Verified solution · Exam P
This Exam P sample reference tests Geometric Distribution. The probability law is geometric on the number of failures before the first success, with success probability 1/5. Conditioning on survival past one merely shifts the variable by two, so its variance remains (4/5)/(1/5)²=20 and choice C is correct.
Independent solution
Setup
Recognize the probability law as a geometric failure count with success probability p=1/5 and failure probability q=4/5.
Model
Given that the failure count exceeds one, at least two failures have occurred. The additional failure count has the original geometric law.
Compute
Use the variance formula for a geometric failure count.
Answer
The conditional variance is 20.
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Original practice · fully worked
Independent inspections find a defect with probability 0.25 on each attempt. Three unsuccessful inspections have already occurred. Let R be the number of additional unsuccessful inspections before the first detection, and let the remaining labor cost be C=10+2R dollars. Calculate Var(C).
Past misses do not change the geometric residual count. Thus Var(R)=0.75/0.25²=12, and the coefficient 2 multiplies variance by 2², giving Var(C)=48 and choice D.
Setup
Describe the remaining miss count using the geometric failure-count model.
Model
Independence makes the three completed misses irrelevant to the future count, while the affine cost transformation scales variance by the square of its slope.
Compute
Compute the residual-count variance and then apply the cost multiplier.
Answer
The remaining labor-cost variance is 48 square dollars.
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ActuaryProof. “SOA Exam P Sample Question #470 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
