This Exam P sample reference tests Conditional Probability. This is a Bayesian update using the zero-count likelihood of each Poisson class. Normalizing the two prior-weighted likelihoods gives 0.749753, which rounds to 0.750 and choice D.
How to solve this Conditional Probability question
Setup
Setup
Let L and H identify the two risk classes, and let Z be the observed zero-count event. A Poisson variable with mean λ is zero with probability exp(-λ).
Pr(Z∣L)=e−0.05
Pr(Z∣H)=e−0.30
Model
Model
Multiply each zero-count likelihood by its class prior and normalize the low-risk weight.
Pr(L∣Z)=0.70e−0.05+0.30e−0.300.70e−0.05
Compute
Compute
Evaluate the two joint weights and their ratio.
wL=0.70e−0.05=0.665860598
wH=0.30e−0.30=0.222245466
Pr(L∣Z)=0.8881060640.665860598=0.749753464
Answer
Answer
The posterior low-risk probability rounds to 0.750.
0.750(D)
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AUsing each Poisson mean itself as its zero-count likelihood gives 0.70(0.05)/[0.70(0.05)+0.30(0.30)]=0.280. The zero probability is exponential.
BA posterior of 0.666 has odds about 1.99, below the prior odds 7/3. That moves probability away from the lower-rate class even though observing zero favors that class.
CThe value 0.700 is the prior probability of the low-risk class and ignores the information supplied by the observed count.
EReplacing exp(−λ) by the coarse linear approximation 1-λ gives 0.70(0.95)/[0.70(0.95)+0.30(0.70)]=0.760.
Original practice · fully worked
Original variant: identify a kit's workshop from pass results
Sealed three-component kits come from workshops A, B, and C in proportions 0.50, 0.30, and 0.20. Within a kit, components independently pass inspection with probabilities 0.80, 0.50, and 0.20 for the respective workshops. A selected kit has exactly two passing components. Calculate the probability it came from workshop A.
A 0.192
B 0.384
C 0.500
D 0.593
E 0.631
Variant answer in brief
The three prior-weighted exactly-two likelihoods are 0.1920, 0.1125, and 0.0192. Workshop A's normalized share is 0.1920/0.3237=0.593142, so choice D.
Setup
Setup
Let E denote exactly two passes. Compute its binomial likelihood under each possible workshop.
Pr(E∣A)=(23)(0.80)2(0.20)=0.384
Pr(E∣B)=(23)(0.50)2(0.50)=0.375
Pr(E∣C)=(23)(0.20)2(0.80)=0.096
Model
Model
Weight each likelihood by the corresponding production share.
wA=0.50(0.384)=0.1920
wB=0.30(0.375)=0.1125
wC=0.20(0.096)=0.0192
Compute
Compute
Normalize the workshop-A weight over all three ways the evidence can arise.
Pr(A∣E)=0.1920+0.1125+0.01920.1920
Pr(A∣E)=0.32370.1920=0.593141798
Answer
Answer
The posterior probability of workshop A is approximately 0.593.
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