This Exam P sample reference tests Geometric Distribution. Writing q for the probability that one patient does not have the disease gives r=q cubed. Conditioning on reaching patient four leaves eight additional failures before patient twelve, so the answer is q to the eighth power=r⁽⁸⁄³⁾, choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe expression r⁽¹¹⁄³⁾=q¹¹ is the unconditional probability of reaching patient twelve. It does not divide by the event of reaching patient four.
BThe expression r cubed=q⁹ counts nine additional failures. From the fourth tested patient to the twelfth requires eight additional failures.
DThe expression r squared=q⁶ counts only six additional failures after the conditioning point.
EThe expression r⁽¹⁄³⁾=q is the probability of only one further failure, not eight.
Original practice · fully worked
Original variant: a geometric doubling score after a checkpoint
Independent attempts continue until a success. The probability that at least three attempts are required is 0.16. After learning that the first two attempts failed, let K be the number of further failures before success. A score is defined as 2ᴷ. Calculate the expected score.
A 0.600
B 1.587
C 1.667
D 3.000
E 5.000
Variant answer in brief
Two initial failures have probability 0.16, so the failure probability is 0.40 and success probability is 0.60. Memorylessness gives P(K=k)=0.60(0.40)ᵏ, and summing 2ᵏ times this mass yields 3, choice D.
Setup
Setup
Let q be the per-attempt failure probability and recover it from the two known failures.
Pr(T≥3)=q2=0.16
q=0.40,p=1−q=0.60
Model
Model
Memorylessness makes the additional failure count geometric on zero, one, two, and so on.
Pr(K=k)=pqk=0.60(0.40)k,k=0,1,2,…
Compute
Compute
Weight each score by its geometric probability and sum the convergent series.
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