This Exam P sample reference tests Conditional Probability. Integrating the density gives survival S(t)=exp(−t)(1+t). Conditional on lasting beyond one year, the probability of ending by year two is 1-S(2)/S(1)=1-3/(2e)=0.448181, so choice C is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis comes from an incorrect antiderivative or normalization of the density and is not a valid ratio of the two relevant probabilities.
BThis is the unconditional probability of lasting between one and two years, S(1)-S(2), without division by P(T≥1).
DThis is P(T≥1)=2/e, the probability of the conditioning event rather than the conditional interval probability.
EThis uses F(2)/S(1), putting the unconditional event T≤2 in the numerator without intersecting it with the conditioning event T≥1.
Original practice · fully worked
Original variant: remaining filter service
A two-stage filter has total service time T with a gamma distribution of shape 2 and rate 0.5 per hour. The filter is still operating after 4 hours. Calculate its expected additional operating time.
A 2.000 hours
B 2.667 hours
C 4.000 hours
D 6.667 hours
E 8.000 hours
Variant answer in brief
For a shape-2 gamma lifetime with rate 0.5, survival is exp(−0.5t)(1+0.5t). Integrating the conditional survival beyond hour 4 gives mean residual life 8/3=2.6667 hours, so choice B is correct.
Setup
Setup
Write the survival function for the shape-two gamma lifetime.
S(t)=e−0.5t(1+0.5t)
S(4)=3e−2
Model
Model
The expected remaining life after age four is the future survival area divided by survival to age four.
E[T−4∣T>4]=S(4)∫4∞S(t)dt
Compute
Compute
Evaluate the survival integral and simplify the ratio.
∫4∞e−0.5t(1+0.5t)dt=8e−2
E[T−4∣T>4]=3e−28e−2=38
Answer
Answer
The expected additional operating time is approximately 2.667 hours.
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