Worked solution · Exam P

SOA Exam P Sample Question #490 Solution

This Exam P sample reference tests Conditional Probability. The favorable two-break outcome has probability 0.0324. Summing the fragile-fragile, mixed, and nonfragile-nonfragile two-break cases gives 0.0964, so the conditional probability is 0.0324/0.0964=0.3361 and choice D.

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.

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Independent solution

How to solve this Conditional Probability question

Setup

Let F and N be the numbers broken among the two higher-risk and two lower-risk packages. The favorable event is F=2 and N=0.

F\sim\operatorname{Binomial}(2,0.2)

N\sim\operatorname{Binomial}(2,0.1)

Model

Exactly two total breaks can occur in three disjoint category-count configurations.

\Pr(F+N=2)=\Pr(F=2,N=0)+\Pr(F=1,N=1)+\Pr(F=0,N=2)

\Pr(F=2,N=0)=(0.2)^2(0.9)^2

Compute

Evaluate each configuration and normalize the favorable probability by their sum.

\Pr(F+N=2)=0.0324+(0.32)(0.18)+0.0064=0.0964

\Pr(F=2\mid F+N=2)=\frac{0.0324}{0.0964}=\frac{81}{241}

\frac{81}{241}=0.3360995851\ldots

Answer

Given exactly two breaks, the probability that they are the two fragile packages is approximately 0.336.

\boxed{0.336\quad\text{(D)}}

Test the method on unseen problems

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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.

AThe value 0.0324 is the joint favorable probability that both fragile packages break and both others remain intact. It has not been divided by the probability of exactly two breaks.
BThe value (0.2)²=0.040 is the probability that both fragile packages break, without requiring the other two to survive or conditioning on the observed total.
CThe value 1/6=0.167 treats the six possible pairs as equally likely. Pairs involving different package types have different break and survival probabilities.
EThis overweights the one-fragile, one-nonfragile configuration. Its actual conditional mass is 0.0576/0.0964=0.5975, and it is a different event from both broken packages being fragile.

Original practice · fully worked

Original variant: infer a calibration mode from paired pulses

A controller selects fast calibration mode with probability 0.70 and cautious mode with probability 0.30. Conditional on the selected mode, two diagnostic pulses are independent. A pulse passes with probability 0.40 in fast mode and 0.80 in cautious mode. Both pulses pass. Calculate the probability that cautious mode was selected.

A
0.192
B
0.300
C
0.368
D
0.500
E
0.632

Answer and calculation

The joint weights for cautious mode and fast mode with two passes are 0.30(0.80)²=0.192 and 0.70(0.40)²=0.112. Normalizing gives 0.192/(0.192+0.112)=12/19=0.632, so choice E is correct.

Setup

Let C and F denote the cautious and fast modes, and let P2 denote two passing pulses. Record the mode priors and conditional pass likelihoods.

\Pr(C)=0.30,\qquad \Pr(F)=0.70

\Pr(P_2\mid C)=(0.80)^2,\qquad \Pr(P_2\mid F)=(0.40)^2

Model

Weight each two-pass likelihood by the prior probability of its calibration mode.

\Pr(C\cap P_2)=0.30(0.80)^2=0.192

\Pr(F\cap P_2)=0.70(0.40)^2=0.112

Compute

Apply Bayes' rule by normalizing the cautious-mode weight over both ways to observe two passing pulses.

\Pr(C\mid P_2)=\frac{0.192}{0.192+0.112}

\Pr(C\mid P_2)=\frac{12}{19}=0.6315789474\ldots

Answer

The posterior probability of cautious mode is approximately 0.632.

\boxed{0.632\quad\text{(E)}}

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How to solve this Conditional Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: infer a calibration mode from paired pulses

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability