This Exam P sample reference tests Conditional Expectation. Given the second roll is the first 6, the first roll is uniformly one of the five non-6 faces. A first-roll 5 occurs with probability 0.2; otherwise memorylessness gives mean 8, so the result is 6.6, choice D.
How to solve this Conditional Expectation question
Setup
Setup
Given that the second roll is the first 6, the first roll is uniformly distributed over the five non-6 faces.
P(X=1∣Y=2)=51
P(X≥3∣Y=2)=54
Model
Model
With conditional probability 1/5 the first roll is already 5. Otherwise no 5 has appeared by roll 2, and the geometric waiting time resumes after that point.
E[X∣X≥3]=2+1/61=8
Compute
Compute
The residual-case conditional mean is 8. Weighting the immediate and residual cases gives 0.2(1)+0.8(8)=6.6.
E[X∣Y=2]=51(1)+54(8)=6.6
Answer
Answer
The conditional expected waiting time is 6.6 rolls, corresponding to choice D.
6.6(D)
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CThe value 6.0 is the unconditional geometric mean waiting time for a 5. It ignores the information about when the first 6 appears.
Original practice · fully worked
Original variant: waiting for red given blue first appears on draw three
A spinner independently lands red with probability 0.20, blue with probability 0.20, and another color otherwise. Let X be the draw of the first red and Y the draw of the first blue. Find E[X | Y=3].
A 4.600
B 5.000
E 5.125
D 5.800
C 6.200
Variant answer in brief
Conditional on no blue in the first two draws, each is red with probability 0.25. Enumerating first red on draw 1, draw 2, or after draw 3 yields the conditional mean.
Setup
Setup
Conditioning on no blue in the first two draws changes the red probability on each of those draws to 0.2/0.8=0.25.
P(R∣not B)=0.80.2=0.25
Model
Model
Partition on first red at draw 1, at draw 2, or after the forced blue at draw 3.
P(X=1∣Y=3)=0.25
P(X=2∣Y=3)=0.75(0.25)
P(X≥4∣Y=3)=0.752
Compute
Compute
The three case probabilities are 0.25, 0.1875, and 0.5625. Their conditional waiting-time means combine to 5.125.
E[X∣Y=3]=0.25(1)+0.1875(2)+0.5625(3+0.21)=5.125
Answer
Answer
Therefore E[X | Y=3]=5.125, corresponding to choice E.
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