Verified solution · Exam P

SOA Exam P Sample Question #349 Solution

This Exam P sample reference tests Law of Total Probability. Let the February pass probability after a January failure be q; after a January pass it is twice q. The overall February pass rate gives q equal to 5/17, so the probability of passing both exams is 7/17, approximately 0.41 and choice B.

Independent analysis by ActuaryProof · reviewed August 4, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Law of Total Probability question

Setup

Represent the two February conditional pass rates with one unknown.

\Pr(F\mid J^c)=q,\qquad \Pr(F\mid J)=2q

Model

Apply total probability to the reported February pass rate.

0.50=0.70(2q)+0.30(q)=1.70q

Compute

Solve for the conditional rate and form the requested joint probability.

q=\frac5{17},\qquad \Pr(F\mid J)=\frac{10}{17}

\Pr(J\cap F)=0.70\frac{10}{17}=\frac7{17}=0.4117647\ldots

Answer

The probability of passing both exams rounds to 0.41.

\boxed{0.41\quad\text{(B)}}

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis multiplies the two marginal pass rates and assumes the exam results are independent, contrary to the stated conditional-rate relationship.
CThis would require a February pass rate of about 0.643 after a January pass. Substitution into the stated two-to-one relationship produces an overall February rate above 0.50.
DThis is the marginal February pass probability. It does not also require a January pass.
EThis is the February pass probability conditional on a January pass, approximately 0.59, before multiplication by the January pass probability.

Original practice · fully worked

Original variant: recover a production-mode share

A production line runs in a precision mode for part of each day and a rapid mode for the rest. The defect probability is 0.10 in precision mode and 0.40 in rapid mode. Across the full day, 0.25 of items are defective. Calculate the probability that a randomly selected item was produced in rapid mode and is defective.

A
0.05
B
0.20
C
0.40
D
0.25
E
0.50

Variant answer in brief

The overall defect rate lies halfway between the two mode-specific rates, so each mode has probability one-half. Rapid-mode defects therefore have joint probability 0.50 × 0.40, or 0.20 and choice B.

Setup

Let r be the probability that an item is produced in rapid mode.

\Pr(R)=r,\qquad \Pr(P)=1-r

Model

Express the overall defect rate as a mixture of the two operating modes.

0.25=(1-r)(0.10)+r(0.40)

Compute

Recover the mode share and then calculate the rapid-and-defective joint probability.

r=0.50

\Pr(R\cap D)=0.50(0.40)=0.20

Answer

Twenty percent of all items are both rapid-mode and defective.

\boxed{0.20\quad\text{(B)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #349 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Law of Total Probability question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: recover a production-mode share

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution