This Exam P sample reference tests Exponential Distribution. The lower-threshold CDF value implies a survival probability of 0.85 at two units. Ten units is five such intervals, so the new CDF is 1-0.85⁵=0.556295 and choice D is correct.
How to solve this Exponential Distribution question
Setup
Setup
Let X have an exponential distribution with rate λ. Translate the given complete-coverage probability into its CDF and survival values at the lower threshold.
FX(t)=1−e−λt
FX(2)=0.15⟹e−2λ=0.85
Model
Model
The larger threshold is five times the smaller one, so its survival probability is the fifth power of the lower-threshold survival probability.
e−10λ=(e−2λ)5
Compute
Compute
Raise 0.85 to the fifth power, then convert survival back to the CDF.
FX(10)=1−(0.85)5
FX(10)=1−0.4437053125=0.5562946875
Answer
Answer
The larger-threshold complete-coverage probability rounds to 0.556.
0.556(D)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.200 is only the threshold ratio 2/10. It does not use the exponential CDF or the supplied probability.
BIf 1-0.85ʳ were 0.351, then r=log(0.649)/log(0.85)=2.66014. The actual threshold multiplier is 10/2=5, so 0.351 uses an incompatible rescaling.
CThis choice is near (0.85)⁵=0.443705, the survival probability Pr(X>10) at the larger threshold. The requested event requires the complementary probability, 0.556295.
EThe value 0.750=5(0.15) linearly scales a CDF probability by the threshold ratio. Exponential survival probabilities multiply across equal intervals; CDF values do not add this way.
Original practice · fully worked
Original variant: paired exponential component lifetimes
Two independent components have identical exponential lifetimes. The probability that at least one component is still operating after three hours is 0.75. Calculate the probability that both components are still operating after six hours.
A 0.1250
B 0.2500
C 0.5000
D 0.0625
E 0.7500
Variant answer in brief
The observation makes each component's three-hour survival probability 0.5. Exponential scaling gives six-hour survival 0.25 per component, and independence gives 0.25²=0.0625, so choice D is correct.
Setup
Setup
Let S(t) be the common component survival function. Complement the observed at-least-one-survivor event.
Pr(both fail by 3)=1−0.75=0.25
Model
Model
Independence makes the probability that both have failed by time three the square of the individual failure CDF.
(1−S(3))2=0.25
1−S(3)=0.5⟹S(3)=0.5
Compute
Compute
Use exponential survival over two consecutive three-hour intervals, then require both independent components to survive.
S(6)=S(3)2=(0.5)2=0.25
Pr(both survive to 6)=S(6)2=(0.25)2=0.0625
Answer
Answer
The probability that both components remain operating after six hours is 0.0625.
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