SOA Exam P Sample Question #717. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #717 — Verified Solution

Independent solution

How to solve this Conditional Distributions question

Answer in brief

Restricting the joint distribution to the stated event gives conditional masses 0.125, 0.625, and 0.250 at values 0, 25, and 50, respectively. The resulting conditional variance is 224.609375 in millions of squared monetary units, which rounds to choice A.

Setup

Let A denote the conditioning event and express the monetary variable in thousands so that its possible values are 0, 25, and 50.

A=\{X\le 25{,}000\},\qquad Y_{\mathrm{k}}=Y/1000

\Pr(A)=0.80

Model

Aggregate the joint masses inside A for each possible value of the target variable, then normalize by the probability of A.

\Pr(Y_{\mathrm{k}}=0\mid A)=\frac{0.10}{0.80}=0.125

\Pr(Y_{\mathrm{k}}=25\mid A)=\frac{0.50}{0.80}=0.625

\Pr(Y_{\mathrm{k}}=50\mid A)=\frac{0.20}{0.80}=0.250

Compute

Calculate the first two conditional moments and subtract the square of the conditional mean.

\mathbb{E}[Y_{\mathrm{k}}\mid A]=25(0.625)+50(0.250)=28.125

\mathbb{E}[Y_{\mathrm{k}}^2\mid A]=25^2(0.625)+50^2(0.250)=1015.625

\operatorname{Var}(Y_{\mathrm{k}}\mid A)=1015.625-28.125^2=224.609375

Answer

Because one squared thousand equals one million in squared base monetary units, the requested variance rounds to 225 million.

\boxed{\operatorname{Var}(Y\mid A)\approx 225\text{ million}\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.

BIgnoring the conditioning event gives the unconditional variance of Y as 243.75 million, which rounds to 244.
CConditioning on the single value X=25,000 instead of the full event X at most 25,000 gives a variance of exactly 250 million.
DInterchanging the table axes and conditioning the wrong variable on the analogous two-level event gives 288.8889 million, which rounds to 289.
EComputing the unconditional variance of X rather than the requested conditional variance of Y gives 306.25 million, which rounds to 306.

Original practice · fully worked

Original variant: submersible diagnostic flag

An underwater drone's emergency repair burden C, in hundreds of dollars, is 0, 20, or 50 with probabilities 0.50, 0.30, and 0.20. A diagnostic flag appears with conditional probabilities 0.20, 2/3, and 1.00 at those respective burden levels. Given that the flag appears, calculate the conditional variance of C.

A
28
B
364
C
376
D
784
E
1160

Variant answer in brief

Bayes weighting gives conditional probabilities 0.20, 0.40, and 0.40 for burden levels 0, 20, and 50 after the flag appears. Their conditional variance is 376, so choice C is correct.

Setup

Multiply each prior burden probability by its corresponding flag likelihood.

\Pr(C=0,F)=0.50(0.20)=0.10

\Pr(C=20,F)=0.30(2/3)=0.20

\Pr(C=50,F)=0.20(1.00)=0.20

Model

Add the joint flag masses and normalize them to obtain the posterior burden distribution.

\Pr(F)=0.10+0.20+0.20=0.50

\Pr(C=0\mid F)=0.20,\quad \Pr(C=20\mid F)=0.40,\quad \Pr(C=50\mid F)=0.40

Compute

Calculate the first two moments of the posterior distribution and form the variance.

\mathbb{E}[C\mid F]=20(0.40)+50(0.40)=28

\mathbb{E}[C^2\mid F]=20^2(0.40)+50^2(0.40)=1160

\operatorname{Var}(C\mid F)=1160-28^2=376

Answer

The flag-conditioned variance is 376 squared hundreds of dollars.

\boxed{\operatorname{Var}(C\mid F)=376\quad\text{(C)}}

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 #717 — Verified Solution.” Verified 2026-07-30.

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