Worked solution · Exam P

SOA Exam P Sample Question #627 Solution

This Exam P sample reference tests Conditional Distributions. This is a conditional-expectation calculation from a joint probability table. Restricting the table to the qualifying rows gives probability 0.70 and a weighted numerator of 0.99, so the conditional mean is 99/70, or 1.414286, and choice C is correct.

Independent analysis by ActuaryProof · reviewed July 30, 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 Conditional Distributions question

Setup

Retain the two rows allowed by the conditioning event. Their combined probability is the denominator of the conditional distribution.

\Pr(X<2)=0.30+0.40=0.70

Model

For each possible value of Y, add the retained joint masses and divide by 0.70. The conditional mean can be computed by dividing the corresponding weighted joint sum by the same denominator.

\operatorname{E}[Y\mid X<2]=\frac{\sum_y y\,\Pr(Y=y,\,X<2)}{\Pr(X<2)}

Compute

The retained column masses for Y equal 0.14, 0.21, 0.27, and 0.08. Weighting them by 0, 1, 2, and 3 produces the numerator.

\sum_y y\,\Pr(Y=y,\,X<2)=1(0.21)+2(0.27)+3(0.08)=0.99

\operatorname{E}[Y\mid X<2]=\frac{0.99}{0.70}=\frac{99}{70}=1.414285714\ldots

Answer

The conditional expected count rounds to 1.41.

\boxed{1.41\quad\text{(C)}}

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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.

AUsing a multiplier of two rather than three for the Y=3 column gives numerator 0.91 and 0.91/0.70=1.30.
BMisadding the retained Y=3 column as 0.07 instead of 0.08 gives numerator 0.96 and 0.96/0.70=1.3714.
DThis is the unconditional mean of Y from all four rows. It ignores the condition on X and therefore uses the wrong distribution.
EUsing a multiplier of four rather than three for the Y=3 column gives numerator 1.07 and 1.07/0.70=1.5286.

Original practice · fully worked

Original variant: microfluidic completion time

Within a normalized one-minute microfluidic cycle, U records when a sample enters a channel and V records when processing finishes. Their joint density equals 2 on the triangular region 0<U<V<1 and is zero elsewhere. For samples that enter after time 1/2, calculate the expected completion time V.

A
0.2083
B
0.6667
C
0.7500
D
0.8333
E
1.0000

Answer and calculation

The event U>1/2 has probability 1/4, while the corresponding first-moment integral for V is 5/24. Their ratio is 5/6, or 0.8333, so choice D is correct.

Setup

The conditioning event restricts U to the interval from one-half to one. For each such entry time, V ranges from U to one.

A=\{U>1/2\}

f_{U,V}(u,v)=2,\qquad \frac12<u<v<1

Model

A conditional expectation is the first-moment integral over the restricted region divided by the probability of that region.

\operatorname{E}[V\mid A]=\frac{\displaystyle\int_{1/2}^{1}\int_u^1 2v\,dv\,du}{\displaystyle\int_{1/2}^{1}\int_u^1 2\,dv\,du}

Compute

Evaluate the restricted probability and first moment separately, then normalize.

\Pr(A)=\int_{1/2}^{1}2(1-u)\,du=\frac14

\operatorname{E}[V\mathbf{1}_A]=\int_{1/2}^{1}(1-u^2)\,du=\frac5{24}

\operatorname{E}[V\mid A]=\frac{5/24}{1/4}=\frac56=0.833333\ldots

Answer

The expected normalized completion time is approximately 0.8333.

\boxed{0.8333\quad\text{(D)}}

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How to solve this Conditional Distributions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: microfluidic completion time

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables