SOA Exam P Sample Question #647. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #647 — Verified Solution

Independent solution

How to solve this Law of Total Probability question

Answer in brief

This is a two-branch mixture whose unknown branch probability is determined by a cumulative probability. Solving the mixture equation gives a weighted-die branch probability of 0.666, after which the probability of rolling a four is 0.203667 and choice C.

Setup

Let h be the probability of selecting the weighted die. Normalizing its face weights gives probability 1/9 on each odd face and 2/9 on each even face.

h=\Pr(\text{weighted-die branch})

\Pr(X=k\mid\text{weighted})=\begin{cases}1/9,&k\text{ odd},\\2/9,&k\text{ even}.\end{cases}

Model

Apply the law of total probability to the event that the displayed value is at most three.

\Pr(X\le3\mid\text{weighted})=\frac{1+2+1}{9}=\frac49

\Pr(X\le3\mid\text{fair})=\frac12

0.463=h\left(\frac49\right)+(1-h)\left(\frac12\right)

Compute

Solve for the mixing probability, then mix the two conditional probabilities for face four.

h=18(0.500-0.463)=0.666

\Pr(X=4)=0.666\left(\frac29\right)+0.334\left(\frac16\right)

\Pr(X=4)=0.203666\ldots

Answer

The requested probability rounds to 0.204 at the precision of the choices.

\boxed{\Pr(X=4)\approx0.204\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.

AThe value 0.167 is the fair-die probability of a four. It ignores the branch using the weighted die.
BThe value 0.194 is the simple average of 1/6 and 2/9, which incorrectly assumes that the two branches are equally likely.
DThe value 0.537 is 1-0.463, the probability of a value above three; it includes faces five and six as well as face four.
EThe value 0.667 is the inferred probability of selecting the weighted branch, not the unconditional probability of the requested face.

Original practice · fully worked

Original variant: adaptive severity labels

A diagnostic platform routes a record to an adaptive classifier with probability r and to a legacy classifier otherwise. The adaptive classifier assigns severity labels 1 through 5 with probabilities 0.10, 0.10, 0.20, 0.20, and 0.40. The legacy classifier assigns the five labels uniformly. Historical output has P(S at most 3)=0.47. Calculate P(S=5).

A
0.200
B
0.260
C
0.300
D
0.330
E
0.650

Variant answer in brief

The adaptive and legacy cumulative probabilities through label three are 0.40 and 0.60. Their mixture equals 0.47 only when r=0.65, and the resulting probability of label five is 0.33, selecting choice D.

Setup

Compute the conditional cumulative probabilities through label three for the two classifiers.

\Pr(S\le3\mid\text{adaptive})=0.10+0.10+0.20=0.40

\Pr(S\le3\mid\text{legacy})=3(0.20)=0.60

Model

Use the observed cumulative probability to identify the routing proportion.

0.47=0.40r+0.60(1-r)

Compute

Solve the mixture equation and then combine the two label-five probabilities.

r=\frac{0.60-0.47}{0.20}=0.65

\Pr(S=5)=0.65(0.40)+0.35(0.20)=0.33

Answer

The platform assigns severity label five with probability 0.330.

\boxed{\Pr(S=5)=0.330\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #647 — Verified Solution.” Verified 2026-07-30.

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