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Exam P formula sheet PDF

Download a four-page probability formula sheet covering events, named distributions, joint discrete models, order statistics, normal approximations, and insurance payments. Read the formulas below or save the printable PDF.

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4 pages · No email or account · Updated September 14, 2026

Independent ActuaryProof reference, checked against the September 2026 Exam P syllabus. This is a study aid, not an official SOA handout or a personal reference permitted in the exam.

First page of the free Exam P formula sheet — Page 1 of the four-page PDF

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Know what is supplied during the exam	Official Tables for Exam P	The normal-distribution table named by the current syllabus and available through the CBT Exhibit button. It is not a general formula sheet, and candidates may not bring a personal copy.
Retrieve formulas quickly for free	Download this free four-page Exam P formula sheet	No email or account. It covers probability, named distributions, moments, joint models, order statistics, normal approximations, and insurance payments, with notation and conditions beside the formulas.
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Notation: p_X is a probability mass function, f_X a density, F_X a cumulative distribution function, and S_X its survival function. Exponential and gamma entries use scale θ, so a rate parameter λ equals 1/θ. Geometric and negative binomial entries count trials including the successful trial.

Events, counting, and conditional probability

PDF page 1

Union and complement

P(A\cup B)=P(A)+P(B)-P(A\cap B),\quad P(A^c)=1-P(A)

Subtract overlap once. The complement has probability 1 − P(A).

Conditional probability

P(A\mid B)=\frac{P(A\cap B)}{P(B)}

Requires P(B) > 0. The denominator is the event already known.

Total probability

P(A)=\sum_j P(A\mid B_j)P(B_j)

The events B_j must be disjoint and cover the sample space.

Bayes’ theorem

P(B_i\mid A)=\frac{P(A\mid B_i)P(B_i)}{\sum_jP(A\mid B_j)P(B_j)}

Requires positive total probability of A.

Permutations and combinations

{}_nP_r=\frac{n!}{(n-r)!},\qquad {n\choose r}=\frac{n!}{r!(n-r)!}

Distinct objects, without replacement. Order matters for permutations.

Independence

P(A\cap B)=P(A)P(B)

Disjoint events with positive probabilities are not independent.

Expectation, variance, and distribution functions

PDF page 1

Expected value

E[g(X)]=\sum_xg(x)p_X(x)\quad\text{or}\quad\int g(x)f_X(x)\,dx

Sum or integrate over the support, with finite expectation when required.

Variance

\operatorname{Var}(X)=E[X^2]-E[X]^2

Variance is nonnegative and has squared units.

Linear transformation

E[aX+b]=aE[X]+b,\quad\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)

Adding a constant changes the mean but not the variance.

CDF and survival

F_X(x)=P(X\le x),\quad S_X(x)=1-F_X(x)

S_X(x) means P(X > x). Keep endpoint mass when X is discrete.

Continuous interval probability

P(a<X\le b)=F_X(b)-F_X(a)=\int_a^b f_X(x)\,dx

For a continuous variable, including a single endpoint changes nothing.

Discrete distributions

PDF page 2

Binomial

p(k)={n\choose k}p^k(1-p)^{n-k}

k = 0,…,n; independent trials. Mean np; variance np(1−p).

Geometric: trials through first success

p(k)=(1-p)^{k-1}p,\quad E[X]=1/p

k ≥ 1; variance (1−p)/p². For failures before success, subtract 1 from X.

Negative binomial: trials through rth success

p(k)={k-1\choose r-1}p^r(1-p)^{k-r}

k ≥ r; mean r/p; variance r(1−p)/p². Failures equal X−r.

Hypergeometric

p(k)=\frac{{K\choose k}{N-K\choose n-k}}{{N\choose n}}

n draws without replacement from N objects, K successes. Mean nK/N; variance n(K/N)(1−K/N)(N−n)/(N−1), N > 1.

Poisson

p(k)=e^{-\lambda}\frac{\lambda^k}{k!},\quad E[X]=\operatorname{Var}(X)=\lambda

k = 0,1,…; λ is the expected count over the stated exposure.

Discrete uniform

E[X]=\frac{a+b}{2},\quad\operatorname{Var}(X)=\frac{(b-a+1)^2-1}{12}

Integers a,…,b equally likely, each with probability 1/(b−a+1).

Continuous distributions

PDF page 2

Uniform on (a,b)

f(x)=\frac1{b-a},\quad E[X]=\frac{a+b}{2},\quad\operatorname{Var}(X)=\frac{(b-a)^2}{12}

a < x < b, a < b. Mean (a+b)/2; density is zero outside the interval.

Exponential: scale θ

f(x)=\frac1\theta e^{-x/\theta},\quad S(x)=e^{-x/\theta}

x > 0, θ > 0. Mean θ; variance θ². Rate λ = 1/θ.

Gamma: shape α, scale θ

f(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha}

x > 0; α,θ > 0. Mean αθ; variance αθ².

Beta on (0,1)

f(x)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}

α,β > 0; B(α,β)=Γ(α)Γ(β)/Γ(α+β). Mean α/(α+β); variance αβ/[(α+β)²(α+β+1)].

