Verified solution · Exam P

SOA Exam P Sample Question #400 Solution

This Exam P sample reference tests Joint Distributions. This problem complements the three joint cells whose combined count is below two. Those cells have total mass 21/54, leaving 33/54=0.611111 and choice E.

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 Joint Distributions question

Setup

Let S=X+Y be the combined count. It is shorter to complement the outcomes with totals zero or one.

\Pr(S\ge2)=1-\Pr(S<2)

Model

Only three support points have a combined count below two.

\{S<2\}=\{(0,0),(0,1),(1,0)\}

Compute

Evaluate the joint mass at those points and subtract their sum from one.

\Pr(S<2)=\frac8{54}+\frac7{54}+\frac6{54}=\frac{21}{54}

\Pr(S\ge2)=1-\frac{21}{54}=\frac{33}{54}=\frac{11}{18}=0.6111111111

Answer

The probability of a combined count of at least two rounds to 0.61.

\boxed{0.61\quad\text{(E)}}

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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 10/54=0.1852 is Pr(X=2), which requires two counts of the first type and ignores qualifying combinations involving smaller X and positive Y.
BThe value 15/54=0.2778 is Pr(X+Y=2), so it omits every support point whose combined count is greater than two.
CThe value 21/54=0.3889 is Pr(X+Y<2), the complement of the requested event.
DThe value 28/54=0.5185 is Pr(X≥2 or Y≥2). It excludes the qualifying cell (1,1), where neither coordinate reaches two but their sum does.

Original practice · fully worked

Original variant: mode change at the next inspection

A machine is currently in operating mode A, B, or C with probabilities 0.20, 0.50, and 0.30. At the next inspection, the probabilities that it remains in its current mode are 0.70, 0.60, and 0.80 for modes A, B, and C, respectively. Calculate the probability that its mode changes.

A
0.20
B
0.30
C
0.32
D
0.40
E
0.68

Variant answer in brief

The mode-specific change probabilities are 0.30, 0.40, and 0.20. Weighting them by the current-mode probabilities gives 0.20(0.30)+0.50(0.40)+0.30(0.20)=0.32 and choice C.

Setup

Take the complement of each mode-specific stay probability.

\Pr(M\mid A)=0.30,\qquad \Pr(M\mid B)=0.40,\qquad \Pr(M\mid C)=0.20

Model

The current modes form a partition, so apply the law of total probability to the change event M.

\Pr(M)=\sum_{j\in\{A,B,C\}}\Pr(j)\Pr(M\mid j)

Compute

Weight the three conditional change probabilities by their current-mode probabilities.

\Pr(M)=0.20(0.30)+0.50(0.40)+0.30(0.20)

=0.06+0.20+0.06=0.32

Answer

The machine changes mode with probability 0.32.

\boxed{0.32\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #400 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Joint Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: mode change at the next inspection

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution