This Exam P sample reference tests Poisson Distribution. For a Poisson count, the second moment is λ squared plus λ, so the rate is 3. Conditioning the upper count event on at least one count gives 0.842813 and choice C.
Let N be Poisson with rate λ and recover that rate from its stated second moment.
E[N2]=Var(N)+E[N]2=λ+λ2
λ2+λ=12
Model
Model
The nonnegative solution is λ equal to 3. The requested conditional event uses the probability of two or more counts divided by the probability of one or more.
λ=3
Pr(N≥2∣N≥1)=Pr(N≥1)Pr(N≥2)
Compute
Compute
Use the Poisson probabilities at zero and one to evaluate both the numerator and denominator.
Pr(N≥2)=1−e−3−3e−3=1−4e−3
Pr(N≥1)=1−e−3
1−e−31−4e−3=0.8428129105…
Answer
Answer
The conditional probability rounds to 0.84.
Pr(N≥2∣N≥1)≈0.84(C)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is below the unconditioned value P(N ≥ 2) = 0.80085, even though conditioning on the smaller space N ≥ 1 must increase the probability.
BThis is the unconditioned numerator P(N ≥ 2) = 1 − 4e⁻³ = 0.80085; it omits division by P(N ≥ 1).
DUsing λ = √12 treats the second moment as only λ²; that wrong rate gives a conditional probability about 0.888.
ESolving the wrong equation λ² − λ = 12 gives λ = 4, whose conditional probability is about 0.925 rather than the value for the correct rate.
Original practice · fully worked
Original variant: overnight support queue
The number Q of urgent support tickets arriving during an overnight shift is Poisson. Its second moment is 20. Calculate the probability that at least two tickets arrive, given that at least one ticket arrives.
A 0.800
B 0.908
C 0.925
D 0.948
E 0.982
Variant answer in brief
The second-moment identity gives a Poisson rate of 4. Dividing the probability of at least two tickets by the probability of at least one gives 0.925371 and choice C.
Setup
Setup
Use the Poisson second-moment formula to identify the overnight rate.
λ2+λ=20
λ=4
Model
Model
The event of two or more tickets is contained in the conditioning event of one or more tickets.
Pr(Q≥2∣Q≥1)=1−Pr(Q=0)1−Pr(Q=0)−Pr(Q=1)
Compute
Compute
Insert the zero-count and one-count probabilities at rate 4.
Pr(Q≥2∣Q≥1)=1−e−41−5e−4
1−e−41−5e−4=0.9253705585…
Answer
Answer
Given that the queue is nonempty, the chance that it contains at least two tickets is about 0.925.
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