This Exam P sample reference tests Joint Distributions. This is a conditional probability obtained by summing selected cells of a joint distribution. The favorable intersection has probability 0.18 and the conditioning event has probability 0.86, giving 0.2093 and choice D.
Let A be the target count event and B the restriction imposed by the condition. The requested probability is the intersection divided by the conditioning total.
Pr(A∣B)=Pr(B)Pr(A∩B)
Model
Model
Add every joint cell allowed by B for the denominator. For the numerator, keep only those allowed cells that also satisfy A.
Pr(B)=0.86
Pr(A∩B)=0.18
Compute
Compute
Normalize the favorable joint mass by the complete conditioning mass.
Pr(A∣B)=0.860.18=439
Pr(A∣B)=0.2093023256
Answer
Answer
The conditional probability is approximately 0.21.
0.21(D)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.14 is the probability outside the conditioning event, 1-0.86. It is not the probability of the target after conditioning.
BRestricting the condition to the zero-count column gives 0.11/0.64=0.171875, near 0.17. The condition also permits the adjacent column.
CThe value 0.18 is the favorable joint probability before normalization. Conditional probability requires division by 0.86.
EUsing exactly one rather than no more than one for the condition gives 0.07/0.22=0.31818, near 0.32.
Original practice · fully worked
Original variant: trace a flagged component to its line
A factory's components come from lines A, B, and C in proportions 0.50, 0.30, and 0.20. Their respective automated-flag probabilities are 0.02, 0.04, and 0.10. One component is selected from the flagged stream. Calculate the probability that it came from line C.
A 0.020
B 0.100
C 0.200
D 0.476
E 0.667
Variant answer in brief
The line-weighted flag contributions are 0.010, 0.012, and 0.020. Line C therefore accounts for 0.020/0.042=10/21=0.4762 of flagged components, so choice D.
Setup
Setup
Let F denote a flag. Multiply each production share by its line-specific flag rate to obtain joint contributions.
Pr(A∩F)=0.50(0.02)=0.010
Pr(B∩F)=0.30(0.04)=0.012
Pr(C∩F)=0.20(0.10)=0.020
Model
Model
The flagged stream is the union of the three mutually exclusive line contributions.
Pr(F)=0.010+0.012+0.020=0.042
Pr(C∣F)=Pr(F)Pr(C∩F)
Compute
Compute
Divide line C's flagged contribution by the total flagged contribution.
Pr(C∣F)=0.0420.020=2110=0.4761904762
Answer
Answer
Approximately 47.6% of flagged components came from line C.
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