This Exam P sample reference tests Conditional Probability. This is a Bayesian posterior-predictive calculation after two observed successes. Updating the selected model and averaging its next-trial success rate gives 0.90, which selects choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is the next-trial success probability under the ordinary model alone and ignores the evidence favoring the always-successful model.
BThe value 9/16 is the unconditional probability of three consecutive successes, not the conditional probability of the third success after two have been observed.
CThe value 5/8 is the unconditional probability of the two observed successes. It is the denominator of the conditional calculation, not its result.
DThe value 4/5 is the posterior probability that the always-successful model was selected. A remaining 1/5 posterior weight still has next-trial success probability 1/2.
Original practice · fully worked
Original variant: prediction after an unrecorded color match
A container holds five silver tokens and three black tokens. Tokens are drawn one at a time without replacement. The first two drawn tokens have matching colors, but their color was not recorded. Calculate the probability that the third token drawn is silver.
A 3/13
B 1/2
C 15/26
D 5/8
E 10/13
Variant answer in brief
Given a match, the first pair is silver with posterior probability 10/13 and black with probability 3/13. Averaging the remaining silver fractions gives 15/26, so choice C.
Setup
Setup
Let M be the event that the first two colors match. Count the possible unordered first pairs by color.
Pr(SS)=(28)(25)=2810
Pr(BB)=(28)(23)=283
Pr(M)=2813
Model
Model
Update the color of the matched pair, then use the corresponding composition of the six remaining tokens.
Pr(SS∣M)=1310,Pr(BB∣M)=133
Pr(S3∣SS)=63,Pr(S3∣BB)=65
Compute
Compute
Average the two possible remaining silver fractions using the posterior pair-color probabilities.
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