This Exam P sample reference tests Conditional Probability. The favorable two-break outcome has probability 0.0324. Summing the fragile-fragile, mixed, and nonfragile-nonfragile two-break cases gives 0.0964, so the conditional probability is 0.0324/0.0964=0.3361 and choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.0324 is the joint favorable probability that both fragile packages break and both others remain intact. It has not been divided by the probability of exactly two breaks.
BThe value (0.2)²=0.040 is the probability that both fragile packages break, without requiring the other two to survive or conditioning on the observed total.
CThe value 1/6=0.167 treats the six possible pairs as equally likely. Pairs involving different package types have different break and survival probabilities.
EThis overweights the one-fragile, one-nonfragile configuration. Its actual conditional mass is 0.0576/0.0964=0.5975, and it is a different event from both broken packages being fragile.
Original practice · fully worked
Original variant: infer a calibration mode from paired pulses
A controller selects fast calibration mode with probability 0.70 and cautious mode with probability 0.30. Conditional on the selected mode, two diagnostic pulses are independent. A pulse passes with probability 0.40 in fast mode and 0.80 in cautious mode. Both pulses pass. Calculate the probability that cautious mode was selected.
A 0.192
B 0.300
C 0.368
D 0.500
E 0.632
Variant answer in brief
The joint weights for cautious mode and fast mode with two passes are 0.30(0.80)²=0.192 and 0.70(0.40)²=0.112. Normalizing gives 0.192/(0.192+0.112)=12/19=0.632, so choice E is correct.
Setup
Setup
Let C and F denote the cautious and fast modes, and let P2 denote two passing pulses. Record the mode priors and conditional pass likelihoods.
Pr(C)=0.30,Pr(F)=0.70
Pr(P2∣C)=(0.80)2,Pr(P2∣F)=(0.40)2
Model
Model
Weight each two-pass likelihood by the prior probability of its calibration mode.
Pr(C∩P2)=0.30(0.80)2=0.192
Pr(F∩P2)=0.70(0.40)2=0.112
Compute
Compute
Apply Bayes' rule by normalizing the cautious-mode weight over both ways to observe two passing pulses.
Pr(C∣P2)=0.192+0.1120.192
Pr(C∣P2)=1912=0.6315789474…
Answer
Answer
The posterior probability of cautious mode is approximately 0.632.
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