This Exam P sample reference tests Exponential Distribution. Exponential memorylessness converts the supplied conditional probability into a ten-unit survival probability of 0.027. The requested additional duration is twice as long, so its survival probability is 0.027²=0.000729 and choice A is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThis treats 0.027 as an unconditional fifteen-unit survival and computes 0.027⁽²⁵⁄¹⁵⁾, instead of using conditional elapsed durations.
CThis linearly halves 0.027 because the additional duration doubles; exponential survival compounds multiplicatively, not linearly.
DThis scales 0.027 by the time ratio 15/25, which is not a valid survival transformation.
EThis applies the inverse linear factor 25/15 to 0.027, making survival increase with a longer duration.
Original practice · fully worked
Original variant: remaining filter-life median
A filter's lifetime T is exponential. It is known that Pr(T>8 given T>3)=0.25. A filter has already lasted 3 months. Calculate the median of its remaining lifetime, in months.
A 1.25
B 2.50
C 3.00
D 5.00
E 8.00
Variant answer in brief
Memorylessness gives five-month survival S(5)=0.25. Since 0.25=(0.5)², the remaining-lifetime survival reaches 0.5 after 2.5 months, so choice B is correct.
Setup
Setup
Translate the conditional statement into survival over the additional five months.
Pr(T>8∣T>3)=S(5)=0.25
Model
Model
The remaining life is exponential with the same rate. Its median m satisfies survival probability one half.
S(m)=0.5
Compute
Compute
Use the exponential relation between survival probabilities and elapsed time.
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