This guide connects the probability foundations of SOA Exam P to random variables, aggregate risk, and insurance payments. Each concept presents a rule, a worked example, and a specific mistake to avoid. Calculus is assumed where densities, expectations, and transformations require differentiation or integration. Read the distribution assumptions and policy terms carefully: they determine which calculation applies.
Events and Probability Rules
1. Sample Spaces and Equally Likely Outcomes
A sample space lists possible outcomes, while an event collects outcomes satisfying a condition. Counting gives P(A) = |A|/|S| only when the elementary outcomes are equally likely. Outcomes created by grouping elementary results, such as totals, need not have equal probabilities.
Worked example: Two independent balanced spinners each show 1, 2, or 3. Of nine equally likely ordered pairs, three have total 4: (1,3), (2,2), and (3,1). The probability is 1/3.
Mistake to avoid: Treating the five possible totals, 2 through 6, as equally likely.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
2. Translating Conditions into Sets
An intersection requires both conditions, a union requires at least one, and a complement reverses a condition within the sample space. De Morgan's laws give (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ and (A ∩ B)ᶜ = Aᶜ ∪ Bᶜ. Preserve strict and inclusive boundaries when translating inequalities.
Worked example: For a nonnegative loss X, let A mean X > 100 and B mean X ≤ 300. Then A ∩ B means 100 < X ≤ 300, while A ∪ B includes every possible loss.
Mistake to avoid: Interpreting 'not both' as 'neither'; those are different events.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
3. Probability Axioms and Valid Assignments
Probabilities are nonnegative, the entire sample space has probability 1, and probabilities add over disjoint events. For a finite sample space, assigning nonnegative probabilities totaling 1 to elementary outcomes defines a valid model. An event's probability is the sum of the probabilities assigned to its outcomes.
Worked example: Three mutually exclusive claim categories have probabilities 0.20, 0.35, and 0.45. They form a valid exhaustive model, and the probability of either of the last two categories is 0.80.
Mistake to avoid: Using category probabilities that exceed 1 when added, despite claiming the categories are disjoint.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
4. Addition and Inclusion–Exclusion
For two events, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The subtraction removes outcomes counted twice. For three events, subtract each pairwise intersection and add the triple intersection back. This calculation applies whether or not the events are independent.
Worked example: A portfolio has probability 0.35 of a property claim, 0.25 of a liability claim, and 0.10 of both. The probability of at least one of these claim types is 0.35 + 0.25 − 0.10 = 0.50.
Mistake to avoid: Adding overlapping event probabilities without removing their shared outcomes.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
5. Complementary Events and At Least One
The complement rule is P(A) = 1 − P(Aᶜ). For an 'at least one' event, the complement is often the simpler event 'none.' Multiplying the probabilities of no occurrence across trials requires independence; the complement rule itself does not require that assumption.
Worked example: Three independent devices each fail with probability 0.20. The probability that none fails is 0.80³ = 0.512, so the probability that at least one fails is 0.488.
Mistake to avoid: Using 1 − (1 − p)ⁿ for dependent failures without establishing the probability of no failures.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
6. Mutual Exclusivity Versus Independence
Mutually exclusive events cannot occur together, so their intersection has probability zero. Independent events satisfy P(A ∩ B) = P(A)P(B). Consequently, two mutually exclusive events with positive probabilities are dependent: observing one rules out the other.
Worked example: A single claim is classified as either theft or water damage, with probabilities 0.20 and 0.30. Their joint probability is zero, whereas independence would require 0.20 × 0.30 = 0.06.
Mistake to avoid: Calling events independent merely because they describe different categories.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
7. Pairwise and Mutual Independence
Pairwise independence checks every pair of events. Mutual independence additionally requires the product rule for every larger subset. With three events, checking the three pairwise intersections is insufficient; the triple intersection must also equal the product of the three individual probabilities.
Worked example: Choose uniformly from 00, 01, 10, and 11. Let A mean the first bit is 1, B the second is 1, and C the bits match. Every pair intersects with probability 1/4, but P(A ∩ B ∩ C) = 1/4 rather than 1/8.
Mistake to avoid: Inferring mutual independence solely from pairwise checks.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
8. The Multiplication Chain Rule
For sequential conditions, P(A ∩ B ∩ C) = P(A)P(B|A)P(C|A ∩ B), provided the conditioning probabilities are defined. Each factor reflects all information already imposed. Independence simplifies these conditional factors, but the chain rule remains valid without independence.
Worked example: An account enters review with probability 0.40. Given review, an issue is found with probability 0.50; given both, escalation occurs with probability 0.20. The probability of the complete sequence is 0.04.
Mistake to avoid: Replacing a conditional probability with its unconditional counterpart without justification.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
Counting and Conditional Probability
9. The Product Rule for Counting
If a construction has successive stages with a fixed number of choices at each stage, multiply those counts. The choices need not be random. If earlier selections change the available choices, use the appropriate count for each branch rather than assuming one uniform product.
Worked example: A policy configuration allows any of three coverage options, either of two billing plans, and any of four service districts. With no restrictions between choices, there are 3 × 2 × 4 = 24 configurations.
Mistake to avoid: Adding stage counts when each completed configuration requires one choice from every stage.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
10. Permutations for Distinct Roles
When order matters and selections are made without replacement, the number of arrangements of r objects from n distinct objects is n!/(n − r)!. Assigning different roles makes order relevant even if the selected people form the same underlying group.
Worked example: Six analysts are eligible for three distinct roles: reviewer, checker, and recorder. There are 6 × 5 × 4 = 120 assignments because each appointment removes one person from the remaining choices.
Mistake to avoid: Using a combination count when exchanging two selected people changes their assigned roles.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
11. Combinations for Unordered Selections
When order does not matter, choosing r objects from n distinct objects gives C(n,r) = n!/[r!(n − r)!]. Dividing the permutation count by r! removes repeated orderings of each selected group. This assumes selection without replacement and no additional restrictions.
Worked example: A review team contains three people selected from nine analysts, with no distinct positions. The number of teams is C(9,3) = 84. Each team would otherwise appear in six different orderings.
Mistake to avoid: Counting the same group several times because its members were listed in different orders.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
12. Multinomial Counts for Repeated Categories
For n positions filled with category counts n₁, …, nₖ totaling n, the number of distinct sequences is n!/(n₁! ⋯ nₖ!). The denominator removes rearrangements among identical category labels. This counts sequences with specified totals, rather than selecting one category independently at every position.
Worked example: Eight daily records contain three A labels, two B labels, and three C labels. Their possible label sequences number 8!/(3!2!3!) = 560.
Mistake to avoid: Using 8! even though swapping two identical labels produces no new sequence.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
13. Restricted Arrangements Using Blocks
An adjacency restriction can sometimes be handled by treating the required adjacent objects as one block. Count arrangements of the block and remaining objects, then multiply by the allowable arrangements inside the block. Other restrictions may require separate cases rather than this shortcut.
Worked example: Six distinct files are arranged randomly, and files A and B must be adjacent. Treating them as a block gives 5! external arrangements and two internal orders, so the probability is 2 × 5!/6! = 1/3.
Mistake to avoid: Forgetting that the adjacent pair can appear as either AB or BA.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
14. Conditional Probability as a Restricted Universe
For P(B) > 0, conditional probability is P(A|B) = P(A ∩ B)/P(B). Conditioning restricts attention to B and rescales its total probability to 1. The numerator must satisfy both the target condition and the information already given.
Worked example: The probability that a policy has both an inspection and a claim is 0.12, while the probability of a claim is 0.30. Given a claim, the probability of an inspection is 0.12/0.30 = 0.40.
Mistake to avoid: Dividing by the target event's probability instead of the probability of the given information.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
15. The Law of Total Probability
If B₁, …, Bₖ are disjoint and exhaustive groups, P(A) = ΣP(A|Bᵢ)P(Bᵢ). This averages group-specific probabilities using group proportions. The groups must form a partition; overlapping groups would count some outcomes more than once.
Worked example: A portfolio is 70% urban and 30% rural. Annual claim probabilities are 0.02 and 0.08 respectively. The portfolio claim probability is 0.70 × 0.02 + 0.30 × 0.08 = 0.038.
Mistake to avoid: Taking an unweighted average of claim probabilities when the groups have different sizes.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
16. Bayes' Theorem and Updated Group Probabilities
Bayes' theorem reverses a conditional probability: P(B|A) = P(A|B)P(B)/P(A). Obtain the denominator from all ways A can occur, often using total probability. A high event probability within a small group does not by itself determine that group's share among observed events.
Worked example: High-risk accounts form 20% of a portfolio and claim with probability 0.15; other accounts claim with probability 0.025. Given a claim, the high-risk probability is 0.03/(0.03 + 0.02) = 0.60.
Mistake to avoid: Equating the probability of a claim given high risk with the probability of high risk given a claim.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
Discrete Random Variables
17. Probability Mass Functions and Support
A probability mass function assigns p(x) = P(X = x) to each supported value. Its values must be nonnegative and sum to 1. Find a missing normalizing constant by summing over the stated support before calculating any event probability.
Worked example: Suppose p(x) = k(1 + x) for x = 0, 1, 2, and zero elsewhere. Since k(1 + 2 + 3) = 1, k = 1/6. Therefore P(X ≥ 1) = 5/6.
Mistake to avoid: Normalizing over values outside the support or assuming the displayed formula applies everywhere.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
18. Discrete Cumulative Distribution Functions
The cumulative distribution function is F(x) = P(X ≤ x). A discrete CDF jumps at values carrying probability, and each jump equals the mass there. Interval probabilities satisfy P(a < X ≤ b) = F(b) − F(a), so endpoint conventions matter.
Worked example: Let X equal 0, 1, or 3 with probabilities 0.20, 0.50, and 0.30. Then F(2) = 0.70, the jump at 1 is 0.50, and P(1 < X ≤ 3) = 0.30.
Mistake to avoid: Subtracting F(a) when the requested interval includes probability mass at a.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
19. Bernoulli Trials and the Binomial Distribution
A Bernoulli variable records one success or failure. The sum of n independent Bernoulli variables with the same success probability p is binomial: P(X = k) = C(n,k)pᵏ(1 − p)ⁿ⁻ᵏ. Its mean is np and variance is np(1 − p).
Worked example: Five independent inspections each find an issue with probability 0.20. The probability of exactly two issues is C(5,2)(0.20)²(0.80)³ = 0.2048.
Mistake to avoid: Applying a binomial model when success probabilities vary across trials or sampling creates dependence.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
20. Geometric Waiting Times
If T counts trials through the first success in independent trials with constant success probability p, P(T = t) = (1 − p)ᵗ⁻¹p for t = 1, 2, …. Its mean is 1/p. A failures-before-success variable instead starts at zero and equals T − 1.
Worked example: A review succeeds independently with probability 0.25 on each attempt. Success first occurs on attempt three with probability 0.75² × 0.25 = 0.140625; the expected number of attempts is four.
Mistake to avoid: Using the failures-count formula for a variable that includes the successful trial.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
21. Negative Binomial Counts
If K counts failures before the rth success, P(K = k) = C(k + r − 1,r − 1)pʳ(1 − p)ᵏ. The final trial must be a success; the earlier trials contain the remaining successes and all failures. The mean failure count is r(1 − p)/p.
Worked example: With success probability 0.40, exactly two failures before the third success has probability C(4,2)(0.40)³(0.60)² = 0.13824. This outcome uses five total trials.
Mistake to avoid: Allowing the final trial to be a failure when counting sequences ending at the rth success.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
22. Hypergeometric Sampling Without Replacement
For a population of N objects containing K successes, an equally likely sample of n distinct objects has P(X = x) = C(K,x)C(N − K,n − x)/C(N,n). Sampling without replacement changes the success probability as objects are removed, producing dependent selections.
Worked example: Ten components include four damaged ones. A sample of three is selected without replacement. The probability of exactly two damaged components is C(4,2)C(6,1)/C(10,3) = 36/120 = 0.30.
Mistake to avoid: Using a binomial formula as an exact calculation for this finite-population sample.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
23. Poisson Counts and Exposure Scaling
A Poisson variable with parameter λ has P(X = k) = e⁻ˡᵃᵐᵇᵈᵃλᵏ/k!, with mean and variance both λ. Under a constant-rate Poisson model, multiply the rate by the exposure duration to obtain the parameter for that interval. Keep rate units and duration units consistent.
Worked example: Calls follow a Poisson model at two per hour. Over 1.5 hours, λ = 3, so the probability of exactly two calls is e⁻³3²/2! ≈ 0.2240.
