This Exam P sample reference tests Geometric Distribution. If p is the per-trial event probability, the supplied second-trial probability gives (1-p)p=0.16. The stated parameter restriction selects p=0.2, after which (0.8)³(0.2)=0.1024 and choice D is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AChoosing the excluded root p=0.8 gives q=0.2 and q³p=(0.2)³(0.8)=0.0064. The parameter restriction is what rules out this root.
BThe value 0.0256=(0.16)² squares the given second-trial probability. First-event masses at different times are not independent events that can be multiplied this way.
CUsing 1-3p=0.4 as an approximation to the probability of no event in the first three trials gives 0.4(0.2)=0.0800. Independence requires q³=0.8³=0.512 instead.
EThe value 0.2000 is the per-trial event probability p. A first event on trial four also requires three preceding non-events, contributing the factor q³.
Original practice · fully worked
Original variant: retry-log probability ratio
A server repeats independent transmission attempts until the first acknowledgement, with the same acknowledgement probability on every attempt. The probability that the first acknowledgement occurs on attempt five is 0.343 times the probability that it occurs on attempt two. Given that attempt one failed, calculate the probability of an acknowledgement by the end of attempt four.
A 0.300
B 0.343
C 0.490
D 0.657
E 0.700
Variant answer in brief
The ratio of the two first-acknowledgement masses is q³=0.343, so the failure probability is q=0.7. Conditional on the first failure, success within the next three attempts has probability 1-q³=0.657, so choice D is correct.
Setup
Setup
Let p be the acknowledgement probability and q=1-p the failure probability on each independent attempt.
Pr(T=k)=qk−1p
Model
Model
Form the ratio of the two supplied first-acknowledgement probabilities; the common factor p cancels.
Pr(T=2)Pr(T=5)=qpq4p=q3=0.343
Compute
Compute
Infer q and complement the probability that all three remaining attempts through attempt four fail.
q=30.343=0.7,p=0.3
Pr(T≤4∣T>1)=1−q3=1−0.343=0.657
Answer
Answer
Given the first failed attempt, the chance of an acknowledgement by attempt four is 0.657.
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