This Exam P sample reference tests Bayes Rule. The joint probability of both conditions is 0.20(0.25)=0.05. Dividing by the cholesterol probability 0.30 gives 1/6, choice A.
First obtain the joint probability by multiplying the probability of the first condition by the stated conditional probability of the second condition.
Pr(H)=0.20,Pr(C)=0.30,Pr(C∣H)=0.25
Model
Model
Bayes' rule reverses the conditioning direction by dividing that joint probability by the probability of the observed condition.
Pr(H∩C)=0.20(0.25)=0.05
Compute
Compute
The joint probability is 0.05 and the observed-condition probability is 0.30, so the requested conditional probability is one sixth.
Pr(H∣C)=0.300.05=61
Answer
Answer
The reverse conditional probability is one sixth.
61(A)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThe value one fifth is the marginal probability of the first condition, not the requested probability after conditioning on the second.
CThe value one quarter is the given probability in the original conditioning direction. The question asks for the reverse direction.
DThe value two thirds divides the two marginal probabilities and omits the given conditional probability needed to form the joint event.
EThe value five sixths is the complement of the correct result, although the question asks for the condition itself.
Original practice · fully worked
Original variant: reverse conditioning for two certifications
Among technicians, 30% hold certification A and 40% hold certification B. Half of those with certification A also hold B. Find the probability that a technician holds A given that the technician holds B.
A 0.250
B 0.300
C 0.375
D 0.500
E 0.750
Variant answer in brief
The joint certification probability is 0.30(0.50)=0.15. Dividing by P(B)=0.40 gives 0.375, choice C.
Setup
Setup
The joint certification probability is the probability of certification A multiplied by the probability of certification B among those with A.
Pr(A∩B)=Pr(A)Pr(B∣A)=0.30(0.50)=0.15
Model
Model
To condition on certification B, divide the joint probability by the marginal probability of B.
Pr(A∣B)=Pr(B)Pr(A∩B)
Compute
Compute
The joint probability is 0.15, and division by 0.40 gives 0.375.
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