ACAS BEST PRACTICE KNOWLEDGE EXAMS 1-6 2026/2027 |
Complete Questions and Solutions | CAS Syllabus Aligned |
Actuarial Certification | Pass Guaranteed - A+ Graded
Best Practice Exam 1: Probability & Statistics
Foundational probability theory, distributions, and statistical inference for actuarial
applications
Q1: An actuary is analyzing claim frequencies for a commercial auto portfolio. Given
that 60% of policies are from urban areas and 40% from rural areas, with urban policies
having a 15% claim probability and rural policies having an 8% claim probability, what is
the probability that a randomly selected policy with a claim is from an urban area?
A. 0.52 — this would result from incorrectly weighting the probabilities equally
B. 0.60 — this simply uses the prior probability without applying Bayes' theorem
C. 0.737 — this correctly applies Bayes' theorem: (0.60 × 0.15) / [(0.60 × 0.15) + (0.40 ×
0.08)] [CORRECT]
D. 0.85 — this incorrectly uses the urban claim probability as the answer
Correct Answer: C
,Rationale: Using Bayes' theorem, P(Urban|Claim) = P(Claim|Urban) × P(Urban) /
P(Claim). The numerator is 0.60 × 0.15 = 0.09. The denominator is (0.60 × 0.15) + (0.40
× 0.08) = 0.09 + 0.032 = 0.122. Thus 0.09/0.122 ≈ 0.737. Option A incorrectly averages,
B ignores the likelihood information, and D confuses conditional probabilities.
Q2: For a Poisson distribution with parameter λ = 4, what is the variance of the
distribution?
A. 2 — this is the square root of lambda, confusing standard deviation with variance
B. 4 — this equals the mean, and for Poisson distributions, variance equals mean
[CORRECT]
C. 8 — this incorrectly doubles the parameter
D. 16 — this squares the parameter, confusing with second moment calculation
Correct Answer: B
Rationale: A fundamental property of the Poisson distribution is that both mean and
variance equal λ. Option A gives the standard deviation, C is an arbitrary doubling, and D
confuses Var(X) with E[X²] without subtracting E[X]².
Q3: Let X and Y be independent random variables with E[X] = 5, Var(X) = 4, E[Y] = 3, and
Var(Y) = 9. What is the variance of 2X - 3Y + 7?
A. 17 — this incorrectly adds variances without considering coefficients
,B. 35 — this miscalculates by not squaring the coefficients
C. 97 — this correctly calculates: 4×Var(X) + 9×Var(Y) = 16 + 81 = 97 [CORRECT]
D. 107 — this incorrectly includes the constant 7 in the variance calculation
Correct Answer: C
Rationale: For independent random variables, Var(aX + bY + c) = a²Var(X) + b²Var(Y).
The constant 7 has zero variance. Thus: 4(4) + 9(9) = 16 + 81 = 97. Option A ignores
coefficients, B uses linear instead of squared coefficients, and D erroneously adds
Var(7).
Q4: The moment generating function of a random variable X is M_X(t) = (1 - 2t)^(-3) for t
< 1/2. What is E[X]?
A. 3 — this incorrectly takes the exponent as the mean
B. 6 — this correctly finds M'X(0) = 6(1-2t)^(-4)|{t=0} = 6 [CORRECT]
C. 9 — this computes E[X²] or confuses with second moment
D. 12 — this doubles the correct answer, possibly from miscalculating the derivative
Correct Answer: B
, Rationale: E[X] = M'_X(0). The derivative is M'_X(t) = 6(1-2t)^(-4), so M'_X(0) = 6. This is a
Gamma distribution with α=3, θ=2, where mean = αθ = 6. Option A confuses parameters,
C is likely E[X²], and D contains a calculation error.
Q5: In a portfolio of 100 independent risks, each with a 5% probability of generating a
claim, what is the approximate probability of observing exactly 5 claims using the
Poisson approximation to the binomial?
A. 0.0337 — this uses the exact binomial formula
B. 0.1755 — this correctly applies Poisson with λ=5: e^(-5) × 5^! ≈ 0.1755
[CORRECT]
C. 0.5000 — this incorrectly assumes the mean equals the probability
D. 0.6065 — this calculates e^(-5) only, missing the probability mass function
Correct Answer: B
Rationale: For large n and small p, Binomial(n,p) ≈ Poisson(λ=np). Here λ = 100 × 0.05 =
5. P(X=5) = e^(-5) × 5^ ≈ 0.1755. Option A is correct but not the approximation
asked for, C misunderstands distributions, and D gives only the survival function at zero.
Q6: For a standard normal random variable Z, what is P(-1.96 < Z < 1.96)?
