Free SOA Exam P (Probability) Conditional Probability & Bayes Practice Questions

Conditional probability and Bayes' theorem are among the most frequently tested topics on SOA Exam P. Questions require calculating posterior probabilities, applying the law of total probability, and combining multiple conditioning events in multi-step problems.

98 questions 52 easy 25 medium 21 hard 2026 syllabus

Sample Questions

Question 1 Easy
P(A) = 0.6, P(B|A) = 0.3, P(B|A') = 0.5. Calculate P(B).
Solution
A is correct.

P(B)P(B) = P(B \mid A) P(A) + P(B \mid A') P(A') = 0.3(0.6) + 0.5(0.4) = 0.18 + 0.20 = 0.38.
Question 2 Medium
An urn contains 7 red and 5 blue balls. Three balls are drawn without replacement. Verify P(2nd ball is red) = 7/12 using the law of total probability conditioning on the first draw.
Solution
D is correct.

P(2nd red)=611⋅712+711⋅512=42+35132=77132=712P(\text{2nd red}) = \frac{6}{11} \cdot \frac{7}{12} + \frac{7}{11} \cdot \frac{5}{12} = \frac{42+35}{132} = \frac{77}{132} = \frac{7}{12}
Question 3 Hard
Let X1,X2,X3X_1, X_2, X_3 be iid Uniform[0,θ][0, \theta] random variables, and define M=max⁡(X1,X2,X3)M = \max(X_1, X_2, X_3). The parameter θ\theta has a prior distribution Uniform on [6,24][6, 24]. Given that M=6M = 6, calculate P(θ>12∣M=6)P(\theta > 12 \mid M = 6).
Solution
B is correct.

Step 1: Likelihood. For 3 iid Uniform[0,θ][0,\theta], the PDF of MM is fM∣Θ(m∣θ)=3m2/θ3f_{M|\Theta}(m|\theta) = 3m^2/\theta^3. At m=6m = 6: L(θ)=108/θ3L(\theta) = 108/\theta^3 for θ≥6\theta \geq 6.

Step 2: Posterior. With uniform prior on [6,24][6, 24]:
fΘ∣M(θ∣6)∝θ−3 for θ∈[6,24]f_{\Theta|M}(\theta|6) \propto \theta^{-3} \text{ for } \theta \in [6, 24]

Step 3: Normalizing constant.
∫624θ−3 dθ=[−12θ2]624=−11152+172=151152=5384\int_6^{24} \theta^{-3}\,d\theta = \left[-\frac{1}{2\theta^2}\right]_6^{24} = -\frac{1}{1152} + \frac{1}{72} = \frac{15}{1152} = \frac{5}{384}

Step 4: Numerator.
∫1224θ−3 dθ=−11152+1288=31152=1384\int_{12}^{24} \theta^{-3}\,d\theta = -\frac{1}{1152} + \frac{1}{288} = \frac{3}{1152} = \frac{1}{384}

Step 5: P(θ>12∣M=6)=1/3845/384=1/5=0.200P(\theta > 12 \mid M = 6) = \frac{1/384}{5/384} = 1/5 = 0.200
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