This Exam P sample reference tests Poisson Tail Probability. Take the ratio of the adjacent Poisson masses to identify the rate as 4. The complement through count 2 then gives an upper-tail probability of approximately 0.761897, corresponding to choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is the single mass P(M=3)=0.195367..., not the probability of the whole upper tail.
BThis is the complement P(M ≤ 2)=0.238103..., so it reverses the requested event.
CThis is P(M ≤ 3)=0.433470..., obtained by adding the mass at 3 to choice B rather than taking the upper tail.
DThis is P(M ≥ 4)=0.566530..., caused by shifting the inclusive cutoff from 3 to 4.
Original practice · fully worked
Original variant: telescope calibration alerts
An autonomous telescope models its number of calibration alerts in one night by a Poisson distribution. The probability of exactly two alerts equals the probability of exactly three alerts. Calculate the probability that no more than one alert occurs during a night.
A 0.0498
B 0.1991
C 0.2240
D 0.4232
E 0.8009
Variant answer in brief
Equality of the masses at 2 and 3 forces the Poisson rate to equal 3. Summing the zero- and one-alert probabilities gives approximately 0.1991, so the answer is B.
Setup
Setup
Let N be the nightly alert count with Poisson rate λ.
Pr(N=k)=e−λk!λk
Model
Model
Use the ratio of the two adjacent masses to solve for the rate.
Pr(N=2)Pr(N=3)=3λ=1
λ=3
Compute
Compute
No more than one alert includes exactly zero and exactly one.
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