Worked solution · Exam P

SOA Exam P Sample Question #154 Solution

This Exam P sample reference tests Law of Total Probability. The high-risk share is 0.20, and the three conditional death rates are in the ratio 6:3:1. Calibrating their weighted average to 0.009 gives high-risk rate 0.0200, choice B.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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

Read the solution Open official source

Independent solution

How to solve this Law of Total Probability question

Setup

Let q be the low-risk death probability and express the other rates in terms of q.

p_L=q,\qquad p_M=3q,\qquad p_H=6q

w_H=1-0.45-0.35=0.20

Model

Apply total probability across the three mutually exclusive risk groups.

0.009=(0.45)q+(0.35)(3q)+(0.20)(6q)

Compute

Collect the weighted multipliers and solve for the baseline rate.

0.009=2.70q,\qquad q=0.0033333

p_H=6q=0.0200000

Answer

The high-risk conditional death probability is 0.0200.

\boxed{0.0200\quad\text{(B)}}

Test the method on unseen problems

General Probability Practice Workbook

30 original general probability problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

See the cover and sample pages

Error diagnosis

Why the other letters tempt

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

AIf the high-risk rate were 0.0025, the implied weighted population rate would be 0.45(0.0025)=0.001125 after expressing the other rates as one-half and one-sixth of it.
CA high-risk rate of 0.1215 would imply overall probability 0.45(0.1215)=0.054675, far above 0.009.
DThe value 0.2000 is the proportion of policyholders in the high-risk group. It is a group weight, not a conditional death probability.
EA high-risk rate of 0.3750 would imply a weighted population death probability of 0.16875. It cannot satisfy the supplied aggregate rate.

Original practice · fully worked

Original variant: update a persistent operating mode

At dawn a controller chooses one of three operating modes with probabilities 1/6, 1/3, and 1/2. Within a chosen mode, successive components pass independently, with respective pass probabilities 0.90, 0.75, and 0.60. The first component passed. Find the conditional probability that the next component passes.

A
0.28214
B
0.50250
C
0.70000
D
0.71786
E
0.75000

Answer and calculation

The first-pass probability is 0.70, while the probability that both successive components pass is 0.5025. Their ratio is 0.71786, choice D.

Setup

Let S1 and S2 be the pass events for the first and second components. Average the first pass over the persistent mode.

\Pr(S_1)=\frac16(0.90)+\frac13(0.75)+\frac12(0.60)=0.70

Model

Conditional independence within a mode makes each two-pass contribution the mode weight times the squared pass rate.

\Pr(S_1\cap S_2)=\frac16(0.90)^2+\frac13(0.75)^2+\frac12(0.60)^2

Compute

Evaluate the joint probability and normalize by the observed first pass.

\Pr(S_1\cap S_2)=0.135+0.1875+0.180=0.5025

\Pr(S_2\mid S_1)=\frac{0.5025}{0.70}=0.7178571

Answer

The first success shifts weight toward the more reliable modes, raising the next-pass probability to 0.71786.

\boxed{0.71786\quad\text{(D)}}

Keep practicing

Next steps for Exam P General Probability

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P General Probability practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Law of Total Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: update a persistent operating mode

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability