SOA Exam P Sample Question #661. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #661 — Verified Solution

Independent solution

How to solve this Discrete Random Variables question

Answer in brief

Let N be the given count. The target event within the condition is 1≤N≤3, whose probability is 0.045, while the conditioning event N≤3 has probability 0.965. Their ratio is 9/193=0.046632, which matches choice A.

Setup

Translate the two verbal events into bounds on the count N. At least one affected unit means N is positive, while at least one unaffected unit rules out the maximum count of four.

\{\text{at least one affected}\}=\{N\ge1\}

\{\text{at least one unaffected}\}=\{N\le3\}

Model

Intersect the target with the conditioning event and apply the conditional-probability ratio.

\Pr(N\ge1\mid N\le3)=\frac{\Pr(1\le N\le3)}{\Pr(N\le3)}

Compute

Add the appropriate entries from the supplied probability distribution.

\Pr(1\le N\le3)=0.015+0.010+0.020=0.045

\Pr(N\le3)=0.920+0.015+0.010+0.020=0.965

\frac{0.045}{0.965}=\frac{9}{193}=0.0466321244\ldots

Answer

The conditional probability rounds to 0.0466.

\boxed{0.0466\quad\text{(A)}}

Error diagnosis

Why the other letters tempt

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

BThe value 0.0800 is the unconditional probability P(N≥1)=1-P(N=0), so it fails to impose the condition that N cannot equal four.
CThe value 0.0829 divides 0.080 by 0.965 but leaves the excluded N=4 mass in the numerator after removing it from the conditioning denominator.
DThe value 0.5625 equals 0.045/0.080 and reverses the conditioning, calculating the chance of at least one unaffected unit given at least one affected unit.
EThe value 0.7500 equals 0.015/0.020 and arises by replacing both 'at least one' statements with exactly-one point probabilities before taking an unrelated ratio.

Original practice · fully worked

Original variant: delayed shipment count

Let N be the number of delayed shipments in a five-shipment dispatch. Its distribution is P(N=0)=0.64, P(N=1)=0.16, P(N=2)=0.09, P(N=3)=0.06, P(N=4)=0.03, and P(N=5)=0.02. Given that at least one shipment arrives on time, calculate the probability that N is odd.

A
0.220000
B
0.224490
C
0.240000
D
0.244898
E
0.775510

Variant answer in brief

The condition removes N=5 and has probability 0.98. Within that event, an odd delayed count means N=1 or N=3, with total mass 0.22; dividing gives 0.224490 and choice B.

Setup

At least one on-time shipment means not all five shipments are delayed.

H=\{N\le4\}

\Pr(H)=1-\Pr(N=5)=1-0.02=0.98

Model

Inside the conditioning event, the possible odd values are one and three; the odd value five has been excluded.

\{N\text{ odd}\}\cap H=\{N=1\}\cup\{N=3\}

Compute

Add the two eligible point probabilities and normalize by the conditioning probability.

\Pr(N\text{ odd},H)=0.16+0.06=0.22

\Pr(N\text{ odd}\mid H)=\frac{0.22}{0.98}=0.2244897959\ldots

Answer

The conditional probability of an odd delayed count is approximately 0.224490.

\boxed{0.224490\quad\text{(B)}}

Continue by topic

Build a complete proof notebook

The Probability Proof Manual reorganizes verified solutions by syllabus skill and adds formula proofs, error patterns, and original practice.

Explore study products Get the free formula sheet

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #661 — Verified Solution.” Verified 2026-07-30.

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