Mistake to avoid: Using the hourly parameter unchanged for an interval of a different length.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
24. Discrete Uniform Distributions
A discrete uniform variable assigns equal mass to each value in its finite support. For consecutive integers a through b, there are m = b − a + 1 values, the mean is (a + b)/2, and the variance is (m² − 1)/12. Count endpoints inclusively.
Worked example: A variable is uniform on 2, 3, 4, 5, and 6. Each value has probability 1/5, P(X ≥ 5) = 2/5, the mean is 4, and the variance is 2.
Mistake to avoid: Using b − a as the number of supported integers.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
Continuous Random Variables
25. Densities and Normalization
A probability density is nonnegative and integrates to 1 over its support. Probabilities are areas under the density, not density values at individual points. A density may exceed 1 on a short interval without violating probability axioms, because its total area determines validity.
Worked example: Let f(x) = kx for 0 < x < 2. Normalization gives 2k = 1, so k = 1/2. Integrating x/2 from 1 to 2 gives P(1 < X < 2) = 3/4.
Mistake to avoid: Interpreting f(1) as P(X = 1), which is zero for a continuous variable.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
26. Continuous CDFs and Survival Probabilities
For a continuous variable, F(x) is the integral of its density up to x, and the survival function is 1 − F(x). Where differentiable, F'(x) = f(x). Interval probabilities are CDF differences; including or excluding a single endpoint does not change them.
Worked example: Suppose F(x) = x³ on 0 ≤ x ≤ 1, with the usual extensions outside that interval. Then f(x) = 3x² and P(0.5 < X ≤ 0.8) = 0.8³ − 0.5³ = 0.387.
Mistake to avoid: Forgetting the CDF must equal zero below the support and one above it.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
27. Continuous Uniform Distributions
For X uniform on [a,b], the density is 1/(b − a), and an interval's probability is its overlap length with [a,b] divided by b − a. The mean is (a + b)/2 and variance is (b − a)²/12. Values outside the support contribute no probability.
Worked example: A loss is uniform from 50 to 150. The probability it exceeds 120 is (150 − 120)/(150 − 50) = 0.30, and its expected value is 100.
Mistake to avoid: Using the full requested interval length when part of that interval lies outside the support.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
28. Exponential Lifetimes and Memorylessness
An exponential variable with rate λ has survival probability e⁻ˡᵃᵐᵇᵈᵃᵗ for t ≥ 0 and mean 1/λ. Its memoryless property states P(X > s + t|X > s) = P(X > t). This property belongs to the exponential model, rather than to lifetimes generally.
Worked example: A lifetime has exponential rate 0.20 per day. Given survival through day five, the probability of surviving ten additional days is e⁻² ≈ 0.1353. The unconditional expected lifetime is five days.
Mistake to avoid: Applying memorylessness to an arbitrary lifetime distribution.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
29. Gamma Shape, Scale, and Waiting Times
Using shape α and scale θ, a gamma variable has mean αθ and variance αθ². An integer shape arises from summing that many independent exponential waiting times with common rate 1/θ. Some formulas use a rate instead of a scale, so identify the parameter convention first.
Worked example: Three independent exponential waits each have rate 0.50 per hour. Their total has gamma shape 3 and scale 2 hours, giving mean 6 hours and variance 12 squared hours.
Mistake to avoid: Substituting a rate directly into a formula written for scale.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
30. Beta Distributions for Bounded Proportions
A beta variable lies between 0 and 1 and has density proportional to xᵅ⁻¹(1 − x)ᵝ⁻¹, with positive shape parameters. Its mean is α/(α + β). When both shapes exceed 1, its interior mode is (α − 1)/(α + β − 2); boundary cases require separate analysis.
Worked example: For beta shapes 2 and 3, the density is 12x(1 − x)². Its mean is 2/5, while its mode is 1/3, showing that mean and most likely location can differ.
Mistake to avoid: Applying the interior-mode formula when either shape is at most 1.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
31. Normal Standardization
If X is normal with mean μ and standard deviation σ > 0, then Z = (X − μ)/σ is standard normal. Express the requested event in terms of Z before using Φ, the standard normal CDF. Upper-tail probabilities use 1 − Φ(z), and interval probabilities use CDF differences.
Worked example: A normally distributed amount has mean 80 and standard deviation 12. Exceeding 104 corresponds to Z > 2, so the probability is 1 − Φ(2) ≈ 0.0228.
Mistake to avoid: Dividing by the variance instead of the standard deviation.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
32. Transforming a Continuous Variable
For a monotone transformation Y = g(X), obtain the transformed CDF by rewriting the event g(X) ≤ y on X's support. Differentiate to find the density when appropriate. A decreasing transformation reverses inequalities; a transformation that is not one-to-one may require multiple contributing intervals.
Worked example: Let X be uniform on (0,2) and Y = X². For 0 < y < 4, Fᵧ(y) = √y/2 and fᵧ(y) = 1/(4√y). Thus P(Y ≤ 1) = 1/2.
Mistake to avoid: Reusing the original density without adjusting for the transformation or its support.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
Expectations and Distribution Summaries
33. Expected Values of Functions
Calculate E[g(X)] directly as Σg(x)p(x) for a discrete variable or ∫g(x)f(x)dx for a continuous variable. You do not need to derive the distribution of g(X) first. Expectation is linear, but generally E[g(X)] does not equal g(E[X]) for nonlinear g.
Worked example: A loss takes values 0, 2, and 5 with probabilities 0.50, 0.30, and 0.20. E[X²] = 0 + 4(0.30) + 25(0.20) = 6.20, whereas E[X]² = 1.60² = 2.56.
Mistake to avoid: Moving a square or another nonlinear function outside an expectation.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
34. Variance and Standard Deviation
Variance measures expected squared deviation from the mean: Var(X) = E[X²] − E[X]². Standard deviation is its nonnegative square root and has the same units as X. Calculate the mean and second moment separately before subtracting; a materially negative variance signals an error.
Worked example: A payment is 10 with probability 0.20 and zero otherwise. Its mean is 2 and second moment is 20, giving variance 20 − 4 = 16 and standard deviation 4.
Mistake to avoid: Reporting variance as though it had the same units as the payment.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
35. Raw and Central Moments
A raw moment of order r is E[Xʳ]; a central moment is E[(X − μ)ʳ]. Central moments describe variation around the mean rather than around zero. Expand the power before taking expectations when useful, and verify that the required moment exists.
Worked example: X equals 1 or 3 with equal probabilities, so μ = 2. Its third raw moment is (1³ + 3³)/2 = 14, while its third central moment is [(-1)³ + 1³]/2 = 0.
Mistake to avoid: Treating a raw higher-order moment as a central moment without centering X.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
36. Affine Changes and Coefficient of Variation
For Y = aX + b, E[Y] = aE[X] + b and SD(Y) = |a|SD(X). With a positive mean, the coefficient of variation is SD(X)/E[X]. Positive scaling alone preserves this ratio, while adding a constant generally changes it.
Worked example: A loss has mean 100 and standard deviation 20. For Y = 1.5X + 30, the mean is 180, standard deviation is 30, and coefficient of variation is 1/6 rather than 0.20.
Mistake to avoid: Adding b to the standard deviation when transforming the variable.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
37. Medians and Percentiles
A convenient percentile definition is qₚ = inf{x: F(x) ≥ p}. For a strictly increasing continuous CDF, solve F(qₚ) = p. For a discrete distribution, the CDF may jump past p, so an exact equality may be impossible. The median corresponds to the 50th percentile under this convention.
Worked example: A payment is zero with probability 0.60 and 100 with probability 0.40. Its 50th percentile is zero and its 75th percentile is 100, although its mean is 40.
Mistake to avoid: Interpolating between discrete support values without a stated interpolation convention.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
38. Conditional Univariate Distributions
Conditioning on an event restricts a variable's support and renormalizes the remaining mass or density. For a continuous variable conditioned on X ∈ A, use f(x)/P(X ∈ A) on A and zero elsewhere. Conditional expectations must be calculated under this revised distribution.
Worked example: If X is uniform on (0,10), then given X > 6 it is uniform on (6,10). Consequently E[X|X > 6] = 8 and P(X > 9|X > 6) = 1/4.
Mistake to avoid: Using the original density unchanged after discarding part of its support.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
Discrete Multivariate Distributions
39. Joint Probability Mass Functions
A joint PMF gives p(x,y) = P(X = x, Y = y). Its entries must be nonnegative and sum to 1 over the actual joint support. The support may exclude combinations even when each coordinate value appears individually, so check feasible pairs before summing.
Worked example: For x,y in {0,1}, suppose p(x,y) = k(x + 2y + 1). The four weights are 1, 3, 2, and 4, giving k = 0.10 and P(X = 1,Y = 1) = 0.40.
Mistake to avoid: Normalizing each row separately when the formula defines one joint distribution.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
40. Joint CDFs and Rectangle Probabilities
The joint CDF is F(x,y) = P(X ≤ x,Y ≤ y). For a rectangle (a,b] × (c,d], its probability is F(b,d) − F(a,d) − F(b,c) + F(a,c). The final addition restores the lower-left overlap subtracted twice.
Worked example: A binary pair has probabilities 0.10, 0.20, 0.30, and 0.40 at (0,0), (0,1), (1,0), and (1,1). The upper-right cell has probability 1 − 0.30 − 0.40 + 0.10 = 0.40.
Mistake to avoid: Subtracting two CDF values without restoring their overlapping region.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
41. Marginal Distributions
A marginal distribution removes the other variable by summing: pₓ(x) = Σᵧp(x,y). This retains all ways the chosen value can occur, regardless of the other coordinate. Marginal moments are then calculated from the resulting one-variable PMF.
Worked example: Joint probabilities at (0,0), (0,1), (1,0), and (1,1) are 0.15, 0.25, 0.35, and 0.25. Therefore P(X = 1) = 0.60 and P(Y = 1) = 0.50.
Mistake to avoid: Selecting one joint cell as a marginal probability instead of summing across the other coordinate.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
42. Conditional Discrete Distributions
For pᵧ(y) > 0, the conditional PMF is pₓ|ᵧ(x|y) = p(x,y)/pᵧ(y). The resulting probabilities sum to 1 over x for that fixed y. Conditioning selects a slice of the joint distribution and then normalizes that slice.
Worked example: Suppose P(X = 1,Y = 1) = 0.18 and P(Y = 1) = 0.30. Then P(X = 1|Y = 1) = 0.60, leaving conditional probability 0.40 for X = 0 if X is binary.
Mistake to avoid: Dividing by P(X = 1) when the given information fixes Y.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
43. Independence of Random Variables
Discrete random variables are independent when p(x,y) = pₓ(x)pᵧ(y) for every pair of values. A single matching cell does not establish independence. Under independence, conditioning on a supported value of one variable leaves the other variable's distribution unchanged.
Worked example: A binary joint table has rows (0.12,0.18) and (0.28,0.42). Row totals are 0.30 and 0.70; column totals are 0.40 and 0.60. Every cell equals its row total times its column total, establishing independence.
Mistake to avoid: Checking factorization for only one convenient pair.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
44. Joint Moments and Covariance
Calculate E[XY] by summing xy times the joint PMF. Covariance is E[XY] − E[X]E[Y], measuring the direction of linear co-movement. Independence implies zero covariance when the moments exist, but covariance calculations must use the joint distribution unless independence is established.
Worked example: Binary indicators satisfy P(X = 1) = 0.40, P(Y = 1) = 0.50, and P(X = 1,Y = 1) = 0.30. Since E[XY] = 0.30, covariance is 0.30 − 0.40(0.50) = 0.10.
Mistake to avoid: Replacing E[XY] with E[X]E[Y] for dependent variables.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
45. Correlation and Nonlinear Dependence
Correlation is covariance divided by the product of the two positive standard deviations. It is dimensionless and measures linear association. Zero correlation does not establish independence: a nonlinear relationship may remain. Correlation is undefined if either variable has zero variance.
Worked example: Let X be uniform on {-1,0,1} and Y = X². Symmetry gives E[XY] = E[X] = 0, so correlation is zero. Yet P(Y = 0|X = 0) = 1, compared with P(Y = 0) = 1/3.
Mistake to avoid: Interpreting zero correlation as absence of every kind of dependence.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
46. Conditional Moments and Total Variance
The laws of total expectation and variance give E[X] = E[E[X|Y]] and Var(X) = E[Var(X|Y)] + Var(E[X|Y]). Total variance combines average within-group variation with variation between group means. Both components matter when mixing populations.
