Free SOA Exam P (Probability) Formula Sheet (2026)

Every Exam P formula you need on the test, grouped by topic and rendered with full math notation. 67 formulas across 8 topics, calibrated to the 2026 syllabus. Free forever, no signup required.

67 formulas 8 topics 2026 syllabus Free forever
Print-ready PDF: 1080x1350 portrait, math pre-rendered, fonts embedded. Download once, study anywhere.
Download PDF →

All Exam P Formulas

Probability Fundamentals 5 items
Set complement
P(Ac)=1P(A)P(A^c) = 1 - P(A)
Addition rule (two events)
P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B)
Addition rule (three events)
P(ABC)=P(A)+P(B)+P(C)P(AB)P(AC)P(BC)+P(ABC)P(A \cup B \cup C) = P(A)+P(B)+P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C)
Permutations (ordered)
nPr=n!(nr)!_nP_r = \dfrac{n!}{(n-r)!}
nn objects taken rr at a time, order matters
Combinations (unordered)
(nr)=n!r!(nr)!\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}
nn objects taken rr at a time, order does not matter
Conditional Probability & Bayes 5 items
Conditional probability
P(AB)=P(AB)P(B),P(B)>0P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}, \quad P(B)>0
Multiplication rule
P(AB)=P(AB)P(B)=P(BA)P(A)P(A \cap B) = P(A \mid B)\,P(B) = P(B \mid A)\,P(A)
Law of total probability
P(A)=iP(ABi)P(Bi)P(A) = \sum_{i} P(A \mid B_i)\,P(B_i)
where {Bi}\{B_i\} is a partition of the sample space
Bayes' theorem
P(BiA)=P(ABi)P(Bi)jP(ABj)P(Bj)P(B_i \mid A) = \dfrac{P(A \mid B_i)\,P(B_i)}{\sum_j P(A \mid B_j)\,P(B_j)}
Independence of two events
AA and BB are independent iff P(AB)=P(A)P(B)P(A \cap B) = P(A)\,P(B)
(equivalently: P(AB)=P(A)P(A\mid B)=P(A))
Discrete Distributions 6 items
Bernoulli distribution — PMF, mean, variance
P(X=1)=p,  P(X=0)=1pP(X=1)=p,\; P(X=0)=1-p
E[X]=p,Var(X)=p(1p)E[X]=p,\quad \text{Var}(X)=p(1-p)
Binomial distribution — PMF, mean, variance
P(X=k)=(nk)pk(1p)nkP(X=k)=\binom{n}{k}p^k(1-p)^{n-k}
E[X]=np,Var(X)=np(1p)E[X]=np,\quad \text{Var}(X)=np(1-p)
k=0,1,,nk=0,1,\ldots,n
Poisson distribution — PMF, mean, variance
P(X=k)=eλλkk!P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}
E[X]=λ,Var(X)=λE[X]=\lambda,\quad \text{Var}(X)=\lambda
k=0,1,2,k=0,1,2,\ldots
Geometric distribution — PMF, mean, variance
P(X=k)=(1p)k1pP(X=k)=(1-p)^{k-1}p (number of trials to first success)
E[X]=1p,Var(X)=1pp2E[X]=\dfrac{1}{p},\quad \text{Var}(X)=\dfrac{1-p}{p^2}
k=1,2,k=1,2,\ldots
Negative Binomial distribution — PMF, mean, variance
P(X=k)=(k1r1)pr(1p)krP(X=k)=\binom{k-1}{r-1}p^r(1-p)^{k-r} (trials for rrth success)
E[X]=rp,Var(X)=r(1p)p2E[X]=\dfrac{r}{p},\quad \text{Var}(X)=\dfrac{r(1-p)}{p^2}
Hypergeometric distribution — PMF, mean, variance
P(X=k)=(Kk)(NKnk)(Nn)P(X=k)=\dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}
E[X]=nKN,Var(X)=nK(NK)(Nn)N2(N1)E[X]=\dfrac{nK}{N},\quad \text{Var}(X)=\dfrac{nK(N-K)(N-n)}{N^2(N-1)}
Continuous Distributions 8 items
Uniform distribution — PDF, mean, variance
f(x)=1ba,a<x<bf(x)=\dfrac{1}{b-a},\quad a<x<b
E[X]=a+b2,Var(X)=(ba)212E[X]=\dfrac{a+b}{2},\quad \text{Var}(X)=\dfrac{(b-a)^2}{12}
Exponential distribution — PDF, CDF, mean, variance
f(x)=λeλx,F(x)=1eλx,x>0f(x)=\lambda e^{-\lambda x},\quad F(x)=1-e^{-\lambda x},\quad x>0
E[X]=1λ,Var(X)=1λ2E[X]=\dfrac{1}{\lambda},\quad \text{Var}(X)=\dfrac{1}{\lambda^2}
Memoryless property (Exponential / Geometric)
P(X>s+tX>s)=P(X>t)P(X > s+t \mid X > s) = P(X > t)
Only the Exponential (continuous) and Geometric (discrete) satisfy this.
Normal distribution — PDF
f(x)=1σ2πexp ⁣((xμ)22σ2)f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\exp\!\left(-\dfrac{(x-\mu)^2}{2\sigma^2}\right)
E[X]=μ,Var(X)=σ2E[X]=\mu,\quad \text{Var}(X)=\sigma^2
Gamma distribution — PDF, mean, variance
f(x)=xα1ex/θΓ(α)θα,x>0f(x)=\dfrac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\,\theta^\alpha},\quad x>0
E[X]=αθ,Var(X)=αθ2E[X]=\alpha\theta,\quad \text{Var}(X)=\alpha\theta^2
α\alpha=shape, θ\theta=scale
Weibull distribution — PDF, mean
f(x)=τθ(xθ)τ1e(x/θ)τ,x>0f(x)=\dfrac{\tau}{\theta}\left(\dfrac{x}{\theta}\right)^{\tau-1}e^{-(x/\theta)^\tau},\quad x>0
E[X]=θΓ ⁣(1+1τ)E[X]=\theta\,\Gamma\!\left(1+\tfrac{1}{\tau}\right)
Lognormal distribution — PDF, mean, variance
f(x)=1xσ2πexp ⁣((lnxμ)22σ2),x>0f(x)=\dfrac{1}{x\sigma\sqrt{2\pi}}\exp\!\left(-\dfrac{(\ln x-\mu)^2}{2\sigma^2}\right),\quad x>0
E[X]=eμ+σ2/2,Var(X)=e2μ+σ2(eσ21)E[X]=e^{\mu+\sigma^2/2},\quad \text{Var}(X)=e^{2\mu+\sigma^2}(e^{\sigma^2}-1)
Beta distribution — PDF, mean
f(x)=xα1(1x)β1B(α,β),0<x<1f(x)=\dfrac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)},\quad 0<x<1
E[X]=αα+β,Var(X)=αβ(α+β)2(α+β+1)E[X]=\dfrac{\alpha}{\alpha+\beta},\quad \text{Var}(X)=\dfrac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}
Joint & Marginal Distributions 3 items
Joint PDF — marginal densities
fX(x)=fX,Y(x,y)dyf_X(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy
fY(y)=fX,Y(x,y)dxf_Y(y)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx
Conditional PDF
fYX(yx)=fX,Y(x,y)fX(x)f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}
Independence of continuous RVs
X,YX,Y independent iff fX,Y(x,y)=fX(x)fY(y)f_{X,Y}(x,y)=f_X(x)\,f_Y(y) for all x,yx,y
Expectation & Variance 7 items
MGF definition and moment extraction
MX(t)=E[etX]M_X(t)=E[e^{tX}]
E[Xn]=MX(n)(0)E[X^n]=M_X^{(n)}(0) (nnth derivative at t=0t=0)
Variance shortcut
Var(X)=E[X2](E[X])2\text{Var}(X)=E[X^2]-(E[X])^2
Covariance
Cov(X,Y)=E[XY]E[X]E[Y]\text{Cov}(X,Y)=E[XY]-E[X]\,E[Y]
If X,YX,Y independent: Cov(X,Y)=0\text{Cov}(X,Y)=0
Correlation coefficient
ρXY=Cov(X,Y)σXσY,1ρ1\rho_{XY}=\dfrac{\text{Cov}(X,Y)}{\sigma_X\,\sigma_Y},\quad -1\le\rho\le1
Variance of a linear combination
Var(aX+bY)=a2Var(X)+b2Var(Y)+2abCov(X,Y)\text{Var}(aX+bY)=a^2\text{Var}(X)+b^2\text{Var}(Y)+2ab\,\text{Cov}(X,Y)
Law of total expectation
E[X]=E[E[XY]]E[X]=E[E[X\mid Y]]
Law of total variance
Var(X)=E[Var(XY)]+Var(E[XY])\text{Var}(X)=E[\text{Var}(X\mid Y)]+\text{Var}(E[X\mid Y])
Insurance Applications 5 items
Ordinary deductible — payment per loss
YL=max(Xd,0)Y^L = \max(X-d,\,0)
E[YL]=E[X]E[Xd]E[Y^L]=E[X]-E[X\wedge d]
where dd = deductible, XX = ground-up loss
Limited expected value (LEV)
E[Xu]=0uS(x)dx=0u[1F(x)]dxE[X\wedge u]=\int_0^u S(x)\,dx = \int_0^u [1-F(x)]\,dx
for X0X\ge 0
Payment per payment (excess loss)
e(d)=E[XdX>d]=E[X]E[Xd]1F(d)e(d)=E[X-d\mid X>d]=\dfrac{E[X]-E[X\wedge d]}{1-F(d)}