A. 0.90 — this corresponds to approximately ±1.645 standard deviations
Complete Questions and Solutions | CAS Syllabus Aligned |
Actuarial Certification | Pass Guaranteed - A+ Graded
Best Practice Exam 1: Probability & Statistics
Foundational probability theory, distributions, and statistical inference for actuarial
applications
Q1: An actuary is analyzing claim frequencies for a commercial auto portfolio. Given
that 60% of policies are from urban areas and 40% from rural areas, with urban policies
having a 15% claim probability and rural policies having an 8% claim probability, what is
the probability that a randomly selected policy with a claim is from an urban area?
A. 0.52 — this would result from incorrectly weighting the probabilities equally
B. 0.60 — this simply uses the prior probability without applying Bayes' theorem
C. 0.737 — this correctly applies Bayes' theorem: (0.60 × 0.15) / [(0.60 × 0.15) + (0.40 ×
0.08)] [CORRECT]
D. 0.85 — this incorrectly uses the urban claim probability as the answer
Correct Answer: C
,Rationale: Using Bayes' theorem, P(Urban|Claim) = P(Claim|Urban) × P(Urban) /
P(Claim). The numerator is 0.60 × 0.15 = 0.09. The denominator is (0.60 × 0.15) + (0.40
× 0.08) = 0.09 + 0.032 = 0.122. Thus 0.09/0.122 ≈ 0.737. Option A incorrectly averages,
B ignores the likelihood information, and D confuses conditional probabilities.
Q2: For a Poisson distribution with parameter λ = 4, what is the variance of the
distribution?
A. 2 — this is the square root of lambda, confusing standard deviation with variance
B. 4 — this equals the mean, and for Poisson distributions, variance equals mean
[CORRECT]
C. 8 — this incorrectly doubles the parameter
D. 16 — this squares the parameter, confusing with second moment calculation
Correct Answer: B
Rationale: A fundamental property of the Poisson distribution is that both mean and
variance equal λ. Option A gives the standard deviation, C is an arbitrary doubling, and D
confuses Var(X) with E[X²] without subtracting E[X]².
Q3: Let X and Y be independent random variables with E[X] = 5, Var(X) = 4, E[Y] = 3, and
Var(Y) = 9. What is the variance of 2X - 3Y + 7?
A. 17 — this incorrectly adds variances without considering coefficients
,B. 35 — this miscalculates by not squaring the coefficients
C. 97 — this correctly calculates: 4×Var(X) + 9×Var(Y) = 16 + 81 = 97 [CORRECT]
D. 107 — this incorrectly includes the constant 7 in the variance calculation
Correct Answer: C
Rationale: For independent random variables, Var(aX + bY + c) = a²Var(X) + b²Var(Y).
The constant 7 has zero variance. Thus: 4(4) + 9(9) = 16 + 81 = 97. Option A ignores
coefficients, B uses linear instead of squared coefficients, and D erroneously adds
Var(7).
Q4: The moment generating function of a random variable X is M_X(t) = (1 - 2t)^(-3) for t
< 1/2. What is E[X]?
A. 3 — this incorrectly takes the exponent as the mean
B. 6 — this correctly finds M'X(0) = 6(1-2t)^(-4)|{t=0} = 6 [CORRECT]
C. 9 — this computes E[X²] or confuses with second moment
D. 12 — this doubles the correct answer, possibly from miscalculating the derivative
Correct Answer: B
, Rationale: E[X] = M'_X(0). The derivative is M'_X(t) = 6(1-2t)^(-4), so M'_X(0) = 6. This is a
Gamma distribution with α=3, θ=2, where mean = αθ = 6. Option A confuses parameters,
C is likely E[X²], and D contains a calculation error.
Q5: In a portfolio of 100 independent risks, each with a 5% probability of generating a
claim, what is the approximate probability of observing exactly 5 claims using the
Poisson approximation to the binomial?
A. 0.0337 — this uses the exact binomial formula
B. 0.1755 — this correctly applies Poisson with λ=5: e^(-5) × 5^! ≈ 0.1755
[CORRECT]
C. 0.5000 — this incorrectly assumes the mean equals the probability
D. 0.6065 — this calculates e^(-5) only, missing the probability mass function
Correct Answer: B
Rationale: For large n and small p, Binomial(n,p) ≈ Poisson(λ=np). Here λ = 100 × 0.05 =
5. P(X=5) = e^(-5) × 5^ ≈ 0.1755. Option A is correct but not the approximation
asked for, C misunderstands distributions, and D gives only the survival function at zero.
Q6: For a standard normal random variable Z, what is P(-1.96 < Z < 1.96)?
A. 0.90 — this corresponds to approximately ±1.645 standard deviations