Worked example: Two groups have weights 0.75 and 0.25, means 2 and 6, and variances 1 and 9. The overall mean is 3. Average conditional variance is 3, variance of conditional means is 3, and total variance is 6.
Mistake to avoid: Averaging group variances while omitting differences between their means.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
Order Statistics and Aggregate Distributions
47. Distributions of Minima and Maxima
For independent variables, the maximum is at most x exactly when every variable is at most x; the minimum exceeds x exactly when every variable exceeds x. Thus multiply their CDFs or survival probabilities. For iid variables, these products become F(x)ⁿ and [1 − F(x)]ⁿ.
Worked example: Two independent amounts are uniform on (0,10). Their maximum is at most 7 with probability 0.70² = 0.49, while their minimum exceeds 7 with probability 0.30² = 0.09.
Mistake to avoid: Using these product formulas when the original observations are dependent.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
48. The kth Order Statistic
For an iid sample of size n, the kth smallest observation is at most x exactly when at least k observations are at most x. That count is binomial with success probability F(x). This converts an order-statistic probability into a binomial tail.
Worked example: Four observations are iid uniform on (0,1). The second smallest is at most 0.30 with probability 1 − 0.70⁴ − 4(0.30)(0.70)³ = 0.3483.
Mistake to avoid: Using exactly k observations below the threshold when the event requires at least k.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
49. Joint Distributions of Extreme Order Statistics
For n ≥ 2 iid continuous observations, the joint density of minimum U and maximum V is n(n − 1)[F(v) − F(u)]ⁿ⁻²f(u)f(v), for u < v. The middle observations must lie between the extremes. Ordered values are generally dependent even when the original observations are independent.
Worked example: For three iid uniform observations on (0,1), this density is 6(v − u). The event U > 0.20 and V < 0.80 means all three observations lie in an interval of length 0.60, giving probability 0.216.
Mistake to avoid: Multiplying marginal densities of the minimum and maximum as though they were independent.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
50. Exact Distributions of Discrete Sums
For independent discrete variables, convolution gives P(X + Y = s) = Σₓpₓ(x)pᵧ(s − x). Include every supported pair producing the target sum. For other linear combinations, enumerate the pairs satisfying the corresponding equation or inequality and add their joint probabilities.
Worked example: Independent Bernoulli variables have success probabilities 0.40 and 0.30. Their sum equals 1 through either (1,0) or (0,1), so its probability is 0.40(0.70) + 0.60(0.30) = 0.46.
Mistake to avoid: Counting only one route to a sum that several outcome pairs can produce.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
51. Moments of Linear Combinations
Expectation is linear without an independence requirement. For independent variables, Var(aX + bY) = a²Var(X) + b²Var(Y). For dependent variables, add 2abCov(X,Y). A negative coefficient changes the mean contribution's sign but does not make its individual variance contribution negative.
Worked example: Independent X and Y have means 3 and 5 and variances 4 and 9. For S = 2X − Y, E[S] = 1, Var(S) = 4(4) + 9 = 25, and SD(S) = 5.
Mistake to avoid: Subtracting variances because the linear combination subtracts a variable.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
52. Exact Normal Linear Combinations
A linear combination of independent normal variables is exactly normal. Calculate its mean by applying the coefficients directly and its variance using squared coefficients. Standardize with the resulting standard deviation. This exact distributional result does not require a large number of terms.
Worked example: Independent normal amounts X and Y have means 100 and 70 and variances 16 and 9. Their difference is normal with mean 30 and variance 25. Therefore P(X − Y > 40) = 1 − Φ(2) ≈ 0.0228.
Mistake to avoid: Treating a difference of normal variables as having variance 16 − 9.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
53. Central Limit Theorem for Sums and Means
For iid variables with finite mean μ and positive finite variance σ², the standardized sum approaches a standard normal distribution as sample size increases. Approximate a sum using mean nμ and standard deviation σ√n, or a sample mean using μ and σ/√n. Approximation quality depends on distribution shape and the requested tail.
Worked example: For 64 iid losses with mean 10 and standard deviation 4, the sample mean has standard deviation 0.50. The CLT approximation gives P(sample mean > 11) ≈ 1 − Φ(2) = 0.0228.
