx$.

$E[X]=\sum_x x\,p(x)$.

$E[X]=\int_{-\infty}^{\infty} x\,f(x)\,dx$.

$E[g(X)]=\sum_x g(x)p(x)$ or $\int g(x)f(x)\,dx$ — no need to find the distribution of $g(X)$.

$\text{Var}(X)=E[(X-\mu)^2]=E[X^2]-(E[X])^2$.

$\sigma=\sqrt{\text{Var}(X)}$ — the spread in the units of $X$.

$CV=\dfrac{\sigma}{\mu}$ — a relative (unitless) measure of spread.

For $Y=aX+b$, $E[Y]=aE[X]+b$.

For $Y=aX+b$, $\text{Var}(Y)=a^2\,\text{Var}(X)$; the shift $b$ does not affect spread.

The value $m$ with $F(m)=0.5$; resists outliers more than the mean.

The value of $x$ where the pmf/pdf is largest.

The $p$th percentile is the value $x$ with $F(x)=p$ (e.g. 90th percentile: $F(x)=0.90$).

$M(t)=E[e^{tX}]$; $M'(0)=E[X]$, $M''(0)=E[X^2]$; it uniquely identifies a distribution.

$E[X^k]$ — the $k$th raw moment; $M^{(k)}(0)=E[X^k]$.

Based on the third central moment: positive = long right tail (mode < median < mean).

Based on the fourth central moment: measures tail heaviness and peakedness.

One trial: $X=1$ with prob $p$. Mean $p$, variance $p(1-p)$.

Successes in $n$ trials: $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$; mean $np$, variance $np(1-p)$.

$E[X]=np$, $\text{Var}(X)=np(1-p)$.

Trials until the first success; mean $1/p$, and it is memoryless.

Trials needed to reach $r$ successes — a generalization of the geometric distribution.

Successes when drawing $n$ items without replacement from a finite two-type population.

Values $1,\dots,n$ equally likely; mean $\dfrac{n+1}{2}$.

$P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}$ for $k=0,1,2,\dots$

Both equal the rate parameter $\lambda$.

Poisson with rate equal to the sum of the rates: $\lambda_1+\lambda_2$.

For large $n$, small $p$, binomial$(n,p)\approx$ Poisson$(\lambda=np)$.

On $[a,b]$: density $\dfrac{1}{b-a}$, mean $\dfrac{a+b}{2}$, variance $\dfrac{(b-a)^2}{12}$.

$E[X]=\dfrac{a+b}{2}$, $\text{Var}(X)=\dfrac{(b-a)^2}{12}$.

Waiting time at rate $\lambda$: $f(x)=\lambda e^{-\lambda x}$; mean and sd $1/\lambda$, variance $1/\lambda^2$.

$E[X]=1/\lambda$, $\text{Var}(X)=1/\lambda^2$.

$P(X>s+t\mid X>s)=P(X>t)$; held by the exponential and geometric distributions.

$h(x)=\dfrac{f(x)}{1-F(x)}$; constant $\lambda$ for the exponential.

A sum of $\alpha$ independent exponential waits; generalizes the exponential.

A continuous distribution on $[0,1]$ used to model proportions and probabilities.

Symmetric bell curve $N(\mu,\sigma^2)$; standardize with $z=\dfrac{x-\mu}{\sigma}$.

$Z\sim N(0,1)$: mean 0, standard deviation 1.

$z=\dfrac{x-\mu}{\sigma}$ converts $X$ to the standard normal for table lookup.

For a normal: $\approx 68\%$ within $1\sigma$, $95\%$ within $2\sigma$, $99.7\%$ within $3\sigma$.

$P(Z\le -z)=1-P(Z\le z)$, and $P(Z>0)=0.5$.

$X$ is lognormal if $\ln X$ is normal — used for positive, right-skewed losses.

$S(x)=P(X>x)=1-F(x)$.

With deductible $d$, payment per loss is $\max(X-d,\,0)$.

The insurer pays a fraction $\alpha$ of the covered loss above the deductible.

Caps the payment at a maximum $u$; the insurer never pays more.

Apply deductible first, then coinsurance, then the policy limit.

The loss is the full damage; the payment is what the insurer pays after deductible, coinsurance, and limits.

$E[\max(X-d,0)]<E[X]$ — small losses are zeroed out.

Multiplying every loss by $1+r$ scales the mean by $1+r$ and the standard deviation by $1+r$.

$\text{Var}(X)=E[X^2]-(E[X])^2$.

