This Exam P sample reference tests Conditional Distributions. Conditional on a first-task duration h, the second duration is uniform on the integers from 1 through h-1 and therefore has mean h/2. Averaging over the three equally likely h values gives 11/6, choice A.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThe value 2 is only the conditional mean when the first duration is 4; it ignores the equally likely first-duration values 2 and 5.
CThe value 17/8 results from unequal weighting of the conditional means 1, 2, and 2.5, even though the three first-duration cases are equally likely.
DThe value 2.5 is the conditional mean in the longest first-duration branch alone, not the average across all three branches.
EThe value 5 2/3 exceeds 4, the largest second duration that can occur, so it cannot be the expectation of the second duration.
Original practice · fully worked
Original variant: nocturnal species detections
A field recorder is placed with equal probability in one of three habitat classes whose species capacities are 4, 7, and 10. Conditional on capacity C=c, the number R of distinct nocturnal species detected is uniformly distributed over the integers 0, 1, ..., c. Calculate the expected value of R.
A 2.333
B 3.000
C 3.500
D 7.000
E 10.500
Variant answer in brief
For a capacity c, the uniform count from 0 through c has mean c/2. Averaging the conditional means for capacities 4, 7, and 10 gives 3.5, so choice C is correct.
Setup
Setup
Represent the habitat capacity by C and the observed count by R.
Pr(C=4)=Pr(C=7)=Pr(C=10)=31
Model
Model
The mean of a discrete uniform distribution on the integers from 0 through c is the midpoint c/2.
E[R∣C=c]=20+c=2c
Compute
Compute
Average the three conditional means with their equal habitat probabilities.
E[R]=31(24+27+210)
E[R]=32+3.5+5=3.5
Answer
Answer
The recorder detects 3.5 distinct species on average.
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