Normal

f(x)=\frac{e^{-(x-\mu)^2/(2\sigma^2)}}{\sigma\sqrt{2\pi}},\quad Z=\frac{X-\mu}{\sigma}

X has mean μ and variance σ², σ > 0. Standardize with Z=(X−μ)/σ.

Discrete joint models and conditional moments

PDF page 3

Marginal and conditional mass

p_X(x)=\sum_y p(x,y),\quad p_{X\mid Y}(x\mid y)=\frac{p(x,y)}{p_Y(y)}

Condition only on y with p_Y(y) > 0.

Joint expectation

E[g(X,Y)]=\sum_x\sum_y g(x,y)p(x,y)

Sum over allowed pairs. Impossible pairs carry zero mass.

Covariance and correlation

\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y],\quad\rho=\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}

Correlation divides covariance by σ_X σ_Y, requiring both standard deviations positive.

Variance of a linear combination

\operatorname{Var}(aX+bY)=a^2\sigma_X^2+b^2\sigma_Y^2+2ab\operatorname{Cov}(X,Y)

Independence makes covariance zero; zero covariance alone need not imply independence.

Total expectation

E[X]=\sum_y E[X\mid Y=y]p_Y(y)

Average the conditional means with the probabilities of their conditioning groups.

Total variance

\operatorname{Var}(X)=E[\operatorname{Var}(X\mid Y)]+\operatorname{Var}(E[X\mid Y])

Within-group variation plus variation between conditional means.

Transformations and order statistics

PDF page 3

CDF method

F_Y(y)=P(g(X)\le y)

Find the transformed support first. Solve the inequality on every applicable branch.

One-to-one continuous transform

f_Y(y)=f_X(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right|

For differentiable monotone g. Multiple inverse branches require adding their density contributions.

Minimum and maximum

F_{\min}(x)=1-[1-F(x)]^n,\quad F_{\max}(x)=F(x)^n

For n independent, identically distributed observations with common CDF F.

kth order statistic CDF

P(X_{(k)}\le x)=\sum_{j=k}^n{n\choose j}F(x)^j[1-F(x)]^{n-j}

At least k of n iid observations must be at or below x.

kth order statistic density

f_{(k)}(x)=\frac{n!}{(k-1)!(n-k)!}F(x)^{k-1}[1-F(x)]^{n-k}f(x)

For iid continuous observations; 1 ≤ k ≤ n.

Insurance payments and loss modifications

PDF page 4

Deductible, coinsurance, and payment cap

Y=c\min\bigl(((1+r)X-d)_+,u\bigr)

X is the original loss, r inflation, d ordinary deductible, c insurer share. Here u caps the covered loss after the deductible, before coinsurance.

Expected payment per loss

E[Y]=c\int_d^{d+u}S_{(1+r)X}(x)\,dx

For the contract defined above, X ≥ 0. Omit the upper cap by taking u = ∞.

Positive-payment probability

P(Y>0)=S_X\!\left(\frac d{1+r}\right)

Requires c > 0, u > 0, and 1+r > 0.

Payment per payment

E[Y\mid Y>0]=\frac{E[Y]}{P(Y>0)}

Only when positive payments have positive probability. Do not confuse this with payment per loss.

Limited expected value

E[\min(X,u)]=\int_0^u S_X(x)\,dx

For nonnegative X and u ≥ 0. A limit on loss differs from a limit on payment.

Second moment of capped deductible payment

E[Y^2]=2c^2\int_0^u yS_{(1+r)X}(d+y)\,dy

Use Var(Y)=E[Y²]−E[Y]². This uses the same contract and cap convention above.

Exponential deductible

E[(X-d)_+]=\theta e^{-d/\theta},\quad E[X-d\mid X>d]=\theta

Exponential scale θ and d ≥ 0. Conditional excess has mean θ by memorylessness.

Normal sums, quantiles, and approximation

PDF page 4

Normal linear combination

\sum_i a_iX_i\sim N\!\left(\sum_i a_i\mu_i,\sum_i a_i^2\sigma_i^2\right)

For independent normal variables. Add covariance terms when a jointly normal vector is dependent.

Central limit theorem

\frac{\bar X-\mu}{\sigma/\sqrt n}\approx N(0,1)

For iid variables with finite mean and nonzero finite variance, as n becomes large.

Normal quantile and symmetry

q_p=\mu+\sigma\Phi^{-1}(p),\quad\Phi(-z)=1-\Phi(z)

Φ is the standard normal CDF. Divide by standard deviation, not variance.

Normal approximation to binomial

P(X\le k)\approx\Phi\!\left(\frac{k+0.5-np}{\sqrt{np(1-p)}}\right)

For sufficiently large np and n(1−p). The 0.5 corrects the discrete boundary.

Which formula should you use?

These four original examples show the decisions that come before substitution. Each includes a check and a common mistake.

The examples are on this page; the four-page download remains the compact formula reference.

Reverse a conditional probability

When to use it: The group is unknown, but an outcome has been observed: use Bayes with a total-probability denominator.

A portfolio is 10% high risk. The annual claim probability is 0.80 for high-risk policies and 0.04 for all others. Given that a randomly selected policy had a claim, what is the probability it is high risk?