$E[c]=c$ and $\text{Var}(c)=0$.

$P(X=x)=0$ for any single value of a continuous variable.

Mode < median < mean; the long tail points right (positive skew).

Mean < median < mode; the long tail points left (negative skew).

$\text{Var}(X)=p(1-p)$, maximized at $p=0.5$.

$P(X>m+n\mid X>m)=P(X>n)$ for the (discrete) geometric distribution.

By the empirical rule, $\approx 2.5\%$ lies above $\mu+2\sigma$.

$I_A=1$ if $A$ occurs, else 0; $E[I_A]=P(A)$.

The $x$ maximizing the density $f(x)$.

Counts failures (or trials) before the $r$th success; reduces to geometric when $r=1$.

$F(x)=\int_{-\infty}^x f(t)\,dt$ and $f(x)=\dfrac{d}{dx}F(x)$.

The set of values where the pmf/pdf is positive.

For a symmetric distribution, the mean equals the median.

$\sum_x p(x)=1$ (discrete) or $\int f(x)\,dx=1$ (continuous).

$N(\mu,\sigma^2)$ has mean $\mu$ and variance $\sigma^2$.

$E[\min(X,u)]$ — the expected loss capped at a policy limit $u$.

Payment $=\alpha\,(X-d)$ for $X>d$, then capped at the policy limit.

$E[X^2]=\text{Var}(X)+(E[X])^2$.

For trials-until-first-success, $E[X]=1/p$.

$P(X=0)=e^{-\lambda}$.

$F(x)=1-e^{-\lambda x}$ for $x\ge 0$.

$P(X>x)=e^{-\lambda x}$.

$P(Z>0)=0.5$ by symmetry about 0.

For values $1,\dots,n$: $\text{Var}(X)=\dfrac{n^2-1}{12}$.

For $Y=aX+b$, $E[Y]=aE[X]+b$ — apply inflation $a$ and a flat add $b$.

A valid pdf satisfies $f(x)\ge 0$ everywhere and integrates to 1.

Discrete: sum a pmf; continuous: integrate a pdf. Both total to 1.

$E[aX+b]=aE[X]+b$.

A z-score is the number of standard deviations a value is from the mean.

A constant force of failure $\lambda$ uniquely characterizes the exponential distribution.

$P(z_1<Z<z_2)=\Phi(z_2)-\Phi(z_1)$ from the standard-normal table.

With $n=10$, $p=0.3$, $E[X]=np=3$.

$\text{Var}(aX)=a^2\,\text{Var}(X)$; standard deviation scales by $|a|$.

For the standard normal, $P(Z>z)=1-\Phi(z)=\Phi(-z)$.

A Bernoulli$(p)$ is a binomial$(1,p)$.

A valid pmf satisfies $p(x)\ge 0$ and $\sum_x p(x)=1$.

For $N(\mu,\sigma^2)$ the median equals the mean $\mu$.

$\text{Var}(X)=E[X^2]-\mu^2$, the second moment minus the squared mean.

For independent trials, $P(k\text{ successes in a row})=p^k$.

$\Phi(z)=P(Z\le z)$ for the standard normal.

The standardized sum of many i.i.d. variables $\dfrac{S_n-n\mu}{\sigma\sqrt{n}}\to N(0,1)$, regardless of shape.

$\text{Cov}(X,Y)=E[XY]-E[X]\,E[Y]$; 0 for independent variables.

$p(x,y)=P(X=x,\,Y=y)$ for discrete variables; sums to 1 over all pairs.

$f(x,y)\ge 0$ with $\iint f(x,y)\,dx\,dy=1$ for continuous variables.

$F(x,y)=P(X\le x,\,Y\le y)$.

$f_X(x)=\int f(x,y)\,dy$ — integrate (or sum) out the other variable.

$f_{Y\mid X}(y\mid x)=\dfrac{f(x,y)}{f_X(x)}$.

$X,Y$ independent $\iff f(x,y)=f_X(x)\,f_Y(y)$ for all $x,y$.

$\displaystyle\lim_{x,y\to\infty}F(x,y)=1$.

For independent $X,Y$: $F(x,y)=F_X(x)\,F_Y(y)$.

$E[X\mid Y=y]=\sum_x x\,p_{X\mid Y}(x\mid y)$ — the mean of the conditional distribution.

$E[X]=E\!\left[\,E[X\mid Y]\,\right]$.

$\text{Cov}(X,Y)=E[XY]-E[X]E[Y]$; if independent, $E[XY]=E[X]E[Y]$ so covariance is 0.