dd = deductible
Policy limit — payment per loss
YL=min(X,u)=XuY^L=\min(X,\,u)=X\wedge u
E[YL]=E[Xu]E[Y^L]=E[X\wedge u]
uu = policy limit
Stop-loss (aggregate) premium
E[(Sd)+]=E[S]E[Sd]E[(S-d)_+]=E[S]-E[S\wedge d]
SS = aggregate loss, dd = retention (aggregate deductible)
Topic 0 28 items
De Morgan's Laws
(AB)c=AcBc(A \cup B)^c = A^c \cap B^c and (AB)c=AcBc(A \cap B)^c = A^c \cup B^c; A, B = events, c^c = complement, \cup = union, \cap = intersection
Stars and bars distribution count
(n+k1k1)\binom{n+k-1}{k-1}, n = identical objects distributed, k = distinct bins (non-negative integer solutions to x1++xk=nx_1+\cdots+x_k=n); use (n1k1)\binom{n-1}{k-1} if each bin needs at least one
Addition rule for mutually exclusive events
P(AB)=P(A)+P(B)P(A \cup B) = P(A) + P(B) when P(AB)=0P(A \cap B) = 0; A, B = mutually exclusive (disjoint) events that cannot both occur
Feasibility bounds for an intersection probability
max(0,P(A)+P(B)1)P(AB)min(P(A),P(B))\max(0, P(A)+P(B)-1) \leq P(A \cap B) \leq \min(P(A), P(B)), lower bound = Bonferroni, upper bound from subset containment
Chain rule for the probability of an intersection of events
P(A1An)=P(A1)P(A2A1)P(A3A1A2)P(AnA1An1)P(A_1 \cap \cdots \cap A_n) = P(A_1)P(A_2 \mid A_1)P(A_3 \mid A_1 \cap A_2)\cdots P(A_n \mid A_1 \cap \cdots \cap A_{n-1}), each factor conditions on all prior events
Odds form of Bayes' theorem
P(BA)P(BcA)=P(AB)P(ABc)P(B)P(Bc)\frac{P(B \mid A)}{P(B^c \mid A)} = \frac{P(A \mid B)}{P(A \mid B^c)} \cdot \frac{P(B)}{P(B^c)}, posterior odds = likelihood ratio × prior odds; A = evidence, B = hypothesis, B^c = complement
Distinguishable permutations of a multiset
N=n!n1!n2!nk! N = \dfrac{n!}{n_1!\,n_2!\cdots n_k!} , n = total items, nin_i = count of identical items of type i, k = number of distinct types
Survival function in terms of the CDF
SX(x)=P(X>x)=1FX(x)S_X(x) = P(X > x) = 1 - F_X(x), SXS_X = survival function, FX(x)=P(Xx)F_X(x)=P(X\le x) = CDF, x = threshold value
Conditional CDF of a left-truncated distribution
FXX>a(x)=FX(x)FX(a)1FX(a), x>aF_{X|X>a}(x) = \frac{F_X(x) - F_X(a)}{1 - F_X(a)}, \ x > a, F = CDF, a = truncation point, x = value
Law of the Unconscious Statistician (LOTUS)
E[g(X)]=g(x)fX(x)dxE[g(X)] = \int_{-\infty}^{\infty} g(x) f_X(x)\, dx, g = function of X, fXf_X = pdf of X (use sum with pXp_X(x) if discrete)
Chebyshev's inequality
P(Xμkσ)1k2P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}, μ = mean, σ = standard deviation, k = number of SDs; equivalently at least 11/k21 - 1/k^2 lies within k SDs
Expected payment per loss with deductible and benefit limit
E[YL]=E[X(d+u)]E[Xd]E[Y^L] = E[X \wedge (d+u)] - E[X \wedge d], d = deductible, u = benefit limit, X = ground-up loss, XaX \wedge a = loss capped at a
Second moment of the excess loss for an exponential
E[((Xd)+)2]=S(d)2θ2=ed/θ2θ2E[((X-d)_+)^2] = S(d)\cdot 2\theta^2 = e^{-d/\theta}\cdot 2\theta^2, d = deductible, θ = exponential mean, S(d) = survival function at d
Pareto distribution pdf, mean, and variance
f(x)=αθα(x+θ)α+1f(x) = \frac{\alpha\theta^\alpha}{(x+\theta)^{\alpha+1}}, E[X]=θα1E[X]=\frac{\theta}{\alpha-1} (α>1\alpha>1), Var(X)=αθ2(α1)2(α2)\text{Var}(X)=\frac{\alpha\theta^2}{(\alpha-1)^2(\alpha-2)} (α>2\alpha>2); α = shape, θ = scale, x > 0
Finite-population correction factor for hypergeometric variance
Var(X)=nKNNKNNnN1\text{Var}(X) = n\cdot\frac{K}{N}\cdot\frac{N-K}{N}\cdot\frac{N-n}{N-1}, n = draws, N = population, K = successes in population, X = successes drawn
Sum of independent Poisson random variables
Poisson(λ1)+Poisson(λ2)=Poisson(λ1+λ2)\text{Poisson}(\lambda_1) + \text{Poisson}(\lambda_2) = \text{Poisson}(\lambda_1 + \lambda_2), λ₁, λ₂ = rates of independent Poisson counts; combined count is Poisson with summed rate
Discrete uniform distribution pmf mean and variance
P(X=x)=1k, E[X]=a+b2, Var(X)=k2112P(X=x)=\tfrac{1}{k},\ E[X]=\tfrac{a+b}{2},\ \text{Var}(X)=\tfrac{k^2-1}{12}; a, b = min, max integers; k = b-a+1 = count of values; x in {a,...,b}
Gamma-Poisson tail connection for integer shape
P(Xt)=P(Poisson(t/θ)α)P(X \le t) = P(\text{Poisson}(t/\theta) \ge \alpha), equivalently P(X>t)=P(Poisson(t/θ)α1)P(X > t) = P(\text{Poisson}(t/\theta) \le \alpha - 1); X~Gamma(α,θ), α = integer shape, θ = scale, t = time
Beta distribution variance
Var(X)=ab(a+b)2(a+b+1)\text{Var}(X) = \frac{ab}{(a+b)^2(a+b+1)}, a, b = shape parameters of Beta on (0,1)
Rectangular probability from a joint CDF
P(a<Xb, c<Yd)=F(b,d)F(a,d)F(b,c)+F(a,c)P(a < X \leq b,\ c < Y \leq d) = F(b,d) - F(a,d) - F(b,c) + F(a,c), F = joint CDF; a,b = X limits; c,d = Y limits
Conditional PMF of a discrete random variable
pYX(yx)=pX,Y(x,y)pX(x) p_{Y|X}(y|x) = \dfrac{p_{X,Y}(x,y)}{p_X(x)} , pX,Yp_{X,Y} = joint PMF, pXp_X = marginal PMF of X, requires pX(x)>0p_X(x) > 0
Compound mean (Wald's identity for the mean)
E[S]=E[N]E[X]E[S] = E[N] \cdot E[X], S = aggregate loss, N = random claim count, X = i.i.d. severity independent of N
Variance of a compound random sum
Var(S)=E[N]Var(X)+Var(N)(E[X])2 \text{Var}(S) = E[N]\,\text{Var}(X) + \text{Var}(N)\,(E[X])^2 , S = sum of N losses, N = random claim count, X = individual loss severity
Covariance via conditional expectation
Cov(X,Y)=E[XE[YX]]E[X]E[Y] \text{Cov}(X,Y) = E[X \cdot E[Y|X]] - E[X] \cdot E[Y] , E[Y|X] = conditional mean of Y given X, E[X], E[Y] = marginal means
PDF of the k-th order statistic
fX(k)(x)=n!(k1)!(nk)![F(x)]k1[1F(x)]nkf(x)f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!}[F(x)]^{k-1}[1-F(x)]^{n-k}f(x), n = sample size, k = rank, F = cdf, f = pdf
Linear combination of independent normal random variables
aX+bY+cN(aμX+bμY+c,  a2σX2+b2σY2)aX + bY + c \sim N(a\mu_X + b\mu_Y + c,\; a^2\sigma_X^2 + b^2\sigma_Y^2), a, b, c = constants, μ = mean, σ² = variance; X, Y independent normals
Mean of a linear combination of random variables
E[i=1naiXi+b]=i=1naiE[Xi]+bE\left[\sum_{i=1}^n a_i X_i + b\right] = \sum_{i=1}^n a_i E[X_i] + b, aia_i = constant coefficients, XiX_i = random variables, b = additive constant (no independence needed)
Central Limit Theorem normal approximation for a sum
Sn=i=1nXiN(nμ,nσ2)S_n = \sum_{i=1}^n X_i \approx N(n\mu, n\sigma^2), standardize z=snμσnz = \frac{s - n\mu}{\sigma\sqrt{n}}; μ\mu = mean, σ2\sigma^2 = variance, n = number of iid terms
Take the free Exam P diagnostic quiz →
Instant readiness score in about 6 minutes. No signup to start.

