Worked solution · Exam P

SOA Exam P Sample Question #285 Solution

This Exam P sample reference tests Uniform Distribution. One appraisal lies below the price with probability 3/4 and above it with probability 1/4. The price is strictly between the extrema unless all four appraisals lie on the same side, so the probability is 1-(3/4)⁴-(1/4)⁴=0.6796875 and choice D.

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 Uniform Distribution question

Setup

Use the relative lengths on the uniform support to find the two one-appraisal side probabilities.

\Pr(A<\theta)=\frac34,\qquad \Pr(A>\theta)=\frac14

Model

The price fails to lie strictly between the sample minimum and maximum only when every appraisal is below it or every appraisal is above it.

\Pr(L<\theta<H)=1-\Pr(H<\theta)-\Pr(L>\theta)

Compute

Use independence for the two all-on-one-side events.

\Pr(L<\theta<H)=1-\left(\frac34\right)^4-\left(\frac14\right)^4

=1-\frac{81}{256}-\frac1{256}=\frac{174}{256}=0.6796875

Answer

The straddling probability rounds to 0.680.

\boxed{0.680\quad\text{(D)}}

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.

AIf K counts appraisals below the price, the incorrect calculation P(K=1)+one half P(K=2)=0.046875+0.10546875=0.15234375 produces this choice. Every K=2 configuration straddles the price and must receive full weight.
BThe value 0.188 is (3/4)(1/4)=0.1875. It treats the minimum and maximum like one prescribed below-above pair and ignores the other appraisals and their possible orderings.
CRounding the side probabilities prematurely to 0.8 and 0.2 gives 1-(0.8)⁴-(0.2)⁴=0.5888, which becomes 0.6 under coarse one-decimal rounding.
EThe value 0.996 is 1-(1/4)⁴=0.996094, the probability of at least one appraisal below the price. It forgets that at least one appraisal must also be above it.

Original practice · fully worked

Original variant: a batch spanning two quality bands

A polymer assay score X has density f(x)=2x between zero and one, and zero elsewhere. Six independent specimens are tested. Calculate the probability that the batch contains at least one score below 0.30 and at least one score above 0.80.

A
0.363
B
0.391
C
0.402
D
0.432
E
0.931

Answer and calculation

One score is low with probability 0.09, middle with probability 0.55, and high with probability 0.36. Inclusion-exclusion over the missing-low and missing-high events gives 0.391091912, so choice B.

Setup

Integrate the density and convert the two score thresholds into three category probabilities.

F(x)=\int_0^x2t\,dt=x^2,\qquad 0\le x\le1

p_L=F(0.30)=0.09,\qquad p_H=1-F(0.80)=0.36

p_M=F(0.80)-F(0.30)=0.55

Model

Complement the events that the batch has no low score or no high score, restoring their overlap.

\Pr(L\ge1,H\ge1)=1-\Pr(L=0)-\Pr(H=0)+\Pr(L=0,H=0)

Compute

Use independence across the six specimens for each all-in-an-allowed-region probability.

\Pr(L\ge1,H\ge1)=1-(0.91)^6-(0.64)^6+(0.55)^6

=0.391091911848

Answer

The probability that the batch reaches both outer quality bands is approximately 0.391.

\boxed{0.391\quad\text{(B)}}

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How to solve this Uniform Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: a batch spanning two quality bands

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables