This Exam P sample reference tests Geometric Distribution. The recurrence defines a geometric mass function with p0=0.8; the tail beyond one is 0.2²=0.04, choice A.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThe value 0.16 is the one-count mass 0.2 × 0.8, not the tail beyond one.
CThe value 0.20 is the common geometric ratio. It equals Pr(N greater than 0), not Pr(N greater than 1).
DThe value 0.80 is the zero-count mass, not the requested tail.
EThe value 0.96 is the sum of the zero-count and one-count masses, the complement of the requested tail.
Original practice · fully worked
Original variant: memoryless continuation after three failed attempts
Independent connection attempts succeed with probability 0.35. It is known that the first three attempts failed. Find the conditional probability that the first success occurs after attempt five.
A 0.1225
B 0.4225
C 0.5775
D 0.6500
E 0.7250
Variant answer in brief
After the first three failures, the geometric process restarts. Passing attempt five requires two additional failures, probability 0.65²=0.4225, choice B.
Setup
Setup
Let q=0.65 be the probability that a single connection attempt fails.
q=1−0.35=0.65
Model
Model
Given the first three failures, the geometric memoryless property reduces the event T greater than 5 to failures on attempts 4 and 5.
Pr(T>5∣T>3)=Pr(attempts 4 and 5 both fail)
Compute
Compute
Independence of the attempts gives q²=0.65²=0.4225 for the two additional failures.
Pr(T>5∣T>3)=q2=0.4225
Answer
Answer
The conditional probability that the first success occurs after attempt five is 0.4225, which is choice B.
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