This Exam P sample reference tests Order Statistics. The maximum is at most 3 exactly when all three independent losses are at most 3. Taking the complement gives about 0.414, choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AChoice A multiplies the three upper-tail probabilities and therefore represents simultaneous exceedance by all three losses.
BChoice B uses only one component's tail probability.
CChoice C replaces the three different losses by a single representative component.
DChoice D adds the three tail probabilities without subtracting their overlaps.
Original practice · fully worked
Original variant: all three calibration jobs finish by a deadline
Three independent calibration jobs have exponential durations with means 2, 3, and 5 hours. Find the probability that every job finishes within 4 hours.
A 0.2520
B 0.3455
E 0.3506
D 0.5276
C 0.7476
Variant answer in brief
Independence makes the all-finish probability the product (1-exp(−2))(1-exp(−4/3))(1-exp(−0.8)), about 0.3506.
Setup
Setup
Every job finishes within four hours only when each of the three independent exponential durations is at most four.
P(Ti≤4)=1−e−4/μi
Model
Model
Compute the deadline cumulative probability for each mean and multiply the three values.
P(maxTi≤4)=(1−e−2)(1−e−4/3)(1−e−0.8)
Compute
Compute
The product is approximately 0.350635.
P(maxTi≤4)=0.350635
Answer
Answer
The probability that all three jobs finish by hour four is approximately 0.3506, selecting choice E.
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