Frequently Asked Questions

Is the Exam P formula sheet free?
Yes. The full Exam P formula sheet is free, with no signup, no email, and no credit card required. 67 formulas across 8 topics, all rendered with the same KaTeX math notation used in the FreeFellow study app.
Can I download the Exam P formula sheet as a printable PDF?
Yes. A 1080x1350 portrait PDF (Instagram and LinkedIn carousel native size, also great for tablet study) is linked at the top of this page. The PDF is fully self-contained: math is pre-rendered, fonts are embedded, no internet connection needed once downloaded.
What's covered on the Exam P formula sheet?
Every formula is grouped by official syllabus topic, with the formula in math notation plus a one-line note on when to use it (or a watch-out from CAIA, CFA, or other prep-provider commentary). Coverage is calibrated to the 2026 syllabus and refreshed when the corpus changes.
What is FreeFellow's relationship with SOA?
No. FreeFellow is not affiliated with the SOA or any examination body. This is an independent study aid covering the published syllabus.
What else is free at FreeFellow for Exam P candidates?
The full original question bank is free with an account, subject to usage limits. Worked solutions, written lessons, mixed practice, and your readiness score stay free. The formula sheet is free too. Fellow is $39 per month or $79 per quarter, per exam family (USD). Fellow Plus is $49 per month, $99 per quarter, or $199 per year, per exam family (USD). Every annual plan is Fellow Plus. Fellow adds timed mock exams, spaced-repetition flashcards, performance analytics, and a personalized study plan.

About FreeFellow

Jeffrey Ting, founder of FreeFellow
Jeffrey Ting
FSA, CFA · Founder

FreeFellow was built by Jeffrey Ting, a credentialed actuary and CFA charterholder who passed thirteen of the hardest exams in finance on the first attempt, and paid four-figure prep fees for every one. The learning itself was always free. The price was a moat.

So he started writing his own questions, then lessons, then mock exams, until it grew into a full prep platform covering 40 finance credentials with more than 45,000 original practice questions. The name says exactly what it is: the question bank is free, and Fellow is what you become once you pass.

01
Cost shouldn't decide who gets in.

The exam is a fair gate. A four-figure prep course is not. FreeFellow takes the second gate down, so the exam is the only one left.

02
Free should mean free.

No trial clock, no email gate, no credit card. The question bank, worked solutions, lessons, and readiness score stay free, and they are enough to pass.

03
Built by someone who sat where you sit.

He paid for the big-name courses, found nothing he respected, and built the prep he wished had existed. Not a marketing team that has never sat an exam.

Free forever

Put the formulas to work.

Every formula on this sheet shows up in the free Exam P question bank, and every question carries a step-by-step solution.

Practice Exam P questions free →

No credit card. No trial clock.