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

Sharpen your skills with conditional probability and Bayes' theorem for Exam P. Practice calculating posterior probabilities, applying the law of total probability, and solving problems that combine multiple conditioning events.

48 Questions

35 Easy

6 Medium

7 Hard

2026 Syllabus

100% Free

Sample Questions

Question 1 Easy

P(A) = 0.4, P(B) = 0.7, P(A∩B) = 0.3. Find P(A'|B).

Solution

P(A'|B) = (P(E)−P(A∩B))/P(E) = 0.4/0.7 = 4/7.

Question 2 Medium

P(A) = 0.5, P(B) = 0.3, P(A ∪ B) = 0.7. Find P(B | A').

Solution

P(A\cap

B) = P(A) +

P(B)

− P(A\cup

B) = 0.5 + 0.3 − 0.7 = 0.1.

P(B\cap

A') =

P(B)

− P(A\cap

B) = 0.3 − 0.1 = 0.2. P(B \mid A') = 0.2/P(A') =

\frac{0.2}{0.5}

= 0.4.

Question 3 Hard

Three coins are in a bag: one fair, one double-headed, one with P(H) = 0.75. A coin is drawn at random and flipped twice, showing HH. What is the probability the fair coin was drawn?

Solution

P(HH \mid \text{fair}) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}

P(HH \mid \text{double}) = 1

P(HH \mid \text{biased}) = \left(\frac{3}{4}\right)^2 = \frac{9}{16}

. Each coin equally likely (prob

\frac{1}{3}

). Numerator:

\frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}

. Denominator:

\frac{1}{12} + \frac{1}{3} + \frac{9}{48} = \frac{4}{48} + \frac{16}{48} + \frac{9}{48} = \frac{29}{48}

P = \frac{4/48}{29/48} = \frac{4}{29}

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

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