Worked solution · Exam P

SOA Exam P Sample Question #475 Solution

This Exam P sample reference tests Discrete Random Variables. This is a tail-sum expectation for a bounded nonnegative integer count. The four positive tail probabilities add to 5/6=0.8333, so choice A is correct.

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 Discrete Random Variables question

Setup

For an integer-valued count that cannot exceed four, its expectation is the sum of the probabilities of reaching each positive level.

\operatorname{E}[X]=\sum_{r=1}^{4}\Pr(X\ge r)

Model

Evaluate the supplied tail relationship at the four supported positive levels.

\Pr(X\ge1)=\frac49,\qquad \Pr(X\ge2)=\frac14

\Pr(X\ge3)=\frac19,\qquad \Pr(X\ge4)=\frac1{36}

Compute

Place the four fractions over a common denominator and add them.

\operatorname{E}[X]=\frac{16+9+4+1}{36}

\operatorname{E}[X]=\frac{30}{36}=\frac56=0.8333333333

Answer

The expected count is approximately 0.83.

\boxed{0.83\quad\text{(A)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

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

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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.

BSquaring only a denominator rather than the full parenthesized tail expression gives a false tail sum 17/18=0.9444, whose nearest listed value is 0.96.
CThe calculation sum r Pr(X≥r)=25/18=1.3889 incorrectly weights tail probabilities by their thresholds; the tail-sum identity uses coefficient one.
DThe value 1.81 fails the direct tail check: the four valid tails total 0.8333, so reaching 1.81 would require adding probability mass from unsupported levels.
EIncluding Pr(X≥0)=1 in the positive-tail sum gives 1+5/6=1.8333. The identity begins at level one because every count already exceeds or equals zero.

Original practice · fully worked

Original variant: expected relay depth of a probe

A probe can reach at most four relays. It reaches relay 1 with probability 0.80. Conditional on reaching the current relay, its probabilities of reaching the next relay are 0.75 from relay 1, 0.50 from relay 2, and 0.40 from relay 3. Let N be the number of relays reached. Calculate E[N].

A
0.12
B
1.02
C
1.70
D
1.82
E
2.45

Answer and calculation

Multiplying continuation probabilities gives the four tails 0.80, 0.60, 0.30, and 0.12. Their sum is E[N]=1.82, so choice D is correct.

Setup

Express the expectation of the bounded relay count as a sum of reach probabilities.

\operatorname{E}[N]=\sum_{r=1}^{4}\Pr(N\ge r)

Model

A later relay is reached only if every earlier continuation succeeds, so multiply along the route.

\Pr(N\ge1)=0.80

\Pr(N\ge2)=0.80(0.75)=0.60

\Pr(N\ge3)=0.60(0.50)=0.30

\Pr(N\ge4)=0.30(0.40)=0.12

Compute

Add the four tail probabilities.

\operatorname{E}[N]=0.80+0.60+0.30+0.12=1.82

Answer

The probe reaches an average of 1.82 relays.

\boxed{1.82\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Discrete Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: expected relay depth of a probe

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables