Verified solution · Exam P

SOA Exam P Sample Question #189 Solution

This Exam P sample reference tests Binomial Distribution. One entry wins with probability 1/C(30,6)=1/593775. With 800,000 independent entries, the winner count is binomial. The state loses only with at least two winners, whose probability is 1-P(0)-P(1)=0.38984, choice B.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Binomial Distribution question

Setup

Calculate the probability that one entry matches the unique selected six-number set.

p=\frac1{\binom{30}6}=\frac1{593775}

Model

Let W count winning entries. Two prizes exceed the entry-fee revenue, whereas one prize does not.

W\sim\operatorname{Binomial}(800000,p)

\{\text{state loses money}\}=\{W\ge2\}

Compute

Complement the zero- and one-winner outcomes.

\Pr(W\ge2)=1-(1-p)^{800000}-800000p(1-p)^{799999}

=1-0.2599378-0.3502179=0.3898443

Answer

The probability that prize expense exceeds revenue rounds to 0.39.

\boxed{0.39\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 value 0.33 is close to a coarsely approximated one-winner probability. Exactly one winner costs less than the collected entry fees, so it is not a loss event.
CThe value 0.61 is P(W≤1)=0.25994+0.35022, the complement of the loss probability.
DThe value 0.67 is the complement of the coarse 0.33 one-winner figure. That complement combines zero winners with multiple winners instead of isolating W≥2.
EThe value 0.74 is P(W≥1)=1-0.25994. It treats a single 500,000 prize as a loss even though revenue is 800,000.

Original practice · fully worked

Original variant: heterogeneous failures exceeding crew capacity

A data center has 30 type-A drives and 20 type-B drives. During one service interval, each type-A drive fails independently with probability 0.01 and each type-B drive fails independently with probability 0.02. A crew can replace at most one failed drive during the interval. Calculate the probability that the number of failures exceeds the crew's capacity.

A
0.15496
B
0.35121
C
0.49383
D
0.50617
E
0.84504

Variant answer in brief

Capacity is exceeded by at least two failures. The probabilities of zero and one total failure are 0.49383 and 0.35121, so the upper tail is 1-0.49383-0.35121=0.15496, choice A.

Setup

Let X and Y count failures in the two independent drive groups.

X\sim\operatorname{Binomial}(30,0.01),\qquad Y\sim\operatorname{Binomial}(20,0.02)

Model

Complement the total-failure counts zero and one.

\Pr(X+Y\ge2)=1-\Pr(X+Y=0)-\Pr(X+Y=1)

Compute

Evaluate the zero-failure product and the two disjoint one-failure cases.

\Pr(X+Y=0)=0.99^{30}0.98^{20}=0.4938299

\Pr(X+Y=1)=30(0.01)0.99^{29}0.98^{20}+20(0.02)0.98^{19}0.99^{30}

\Pr(X+Y=1)=0.3512086

\Pr(X+Y\ge2)=1-0.4938299-0.3512086=0.1549615

Answer

The probability that at least two drives fail is about 0.15496.

\boxed{0.15496\quad\text{(A)}}

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ActuaryProof. “SOA Exam P Sample Question #189 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Binomial Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: heterogeneous failures exceeding crew capacity

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution