This Exam P sample reference tests Poisson Distribution. This problem expresses an increasing sequence of charges as a triangular-number function of a Poisson count. Using the first two Poisson moments gives an expected total of 12 and choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis evaluates the unordered-pair count at the mean of four, giving six. It uses the wrong charge function and moves the expectation through a nonlinear calculation.
BThis takes half the second factorial moment, giving eight unordered pairs, but omits the additional linear term in the triangular charge.
CThis inserts the mean of four into the triangular formula and gets ten. A nonlinear function evaluated at the mean does not generally equal its expectation.
EThis counts the linear contribution twice, combining the second moment 20 with twice the mean to obtain 14.
Original practice · fully worked
Original variant: expected review cost for alert triples
A monitoring system records a Poisson number N of alerts per shift with mean 3. After each shift, every unordered group of three distinct alerts is reviewed, at a cost of 2 credits per group. Calculate the expected review cost.
A 4.5 credits
B 9 credits
C 18 credits
D 27 credits
E 54 credits
Variant answer in brief
The reviewed groups are the unordered triples of alerts. The third Poisson factorial moment gives an expected 4.5 groups, so the two-credit review cost has mean 9 and choice B.
Setup
Setup
Express the number of unordered three-alert groups as a function of the shift count.
G=(3N)=3!N(N−1)(N−2)
C=2G
Model
Model
Use the third factorial moment of a Poisson variable.
E[N(N−1)(N−2)]=λ3
λ=3
Compute
Compute
Convert ordered factorial triples to unordered groups and apply the cost per group.
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