This Exam P sample reference tests Poisson Distribution. Use the Poisson second-moment identity to identify the rate, then evaluate the upper tail by complement. The verified value is approximately 0.323324, corresponding to choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is the probability of four or more events, so it results from shifting the tail cutoff upward by one.
BThis keeps only the probability of exactly three events instead of adding the entire upper tail.
DThis is produced by using 3 as the Poisson rate before evaluating the same tail.
EThis is the probability of two or more events, so it results from shifting the cutoff downward by one.
Original practice · fully worked
Original variant: observatory alert count
An observatory models the number of false instrument alerts during a 12-hour shift with a Poisson random variable N. The probability of no alerts during a shift is exp(−1.6). Calculate the probability that at most two alerts occur during a shift.
A 0.202
B 0.323
C 0.525
D 0.783
E 0.921
Variant answer in brief
The zero-count probability identifies the rate as 1.6. Summing the masses for zero, one, and two gives approximately 0.783.
Setup
Setup
For a Poisson count, the zero-count probability is the exponential of the negative rate.
Pr(N=0)=e−λ
Model
Model
Match the supplied zero-count probability to the Poisson formula.
e−λ=e−1.6⟹λ=1.6
Compute
Compute
Add the probabilities of zero, one, and two alerts.
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