This Exam P sample reference tests Poisson Distribution. The observation removes the zero-count outcome. Dividing the Poisson upper tail above three by the nonzero probability gives 0.577100, so choice B is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.566530 is the unconditional probability Pr(N>3). It omits the division by Pr(N≥1)=0.981684 required by the observation.
CDividing the correct numerator by Pr(N≥2)=0.908422 gives 0.623642, which rounds to 0.624. That conditions on the wrong lower bound.
DReplacing N>3 by N≥3 gives Pr(N≥3)/Pr(N≥1)=0.776112. This includes the entire N=3 mass even though it is outside the target event.
EUsing both shifted bounds gives Pr(N≥3)/Pr(N≥2)=0.838704, which rounds to 0.839. Neither the numerator nor denominator then matches the required events.
Original practice · fully worked
Original variant: routing sources in a merged alert stream
Two independent monitoring channels generate alerts according to Poisson processes with hourly means 1.2 and 1.8. During an hour, the merged log contains exactly four alerts. Calculate the conditional probability that at least two of those alerts came from the first channel.
A 0.1792
B 0.3456
C 0.4752
D 0.5248
E 0.8704
Variant answer in brief
Conditional on a merged total of four, the first-channel count is binomial with success probability 1.2/3.0=0.4. Its probability of at least two is 0.5248, so choice D is correct.
Setup
Setup
Let K be the number of first-channel alerts among the four merged alerts. Poisson splitting determines the conditional source probability.
r=1.2+1.81.2=0.4
Model
Model
Given the merged total, each alert is assigned to the first channel with probability r, so K has a binomial distribution.
K∣(N1+N2=4)∼Binomial(4,0.4)
Compute
Compute
Use the complement of zero or one first-channel alert.
Pr(K≥2)=1−(04)(0.6)4−(14)(0.4)(0.6)3
Pr(K≥2)=1−0.1296−0.3456=0.5248
Answer
Answer
The conditional probability of at least two first-channel alerts is 0.5248.
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