Worked solution · Exam P

SOA Exam P Sample Question #607 Solution

This Exam P sample reference tests Disjoint Events. This is a cancellation problem involving five overlapping two-unit intervals. Alternating their probabilities cancels every interior one-unit bin and leaves the two requested edge bins, whose combined probability is 0.18, so choice B is correct.

Independent analysis by ActuaryProof · reviewed July 30, 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 Disjoint Events question

Setup

Let q_k be the probability of a duration in the one-unit interval beginning at k, for k from two through seven.

q_k=\Pr(k\le T<k+1)

q_2+q_3=0.20,\ q_3+q_4=0.34,\ q_4+q_5=0.42

q_5+q_6=0.38,\ q_6+q_7=0.28

Model

Take an alternating sum of the five observed window probabilities. Each interior bin appears once with each sign and cancels.

(q_2+q_3)-(q_3+q_4)+(q_4+q_5)-(q_5+q_6)+(q_6+q_7)=q_2+q_7

Compute

Insert the five reported probabilities in the same alternating order.

q_2+q_7=0.20-0.34+0.42-0.38+0.28

q_2+q_7=0.18

Answer

The two disjoint edge intervals together contain 18 percent of the calls.

\boxed{18\%\quad\text{(B)}}

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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 partial alternating calculation 0.20+0.34-0.42 equals 0.12, but it stops before the last two windows can complete the edge-bin cancellation.
CUsing only the final three windows gives 0.34-0.38+0.28=0.24 and omits information needed to recover the first edge bin.
DThis simply reports the probability of the final two-unit window. That window includes both the desired last bin and an unwanted adjacent bin.
EAlternating only the three middle observations gives 0.34-0.42+0.38=0.30; dropping the two outer observations prevents the target edge terms from being isolated.

Original practice · fully worked

Original variant: dual diagnostic flags

A battery pack is evaluated by diagnostic flags A and B. Flag A appears on 55% of packs, flag B appears on 40%, and at least one of the two flags appears on 70%. Given that a pack has at least one flag, calculate the probability that it has exactly one flag.

A
0.250
B
0.450
C
0.550
D
0.643
E
0.700

Answer and calculation

Inclusion-exclusion gives a 0.25 probability of both flags and a 0.45 probability of exactly one. Conditioning on the 0.70 union gives 0.45/0.70=9/14, or 0.642857, so choice D is correct.

Setup

Use the two marginal flag probabilities and their reported union to recover the overlap.

\Pr(A\cap B)=\Pr(A)+\Pr(B)-\Pr(A\cup B)

\Pr(A\cap B)=0.55+0.40-0.70=0.25

Model

Within the union, exactly one flag is the part remaining after removing the intersection.

\Pr(\text{exactly one})=\Pr(A\cup B)-\Pr(A\cap B)

Compute

Find the exactly-one mass and normalize it by the probability of observing at least one flag.

\Pr(\text{exactly one})=0.70-0.25=0.45

\Pr(\text{exactly one}\mid A\cup B)=\frac{0.45}{0.70}=\frac9{14}=0.642857\ldots

Answer

Among flagged packs, approximately 64.3 percent have exactly one flag.

\boxed{0.643\quad\text{(D)}}

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How to solve this Disjoint Events question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: dual diagnostic flags

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P General Probability