$\text{Cov}(X,X)=\text{Var}(X)$.

$\text{Cov}(aX,bY)=ab\,\text{Cov}(X,Y)$.

$\rho=\dfrac{\text{Cov}(X,Y)}{\sigma_X\,\sigma_Y}$, always in $[-1,1]$.

Independence implies zero covariance, but zero covariance does NOT imply independence.

$\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)+2\,\text{Cov}(X,Y)$.

$\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)$ when $X,Y$ are independent.

$\text{Var}(X-Y)=\text{Var}(X)+\text{Var}(Y)-2\,\text{Cov}(X,Y)$.

$\text{Var}(aX+bY)=a^2\text{Var}(X)+b^2\text{Var}(Y)+2ab\,\text{Cov}(X,Y)$.

$E[aX+bY]=aE[X]+bE[Y]$ — holds whether or not $X,Y$ are independent.

$E[XY]=E[X]E[Y]$ only when $X$ and $Y$ are independent (or uncorrelated).

For $S_n=\sum X_i$ i.i.d.: $E[S_n]=n\mu$, $\text{Var}(S_n)=n\sigma^2$.

$E[\bar X]=\mu$, $\text{Var}(\bar X)=\sigma^2/n$, and $\bar X$ is approximately normal for large $n$.

A linear combination of independent normals is normal, with means and variances adding.

A sample sorted smallest to largest: $X_{(1)}\le\dots\le X_{(n)}$.

For $n$ i.i.d. with CDF $F$: $P(X_{(n)}\le t)=[F(t)]^n$.

For $n$ i.i.d. with CDF $F$: $P(X_{(1)}\le t)=1-[1-F(t)]^n$.

$R=X_{(n)}-X_{(1)}$ — the largest minus the smallest order statistic.

The pmf/pdf of $X+Y$ for independent variables is the convolution of the two distributions.

$E[XY]=\sum_x\sum_y x\,y\,p(x,y)$ (discrete) — sum over all pairs.

Row totals give the marginal of one variable; column totals give the marginal of the other.

A cell divided by its row (or column) total gives a conditional probability.

$\text{Var}(X\mid Y=y)$ is the variance of the conditional distribution of $X$ given $Y=y$.

$\text{Cov}(X,Y)=0$ when $X$ and $Y$ are independent.

$\rho=0$ for independent variables (but $\rho=0$ alone does not prove independence).

$X+Y\sim\text{Poisson}(\lambda_1+\lambda_2)$ for independent Poissons.

$Z=\dfrac{S_n-n\mu}{\sigma\sqrt{n}}$ is approximately $N(0,1)$.

The CLT makes sums and averages (e.g. aggregate claims) approximately normal even from skewed components.

If $f(x,y)$ factors into a function of $x$ times a function of $y$ over a rectangle, $X$ and $Y$ are independent.

$E\!\left[\sum a_i X_i\right]=\sum a_i E[X_i]$ — always, regardless of dependence.

$\text{Var}\!\left(\sum_{i=1}^n X_i\right)=n\sigma^2$ when the $X_i$ are independent and identically distributed.

The minimum of independent exponentials with rates $\lambda_i$ is exponential with rate $\sum\lambda_i$.

Variables can be dependent through a joint distribution even with simple-looking marginals — check the joint.

If the conditional distribution of $Y$ given $X$ changes with $X$, then $X$ and $Y$ are dependent.

For odd $n$, the sample median is the middle order statistic $X_{((n+1)/2)}$.

Positive: variables move together; negative: one rises as the other falls; zero: no linear relationship.

$-1\le\rho\le 1$; $|\rho|=1$ means a perfect linear relationship.

Total claims $S=\sum X_i$; by the CLT $S$ is approximately normal for large claim counts.

$E[X]$ uses the marginal of $X$, obtained by summing/integrating out $Y$.

$\sigma_{X+Y}=\sqrt{\sigma_X^2+\sigma_Y^2}$ when independent.

$E\!\left(\sum_{i=1}^n X_i\right)=n\mu$.

$\sum_x\sum_y p(x,y)=1$ or $\iint f(x,y)\,dx\,dy=1$.

$aX+bY$ is normal when $X,Y$ are independent normals.

$\text{Cov}=0$ is necessary but not sufficient for independence.

$\text{Cov}(X,c)=0$ for any constant $c$.

The classic CLT needs i.i.d. variables with finite mean $\mu$ and variance $\sigma^2$.