Worked solution · Exam P

SOA Exam P Sample Question #266 Solution

This Exam P sample reference tests Continuous Random Variables. Integrating the one-claim density gives F(25)=1-10/25=3/5. The largest of three independent claims is below 25 only when all three are below 25, so the probability is (3/5)³=27/125 and choice C.

Independent analysis by ActuaryProof · reviewed July 31, 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 Continuous Random Variables question

Setup

First obtain the one-claim distribution function at the threshold.

F(25)=\int_{10}^{25}\frac{10}{x^2}\,dx

Model

The maximum is below the threshold exactly when every independent claim is below it.

\Pr(M<25)=\Pr(X_1<25,X_2<25,X_3<25)=F(25)^3

Compute

Evaluate the integral and cube the resulting probability.

F(25)=\left[-\frac{10}{x}\right]_{10}^{25}=1-\frac{10}{25}=\frac35

\Pr(M<25)=\left(\frac35\right)^3=\frac{27}{125}=0.216

Answer

The exact listed probability is 27/125.

\boxed{\frac{27}{125}\quad\text{(C)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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.

AThe value 8/125=(2/5)³ uses the one-claim upper-tail probability and requires all three claims to be at least 25. It describes the minimum exceeding the threshold.
BThe value 12/125=(3/5)(2/5)² is one specified ordering with one claim below 25 and two above. It is neither the maximum-below event nor the sum over all three such orderings.
DThe value 2/5 is the probability that one claim is at least 25. It uses the wrong tail and does not combine three claims.
EThe value 3/5 is the probability that one claim is below 25. It omits the independence cube required for the maximum.

Original practice · fully worked

Original variant: covariance of two ordered readings

Two independent calibration readings are uniformly distributed between 2 and 8. Let L be the smaller reading and H the larger reading. Calculate Cov(L,H).

A
1 square unit
B
4 square units
C
6 square units
D
24 square units
E
25 square units

Answer and calculation

After scaling to two unit-uniform readings, the ordered pair has means 1/3 and 2/3, while the product of the minimum and maximum has mean 1/4. Its covariance is 1/36, and scaling both readings by 6 multiplies covariance by 36, giving 1 square unit and choice A.

Setup

Scale the readings to independent unit-uniform variables and order them.

X_i=2+6U_i,\qquad U_i\sim\operatorname{Uniform}(0,1)

A=\min(U_1,U_2),\qquad B=\max(U_1,U_2)

L=2+6A,\qquad H=2+6B

Model

Use the marginal order-statistic means and the fact that ordering does not change the product.

E[A]=\frac13,\qquad E[B]=\frac23

AB=U_1U_2,\qquad E[AB]=E[U_1]E[U_2]=\frac14

Compute

Compute covariance on the unit scale, then apply the two affine scale factors.

\operatorname{Cov}(A,B)=\frac14-\left(\frac13\right)\left(\frac23\right)=\frac1{36}

\operatorname{Cov}(L,H)=6(6)\operatorname{Cov}(A,B)=36\left(\frac1{36}\right)=1

Answer

The covariance of the smaller and larger readings is 1 square unit.

\boxed{1\text{ square unit}\quad\text{(A)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Continuous Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: covariance of two ordered readings

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables