This Exam P sample reference tests Negative Binomial Distribution. This is a truncated waiting-time expectation for the second success in independent trials. The effective success probability is 0.18, and summing the four possible stopping-time contributions gives 4.73117072, which selects choice B.
How to solve this Negative Binomial Distribution question
Setup
Setup
A trial produces the event of interest only when both required conditions occur. Multiply their probabilities and denote the complementary probability by q.
p=(0.90)(0.20)=0.18
q=1−p=0.82
Model
Model
Let T be the number of trials observed, stopping at the second success or at the five-trial cap. For stopping times two through four, the final trial is a success and exactly one earlier trial is a success. Reaching the cap means that the first four trials contain at most one success.
Pr(T=k)=(1k−1)p2qk−2,k=2,3,4
Pr(T=5)=q4+4pq3
Compute
Compute
Evaluate the stopping probabilities and form the finite expectation.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AUsing 0.20 itself as the success probability, and thereby ignoring the other required condition, gives an expectation of 4.6752 and leads to 4.68.
CUsing 0.16 as the joint probability, from multiplying 0.20 by the wrong complementary rate 0.80, gives 4.78299392 and leads to 4.78.
DAn expectation near 4.83 would require an effective success probability near 0.14. That rate is inconsistent with the required product 0.90 × 0.20.
EAn expectation near 4.88 would require an effective success probability near 0.116. This understates the correctly computed 0.18 and makes the stopping time too long.
Original practice · fully worked
Original variant: staged inspection with changing alarm rates
A laboratory checks up to four independent sensors in a fixed order. Their alarm probabilities are 0.10, 0.20, 0.30, and 0.40, respectively. Testing stops at the first alarm; if no earlier sensor alarms, the fourth sensor is still checked. Let T be the number of sensors checked. Calculate E[T].
A 0.504
B 1.914
C 2.124
D 3.124
E 4.000
Variant answer in brief
Use the tail-sum formula with the probability of reaching each stage. The expectation is 1+0.90+(0.90)(0.80)+(0.90)(0.80)(0.70)=3.124, so choice D.
Setup
Setup
A stage is reached only if every earlier sensor remains quiet.
Pr(T≥1)=1,Pr(T≥2)=0.90
Pr(T≥3)=(0.90)(0.80)=0.72
Pr(T≥4)=(0.90)(0.80)(0.70)=0.504
Model
Model
For a positive integer-valued variable capped at four, its expectation is the sum of its four tail probabilities.
E[T]=k=1∑4Pr(T≥k)
Compute
Compute
Add the probabilities of reaching the successive checks.
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