Verified solution · Exam P

SOA Exam P Sample Question #85 Solution

This Exam P sample reference tests Discrete Distributions. Write the six probabilities as an arithmetic sequence and solve the normalization and 40% conditions. The last two probabilities total 4/15 = 0.2667, choice C.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Discrete Distributions question

Setup

Represent the six probabilities for counts 0 through 5 as a decreasing arithmetic sequence with initial value p0 and decrement c.

p_n=p_0-nc,\qquad n=0,1,\ldots,5

Model

Use the stated 40% condition for the first two masses together with the requirement that all six masses sum to one.

2p_0-c=0.4

6p_0-15c=1

Compute

Solving the two linear equations gives p0=5/24 and c=1/60. The masses at counts 4 and 5 then total 4/15.

p_0=\frac5{24},\qquad c=\frac1{60}

p_4+p_5=2p_0-9c=\frac4{15}=0.266667

Answer

The requested probability is 0.266667, which rounds to 0.27 and corresponds to choice C.

\boxed{0.27\quad\text{(C)}}

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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 0.14 is the rounded mass at count 4 alone. The requested event also includes count 5.
DThe value 0.29 is consistent with shifting the two-term sum to nearby sequence positions rather than using counts 4 and 5.

Original practice · fully worked

Original variant: linearly decreasing queue-count probabilities

A service desk handles 0 through 4 escalations daily. The five probabilities form a decreasing arithmetic sequence, and P(N<2)=0.55. Find P(N≥3).

D
0.10
B
0.15
C
0.20
A
0.25
E
0.30

Variant answer in brief

Let p_n=a-nd. Solving 2a-d=0.55 and 5a-10d=1 gives a=0.30 and d=0.05; the last two probabilities total 0.25.

Setup

Write the five probabilities for counts 0 through 4 as an arithmetic sequence with first term a and decrement d.

p_n=a-nd,\qquad n=0,1,2,3,4

Model

The condition for counts below 2 supplies one equation; normalization of all five masses supplies the other.

2a-d=0.55

5a-10d=1

Compute

Solving gives a=0.30 and d=0.05. The masses at counts 3 and 4 are 0.15 and 0.10, totaling 0.25.

a=0.30,\quad d=0.05

p_3+p_4=0.15+0.10=0.25

Answer

Therefore P(N ≥ 3)=0.25, corresponding to choice A.

\boxed{0.25\quad\text{(A)}}

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ActuaryProof. “SOA Exam P Sample Question #85 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Discrete Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: linearly decreasing queue-count probabilities

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution