This Exam P sample reference tests Poisson Distribution. After the first two accident-free years are conditioned away, the first accident must be avoided in three further years and then occur in the next year. The resulting probability is 0.070498, so choice C is correct.
A year with a Poisson mean of 0.10 has a zero-count probability obtained directly from the Poisson mass function.
q=Pr(N=0)=e−0.10
Model
Model
The stated conditioning fixes the first two years. For the first accident to arrive in year six, years three through five must have no accidents and year six must have at least one.
Pr(target∣first two years empty)=q3(1−q)
Compute
Compute
Independence across years permits multiplication of the four yearly factors.
q3(1−q)=e−0.30(1−e−0.10)
e−0.30(1−e−0.10)=0.07049817465
Answer
Answer
The nearest listed probability is 0.0705.
0.0705(C)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis cubes the yearly positive-count probability, (1-exp(−0.1))³=0.000862, as though accidents rather than empty years were required in the intervening period.
BThis ignores the conditioning and requires all five years before year six to be empty: exp(−0.5)(1-exp(−0.1))=0.05772.
DThis treats the Poisson mean 0.10 as a Bernoulli accident probability, giving 0.9³(0.1)=0.0729.
EThis accumulates the probability of at least one accident over four years, 1-exp(−0.4)=0.3297, instead of requiring the first nonempty year to be the last one.
Original practice · fully worked
Original variant: first week with service requests
Requests arrive at a remote support desk according to an independent Poisson count with mean 0.25 per week. No requests arrived during the first four weeks of a monitoring period. Calculate the probability that week eight is the first later week containing at least one request.
A 0.0489
B 0.0814
C 0.1045
D 0.1723
E 0.5276
Variant answer in brief
Weeks five through seven must be empty and week eight nonempty. Independence gives exp(−0.75)(1-exp(−0.25))=0.104487, so choice C is correct.
Setup
Setup
Let q be the probability of an empty week.
q=e−0.25
Model
Model
The earlier observation is already conditioned upon. Three additional empty weeks must precede one nonempty week.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.
