This Exam P sample reference tests Law of Total Probability. Condition on the three equally likely unordered die pairs. Their conditional target probabilities are 1/18, 1/6, and 0, so the total probability is 2/27=0.074074 and choice C is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis value is smaller than the constant-six pair's weighted contribution alone, (1/3)(1/6)=0.0556; it necessarily drops that valid branch and therefore undercounts the mixture.
BThis is 3/108=1/36. It counts the two favorable U-E physical-face outcomes but only one U-S outcome, collapsing the constant die's six equiprobable faces into one instead of counting all six.
DThis is 1/9, obtained by replacing the U-and-E probability 1/18 with 1/6 and then computing (1/3)(1/6+1/6). It ignores the requirement that E also show 6.
EThis is 1/18+1/6=2/9. It adds the two nonzero conditional probabilities but omits the factor 1/3 for selecting each pair.
Original practice · fully worked
Original variant: randomized scoring protocols
A calibration run uses protocol I with probability 1/2, protocol II with probability 1/3, and protocol III with probability 1/6. Under I, independent scores are drawn uniformly from {0,1,2} and {0,2}, then added. Under II, a score drawn uniformly from {1,3} is increased by 2. Under III, a score drawn uniformly from {0,1,2} is increased by 1. Calculate the probability that the final score equals 3.
A 0.1389
B 0.1667
C 0.3056
D 0.3333
E 0.6944
Variant answer in brief
The conditional target probabilities under protocols I, II, and III are 1/6, 1/2, and 1/3. Weighting by the protocol probabilities gives 11/36=0.305556, so choice C is correct.
Setup
Setup
Determine the target probability separately under each protocol.
qI=31⋅21=61
qII=21
qIII=31
Model
Model
The protocol label partitions the experiment, so its stated probabilities are the mixture weights.
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