This Exam P sample reference tests Survival Functions. This is a conditional upper-tail calculation for a power-law lifetime. Its survival function is proportional to the inverse fourth power of time, so conditioning from month 3 to month 7 cancels the unknown scale. Raising 3/7 to the fourth power gives 81/2401, making choice A correct.
Integrate the density above an age t to obtain its survival function. The supplied parameter restriction places both relevant ages above the lower endpoint.
S(t)=∫t∞u54β4du
S(t)=(tβ)4,t≥β
Model
Model
Four more months after an elapsed age of three corresponds to total age seven. Express the requested probability as a survival ratio.
Pr(T>7∣T>3)=S(3)S(7)
Pr(T>7∣T>3)=(β/3)4(β/7)4
Compute
Compute
Cancel the scale parameter before evaluating the power.
(β/3)4(β/7)4=(73)4
(73)4=240181=0.0337359434
Answer
Answer
The conditional probability is 81/2401.
240181(A)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BConditioning from month 4 to month 7 raises 4/7 to the fourth power, giving 256/2401. This treats the requested four-month extension as if it were the already attained age.
CConditioning from month 3 to month 4 raises 3/4 to the fourth power, giving 81/256. That calculates survival for only one additional month.
DThe displayed ratio divides cumulative probabilities at months 3 and 7. The given information and requested event are both upper-tail events.
EThe displayed ratio divides cumulative probabilities at months 4 and 7. It substitutes the four-month increment for the elapsed-age threshold and uses lower tails instead of survivals.
Original practice · fully worked
Original variant: survival through changing weekly risks
A field sensor is active at the end of week 2. Conditional on being active at the start of a week, its failure probabilities during weeks 3, 4, and 5 are 0.10, 0.15, and 0.20, respectively. Calculate the probability that the sensor remains active through the end of week 5.
A 0.003
B 0.388
C 0.450
D 0.550
E 0.612
Variant answer in brief
Convert each age-specific failure probability to a one-week conditional survival probability. Multiplying 0.90, 0.85, and 0.80 gives 0.612, so choice E.
Setup
Setup
The known survival through week two makes week three the first remaining interval. Complement each upcoming conditional failure probability.
q3=1−0.10=0.90
q4=1−0.15=0.85
q5=1−0.20=0.80
Model
Model
Remaining active through week five requires survival in all three successive weeks.
Pr(active through week 5∣active after week 2)=q3q4q5
Compute
Compute
Multiply the conditional survival probabilities in chronological order.
q3q4q5=(0.90)(0.85)(0.80)=0.612
Answer
Answer
The sensor has a 61.2% chance of staying active throughout the three remaining weeks.
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