Write H for high risk and C for a claim. The numerator counts high-risk claims. The denominator includes claims from both groups.

\Pr(C)=0.10(0.80)+0.90(0.04)=0.116

\Pr(H\mid C)=\frac{0.10(0.80)}{0.116}=\frac{20}{29}\approx0.689655

Check: In a model portfolio of 1,000 policies, the expected counts are 80 high-risk claims and 36 other claims: 80/116 gives the same answer.

Common mistake: 0.80 is P(claim | high risk), which reverses the condition you were asked to calculate.

Return to the formulas · Conditional probability problems

Convert the interval before using a Poisson formula

When to use it: A Poisson model gives a rate, while the question uses a different length of time.

Claims follow a Poisson process at 2 per hour. What is the probability of at least one claim in 90 minutes?

The interval mean is rate multiplied by elapsed time in hours. Subtract the zero-claim probability from one.

\lambda=2\times1.5=3

\Pr(N\ge1)=1-\Pr(N=0)=1-e^{-3}\approx0.950213

Check: Three independent half-hour intervals each have zero-claim probability e⁻¹. Their product is e⁻³, agreeing with the 90-minute calculation.

Common mistake: Using λ=2 calculates the probability for one hour, not the requested 90 minutes.

Return to the formulas · Poisson distribution problems

Distinguish payment per loss from payment per paid claim

When to use it: An ordinary deductible creates zero payments; decide whether those zeros belong in the expectation.

A loss X is exponential with mean 1,000. An insurer pays Y=max(X−250, 0), with no other limit. Find E[Y] per loss and E[Y | Y>0] per paid claim.

Integrate the loss survival function above the deductible to include the zero-payment losses. Then divide by the probability that a payment occurs.

E[Y]=\int_{250}^{\infty}e^{-x/1000}\,dx=1000e^{-0.25}\approx778.800783

\Pr(Y>0)=e^{-0.25},\qquad E[Y\mid Y>0]=\frac{1000e^{-0.25}}{e^{-0.25}}=1000

Check: The exponential memoryless property gives a mean remaining loss of 1,000 after the deductible is exceeded. Multiplying by the payment probability recovers 778.800783.

Common mistake: Subtracting 250 from E[X] gives 750 and ignores the fact that payments cannot be negative.

Return to the formulas · Univariate and insurance-payment problems

Compare survival curves and the mean per loss versus per positive payment.

Square coefficients when combining variances

When to use it: The question asks for the distribution of a linear combination of independent normal variables.

Independent normal variables X and Y have means 10 and 3, and variances 4 and 1. Find the mean and variance of Z=2X−Y.

Means are linear. Independence removes the covariance term, but the coefficient of each variance is squared. A linear combination of independent normals is normal.

E[Z]=2(10)-3=17

\operatorname{Var}(Z)=2^2(4)+(-1)^2(1)=17,\qquad Z\sim N(17,17)

Check: Write X=10+2U and Y=3+V for independent standard normals. Then Z=17+4U−V, whose variance is 16+1.

Common mistake: 2×4−1=7 is not a variance calculation. A minus sign on a random variable does not create a negative variance contribution. N(17,17) here uses variance as the second parameter.

Return to the formulas · Variance problems

What is included in this Exam P cheat sheet?

The reference follows the current probability curriculum: general probability; univariate discrete and continuous distributions; discrete joint and conditional models; normal linear combinations; order statistics; and insurance loss modifications. Hypergeometric, discrete uniform, and beta distributions are included. Moment-generating functions and general continuous joint-density methods are outside the scope of this sheet.

Does SOA provide this formula sheet on exam day?

No. This downloadable sheet is an independent preparation resource. Use the official syllabus and exam instructions to check which tables and materials are supplied. The Exam P syllabus links a normal distribution table; that does not make a personal formula sheet an allowed reference.

Free reference or a paid study PDF?

The free four-page PDF gives formulas, notation, and conditions. It contains no full sample-solution manual or timed practice exam. Use it to find an identity quickly; use a worked solution when you need the reasoning that leads to that identity.

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Formula and distribution cards

Probability Formula & Distribution Cards, USD 19. The product page shows the card format and included material.

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Choosing your first exam? Read Exam P vs FM: which should you take first? Or open the Exam FM formula sheet.

For a learning sequence and help choosing study materials, read the Exam P study guide.

Source and format comparison reviewed September 26, 2026; worked examples added September 24, 2026. Published by ActuaryProof; report a correction.

Sources and accountability

Primary sources and correction path

SOA Exam P study pageLists current syllabus, updates, study materials, and sample-exam resources.
September 2026 Exam P syllabusControls the tested scope and identifies the normal-distribution table supplied during the exam.
Official Tables for Exam PThe normal-distribution exhibit supplied in the exam; it is not a complete formula sheet.

ActuaryProof publishes the independent explanations on this page. Official exam facts on this page come from the sources above; schedules, study ledgers, and checking protocols are ActuaryProof editorial tools unless stated otherwise. Read about the publisher or report a suspected error.

Primary sources last checked: September 26, 2026.