This Exam P sample reference tests Law of Total Probability. Let x be the territory-X share. The overall no-claim rate gives 0.20=0.15x+0.40(1-x), so x=0.80. The no-claim mass from X is then 0.15(0.80)=0.12, and 0.12/0.20=0.60 gives choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AAfter solving Pr(X)=0.80, the territory-Y no-claim contribution is 0.20(0.40)=0.08, near this small choice. It is a joint mass for the wrong territory and is not normalized.
BThe value 0.27 is 0.15/(0.15+0.40)=0.2727. It assumes the two territories have equal prior shares instead of solving the supplied overall no-claim rate.
CThe value 0.50 assumes equal territory shares and ignores both the different no-claim rates and the overall rate.
EThe value 0.80 is the recovered prior probability Pr(X). It stops before conditioning on no claims.
Original practice · fully worked
Original variant: format of a uniformly sampled archive byte
In a large digital archive, 60% of files use format S and the remaining 40% use format L. The average file sizes are 2 megabytes for format S and 5 megabytes for format L. One stored byte is selected uniformly from all bytes in the archive. Calculate the probability that its containing file uses format L.
A 0.3750
B 0.4000
C 0.5000
D 0.6250
E 0.7143
Variant answer in brief
Uniform byte sampling weights each file format by both its file share and its average size. The L and S byte-volume weights are 0.40(5)=2.0 and 0.60(2)=1.2, so the L share is 2.0/3.2=0.6250 and choice D.
Setup
Setup
Form a byte-volume weight for each format by multiplying file prevalence by average file size.
wS=0.60(2)=1.20
wL=0.40(5)=2.00
Model
Model
A uniformly sampled byte belongs to a format in proportion to that format's total byte volume.
Pr(L∣sampled byte)=wS+wLwL
Compute
Compute
Normalize the L byte-volume weight across both formats.
Pr(L∣sampled byte)=1.20+2.002.00
=85=0.6250
Answer
Answer
The sampled byte comes from format L with probability 0.6250.
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