This Exam P sample reference tests Poisson Distribution. Independent Poisson increments make the observed early count irrelevant to the remaining interval except for fixing the number still needed. The final three years must contain exactly two events, whose probability is 0.22404, so choice E is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AUsing exp(−3) calculates no events in the remaining three years. Two additional events, not zero, are required to reach the specified total.
BThis is approximately the unconditional probability of ten events over all ten years, using a Poisson mean of ten. It discards the observed seven-year count instead of conditioning on it.
CThis is approximately the probability of eight events in the first seven years, the event already supplied as information. The task is to evaluate what happens afterward.
DThis is close to a two-event probability under a one-year Poisson mean. The remaining exposure lasts three years, so its mean is three rather than one.
Original practice · fully worked
Original variant: downlink timing within a pass
Telemetry downlinks arrive according to a homogeneous Poisson process. During a four-hour satellite pass, exactly six downlinks are recorded. Conditional on that total, calculate the probability that exactly two of the downlinks occurred during the first hour.
A 0.234
B 0.251
C 0.297
D 0.356
E 0.466
Variant answer in brief
Given six arrivals in four hours, each arrival independently falls in the first hour with probability 1/4. The resulting binomial probability is 0.29663, so choice C is correct.
Setup
Setup
Condition on the observed total of six arrivals over the complete four-hour pass.
N(4)=6
Model
Model
Conditional on the total, homogeneous Poisson arrival times are uniformly distributed across the pass. Each arrival therefore lands in the first one-quarter of the interval with probability one-quarter.
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