Verified solution · Exam P

SOA Exam P Sample Question #600 Solution

This Exam P sample reference tests Cumulative Probabilities. Evaluating the loss CDF at the reimbursement limit gives probability 0.65 of full reimbursement and 0.35 of a partial shortfall. Independence makes the probability of two full reimbursements followed by a shortfall equal to 0.65 squared times 0.35, or 0.147875, so choice D is correct.

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 Cumulative Probabilities question

Setup

A loss is fully reimbursed exactly when it does not exceed the payment limit.

p=\Pr(X\le0.80)=F(0.80)

Model

Evaluate the supplied CDF and use its complement for a partially unreimbursed loss.

p=\frac{0.80+15(0.80)^2}{16}=0.65

q=1-p=0.35

Compute

The first shortfall occurs on the third event only when the first two are fully reimbursed and the third exceeds the limit.

\Pr(T=3)=p^2q

\Pr(T=3)=(0.65)^2(0.35)=0.147875

Answer

The required probability rounds to 0.148.

\boxed{0.148\quad\text{(D)}}

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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 uses the four-event pattern 0.65(0.35)³, inserting three shortfalls after one fully reimbursed loss instead of locating the first shortfall on event three.
BThis computes 0.65(0.35)², requiring one full reimbursement followed by two shortfalls rather than two full reimbursements followed by one shortfall.
COmitting the square on 0.80 inside the CDF gives p=0.80 and then p squared times (1-p)=0.128.
EThis is the probability 0.35 that one loss exceeds the limit and ignores the required position of the first such loss.

Original practice · fully worked

Original variant: polymer curing flags

Four independent polymer samples have curing index Y with distribution function F(y)=y² for 0≤y≤1. A scanner flags a sample when its curing index exceeds 0.50. Calculate the probability that exactly two of the four samples are flagged.

A
0.035
B
0.211
C
0.262
D
0.422
E
0.750

Variant answer in brief

A sample is flagged with probability 1-F(0.5)=0.75. The number flagged among four independent samples is binomial, and its mass at two is 0.2109375, which rounds to choice B.

Setup

Use the distribution function to find the flag probability for one sample.

q=\Pr(Y>0.50)=1-F(0.50)=1-(0.50)^2=0.75

Model

Independence makes the number of flagged samples a binomial count.

K\sim\operatorname{Binomial}(4,0.75)

Compute

Evaluate the binomial mass for exactly two flags.

\Pr(K=2)=\binom42(0.75)^2(0.25)^2

\Pr(K=2)=0.2109375

Answer

The probability rounds to 0.211.

\boxed{0.211\quad\text{(B)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #600 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.

How to solve this Cumulative Probabilities question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: polymer curing flags

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution