Verified solution · Exam P

SOA Exam P Sample Question #260 Solution

This Exam P sample reference tests Binomial Distribution. The affected-applicant count is Binomial(200,0.01). Summing its masses from zero through five gives 0.9839770931, which rounds to 0.98 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 Binomial Distribution question

Setup

Let X count affected applicants in the independent sample.

X\sim\operatorname{Binomial}(200,0.01)

Model

The requested lower tail contains the six disjoint counts from zero through five.

\Pr(X\le5)=\sum_{k=0}^{5}\binom{200}{k}(0.01)^k(0.99)^{200-k}

Compute

Evaluate the finite binomial sum without rounding its terms prematurely.

\Pr(X\le5)=0.9839770930908225

Answer

The probability rounds to 0.98.

\boxed{0.98\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 0.85 is the sum of the masses from one through five, 0.849997. It omits the zero-count outcome from a lower tail that begins at zero.
BThe value 0.88 results from omitting the four-count mass and truncating the retained masses to two decimals: 0.13+0.27+0.27+0.18+0.03=0.88. A cumulative sum cannot skip an interior count, and termwise truncation compounds the error.
CUsing an incorrect Poisson mean of 3 instead of np=2 gives Pr(Poisson(3)≤5)=0.916082, leading to the 0.91 choice.
DThe value 0.95 is Pr(X≤4)=0.948254 rounded. It omits the probability of exactly five affected applicants.

Original practice · fully worked

Original variant: flags from two sensor generations

A gateway monitors four legacy sensors and six upgraded sensors. During a diagnostic window, each legacy sensor independently flags with probability 0.30 and each upgraded sensor independently flags with probability 0.10. All ten flag events are independent. Calculate the probability that at most one sensor flags.

A
0.1276
B
0.2127
C
0.3038
D
0.3463
E
0.4314

Variant answer in brief

No flags has probability 0.7⁴ times 0.9⁶. Exactly one flag can come from either sensor generation. Adding those three disjoint contributions gives 0.4314060891, so choice E.

Setup

Separate the zero-flag event from the two ways to receive exactly one flag.

\Pr(N=0)=0.7^4 0.9^6=0.1275989841

Model

For exactly one flag, either one legacy sensor flags or one upgraded sensor flags.

\Pr(N=1,\text{ legacy})=\binom41(0.3)(0.7)^3(0.9)^6=0.2187411156

\Pr(N=1,\text{ upgraded})=\binom61(0.1)(0.9)^5(0.7)^4=0.0850659894

Compute

Add the mutually exclusive count cases.

\Pr(N\le1)=0.1275989841+0.2187411156+0.0850659894

=0.4314060891

Answer

The probability of at most one flag rounds to 0.4314.

\boxed{0.4314\quad\text{(E)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #260 — 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: flags from two sensor generations

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution