This Exam P sample reference tests Independence. Equating the same-color probability to 0.44 gives a linear equation in the unknown count; its unique nonnegative solution is 4, choice A.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BSubstituting x=20 into the same-category formula gives (0.4)(16/36)+(0.6)(20/36)=0.5111, not 0.44.
CSubstituting x=24 gives (0.4)(16/40)+(0.6)(24/40)=0.52, so it does not satisfy the observed probability.
DSubstituting x=44 gives (0.4)(16/60)+(0.6)(44/60)=0.5467, not 0.44.
EThe value 64 is the constant term obtained before the linear equation is solved. Substitution x=64 gives a same-category probability of 0.56.
Original practice · fully worked
Original variant: infer a routing share from an aggregate success rate
A request router sends an unknown proportion p of jobs through a fast path whose success probability is 0.80; all remaining jobs use a fallback path whose success probability is 0.50. The observed overall success probability is 0.62. Determine p.
A 0.40
B 0.48
C 0.50
D 0.60
E 0.75
Variant answer in brief
The aggregate rate is 0.80p+0.50(1-p). Solving 0.50+0.30p=0.62 gives p=0.40, choice A.
Setup
Setup
Let p be the proportion routed through the fast path; the remaining proportion 1-p uses the fallback path.
p=Pr(fast path)
Model
Model
Use the law of total probability: the overall success rate is the route-share-weighted average of the two conditional success rates.
0.62=0.80p+0.50(1−p)
Compute
Compute
Expanding 0.80p+0.50(1-p) and equating it to 0.62 gives 0.30p=0.12, hence p=0.40.
0.12=0.30p,p=0.40
Answer
Answer
The fast path receives 40% of jobs, which is choice A.
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