SOA Exam P Sample Question #703. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #703 — Verified Solution

Independent solution

How to solve this Binomial Distribution question

Answer in brief

Model the independent claim indicators as Bernoulli trials and equate the two specified waiting-time probabilities. Cancellation leaves 7p(1-p)=1, whose admissible root is approximately 0.172673 and therefore choice C.

Setup

Let p be the claim probability and let q=1-p. Translate each stopping event into a Bernoulli-sequence probability.

q=1-p

\Pr(\text{first claim at trial }6)=q^5p

\Pr(\text{second claim at trial }8)=\binom{7}{1}p^2q^6

Model

Equate the probabilities and cancel factors that are positive for an interior probability.

q^5p=7p^2q^6

1=7pq=7p(1-p)

Compute

Solve the resulting quadratic and apply the stated restriction that p is below one half.

7p^2-7p+1=0

p=\frac{7\pm\sqrt{21}}{14}

p=\frac{7-\sqrt{21}}{14}=0.1726731646\ldots

Answer

The smaller root rounds to 0.173 and matches choice C.

\boxed{p\approx0.173\quad\text{(C)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

ADropping the remaining factor 1-p after cancellation gives 7p=1 and therefore p=1/7, which rounds to 0.143.
BCounting only six possible positions for the earlier claim and also dropping the remaining failure factor gives p=1/6, or 0.167.
DUsing six rather than seven possible positions but retaining the failure factor leads to 6p(1-p)=1; its smaller root is approximately 0.211.
EConfusing the five failures before the first claim with the number of possible earlier-claim positions gives 5p(1-p)=1 and the smaller root 0.276.

Original practice · fully worked

Original variant: seal-inspection calibration

A packaging line produces independently sealed cartons, each of which fails an inspection with an unknown probability p below 0.5. The chance that exactly two of the next five cartons fail is equal to the chance that none of the next three cartons fail. Determine p.

A
0.100
B
0.224
C
0.258
D
0.316
E
0.408

Variant answer in brief

The two probabilities share a factor of (1-p) cubed. Cancelling it leaves 10p squared equal to one, so p is approximately 0.316228 and choice D is correct.

Setup

Write each inspection statement as a binomial probability, using q=1-p for a passing carton.

\Pr(\text{two failures in five})=\binom{5}{2}p^2q^3

\Pr(\text{no failures in three})=q^3

Model

Equate the two expressions and cancel q cubed, which is positive because p is below one half.

10p^2q^3=q^3

10p^2=1

Compute

Take the admissible positive square root.

p=\frac{1}{\sqrt{10}}=0.3162277660\ldots

Answer

Rounding to three decimals selects choice D.

\boxed{p\approx0.316\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #703 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.