Mistake to avoid: Presenting the approximation as exact for a nonnormal underlying distribution.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
54. Continuity Correction in Normal Approximations
When approximating an integer-valued distribution by a continuous normal distribution, place boundaries halfway between integers. For example, X ≥ k becomes a normal event above k − 0.5. This aligns the continuous area with the included integer masses; it does not guarantee that the underlying normal approximation is accurate.
Worked example: For X binomial with n = 100 and p = 0.40, approximate P(X ≥ 45) using a normal mean of 40, variance 24, and boundary 44.5. The standardized boundary is about 0.919, giving a probability near 0.179.
Mistake to avoid: Moving the boundary to 45.5 for an event that includes X = 45.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
Insurance Losses and Payments
55. Ordinary Deductibles and Zero-Payment Mass
With loss X and an ordinary deductible d, payment is Y = max(X − d,0). Losses at or below d become zero payments, creating probability mass at zero even when X is continuous. For y ≥ 0, the payment CDF is P(Y ≤ y) = Fₓ(d + y).
Worked example: A loss is uniform on (0,400), with deductible 100. The probability of zero payment is 100/400 = 0.25. A loss of 260 produces payment 160.
Mistake to avoid: Assuming a continuous loss always creates a purely continuous payment distribution.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
56. Payment per Loss Versus Payment per Payment
Payment per loss includes zero payments; payment per payment conditions on a positive payment. For nonnegative Y with P(Y > 0) > 0, E[Y|Y > 0] = E[Y]/P(Y > 0). Distinguish the population being averaged before calculating a mean or distribution.
Worked example: For a uniform loss on (0,600) and deductible 200, positive payments are uniform on (0,400), averaging 200. Payment occurs with probability 2/3, so expected payment per loss is 400/3.
Mistake to avoid: Using the average positive payment as the expected payment across all losses.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
57. Coinsurance and the Order of Policy Terms
If the contract specifies an ordinary deductible first and insurer coinsurance share c afterward, payment without a limit is Y = c max(X − d,0). The percentage must refer to the insurer's share. Different stated ordering or definitions require a different formula.
Worked example: A contract pays 75% of loss above a deductible of 200. For loss 1,000, the insurer pays 0.75(1,000 − 200) = 600; the policyholder bears the remaining 400.
Mistake to avoid: Calculating 0.75X − d, which applies the deductible after coinsurance.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
58. Benefit Limits and Their Loss Thresholds
For a per-claim benefit limit L applied after deductible and insurer share c > 0, payment is min(c max(X − d,0),L). The cap begins at loss d + L/c. A limit on insurer payment differs from a limit on covered loss before coinsurance.
Worked example: With deductible 100, insurer share 0.80, and payment limit 240, the cap starts at loss 400. A loss of 500 gives an uncapped payment of 320 but an actual payment of 240.
Mistake to avoid: Using d + L as the cap threshold when coinsurance also applies.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
59. Inflation with Fixed Policy Terms
When losses inflate by factor a while deductibles and limits remain fixed, transform the loss first and then apply the unchanged policy terms. Payment generally does not scale by a because deductibles and caps introduce nonlinearities. Scale policy terms only when the stated contract also adjusts them.
Worked example: A base loss of 800 increases by 10%. With fixed deductible 100 and insurer share 0.75, payment becomes 0.75(880 − 100) = 585. Inflating the previous payment of 525 would instead give 577.50.
Mistake to avoid: Inflating an old payment directly despite a fixed deductible.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
60. Moments of Deductible-and-Limited Payments
For a bounded nonnegative payment Y, E[Y] = ∫P(Y > y)dy and E[Y²] = 2∫yP(Y > y)dy, integrating from zero to its maximum. These survival formulas accommodate payment masses at zero and at a cap. Then obtain variance by subtracting the squared mean.
Worked example: Let X be uniform on (0,100) and Y = min(max(X − 20,0),50). For 0 ≤ y < 50, P(Y > y) = 0.80 − y/100. Thus E[Y] = 27.50, E[Y²] = 3500/3, variance = 4925/12, and standard deviation ≈ 20.26.
Mistake to avoid: Integrating only the interior payment density and omitting the probability mass at the limit.
Reference: SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
Sources
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- SOA Probability Exam November 2026 syllabus, format, learning outcomes, weights, and readings
- Probability (P) Exam