SOA Exam P Sample Question #705. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #705 — Verified Solution

Independent solution

How to solve this Continuous Random Variables question

Answer in brief

The policyholder retains 40% of each loss, so the payment threshold corresponds to an original loss below 3. Integrating the given density from 0 to 3 gives 13/56, approximately 0.232143, and selects choice E.

Setup

Let X denote the original loss and U the unreimbursed amount. Convert the payment condition back to a condition on X.

U=(1-0.60)X=0.40X

U<1.20\iff X<\frac{1.20}{0.40}=3

Model

Use the loss density over the transformed interval.

\Pr(U<1.20)=\int_0^3\frac{1}{42}\left(\frac{x}{2}+1\right)^2\,dx

Compute

Integrate after recognizing the linear inner function and evaluate at both bounds.

\int\frac{1}{42}\left(\frac{x}{2}+1\right)^2dx=\frac{1}{63}\left(\frac{x}{2}+1\right)^3

\Pr(U<1.20)=\frac{1}{63}\left[\left(\frac{5}{2}\right)^3-1\right]

\Pr(U<1.20)=\frac{13}{56}=0.2321428571\ldots

Answer

Rounding to three decimals gives 0.232.

\boxed{\Pr(U<1.20)\approx0.232\quad\text{(E)}}

Error diagnosis

Why the other letters tempt

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

AUsing 60% as the retained share gives a cutoff of 2, and then missing the factor of two from substitution halves the resulting 1/9 to about 0.056.
BTreating the reimbursed 60% as the policyholder's retained share changes the cutoff to 2 and gives F_X(2)=1/9, approximately 0.111.
CThe value 0.116 is one half of the correct probability; it results from omitting the derivative factor when substituting u=x/2+1.
DEvaluating the density at the cutoff gives f_X(3)=6.25/42, approximately 0.149, but a probability requires integrating over the interval.

Original practice · fully worked

Original variant: ceramic restoration credit

The replacement cost X, in thousands of dollars, for a custom ceramic installation has density f(x)=(x+2)/30 for 0<x<6 and zero elsewhere. A restoration program credits one third of the replacement cost, leaving the buyer to pay the remaining two thirds. Calculate the probability that the buyer pays less than 2 thousand dollars.

A
0.175
B
0.200
C
0.350
D
0.650
E
1.000

Variant answer in brief

The buyer pays two thirds of X, so the payment is below 2 precisely when X is below 3. Integrating (x+2)/30 to 3 gives 0.350, which is choice C.

Setup

Translate the buyer's out-of-pocket threshold into a replacement-cost threshold.

Y=\frac{2}{3}X

Y<2\iff X<3

Model

Integrate the replacement-cost density over the interval selected by the payment condition.

\Pr(Y<2)=\int_0^3\frac{x+2}{30}\,dx

Compute

Evaluate the elementary antiderivative.

\Pr(Y<2)=\left.\frac{x^2/2+2x}{30}\right|_0^3

\Pr(Y<2)=\frac{4.5+6}{30}=0.35

Answer

The buyer's payment is below the threshold with probability 0.350.

\boxed{\Pr(Y<2)=0.350\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #705 — Verified Solution.” Verified 2026-07-30.

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