Verified solution · Exam P
This Exam P sample reference tests Continuous Distributions. This is a conditional interval probability under a linearly increasing density. The relevant density mass is proportional to 10²-5² in the numerator and 20²-5² in the denominator, giving 75/375=0.20 and choice B.
Independent solution
Setup
The normalizing constant cancels from the conditional ratio, so integrate the linear density only over the relevant intervals.
Model
An antiderivative of kt is kt squared over two. Apply it to the target interval and the full conditioned interval.
Compute
Cancel the common factor and simplify the squared-endpoint differences.
Answer
The conditional probability is 0.20.
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Original practice · fully worked
A point (X,Y) is selected uniformly from the unit square. An observer is told that X+Y<1. Given this information, calculate the probability that both coordinates are less than 3/4.
The conditioning triangle has area 1/2. Requiring both coordinates below 3/4 removes two disjoint corner triangles, each of area 1/32, leaving area 7/16; the conditional ratio is (7/16)/(1/2)=7/8, choice D.
Setup
Uniform selection turns probability ratios into area ratios. The condition selects the lower-left half of the unit square.
Model
Within C, failure occurs when either coordinate is at least 3/4. Each failure region is a right triangle with leg length 1/4, and the two regions cannot overlap.
Compute
Remove both corner regions from the conditioning triangle, then divide by the conditioning area.
Answer
The requested conditional probability is 0.875.
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These references share syllabus concepts or sit nearby in the official sequence. All 388 Exam P univariate random variables practice problems.
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ActuaryProof. “SOA Exam P Sample Question #477 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
