This Exam P sample reference tests Exponential Distribution. This is an exponential residual-loss calculation after a deductible has been crossed. The stated median fixes the residual survival scale, and the probability between the two requested bounds is 0.3536, giving choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
ARounding the inferred exponential scale 8656 to 9000 before evaluating the lower tail gives 1-exp(-3000/9000)=0.2835, producing 0.28. Besides the coarse parameter rounding, this is the probability below the lower boundary.
CThe value 0.50 is the supplied probability F_R(6000). It calibrates the exponential scale but does not answer the different two-boundary question.
DThe value 0.65 is F_R(9000)=1-2⁻³⁄²=0.6464 rounded to a listed choice. It includes all residual amounts below 3000.
EUsing the same prematurely rounded scale 9000 gives Pr(R>3000)=exp(-3000/9000)=0.7165, producing 0.72. This tail also includes amounts above 9000 that must be removed.
Original practice · fully worked
Original variant: surviving components at a later checkpoint
At a maintenance check, three independent components with exponential lifetimes are all operating. For any one component, the chance of continuing for at least five more hours is 0.70. Ten hours later, what is the chance that the checkpoint shows precisely two components still operating?
A 0.1176
B 0.2401
C 0.3674
D 0.3823
E 0.4900
Variant answer in brief
Exponential survival over ten hours is the square of the stated five-hour survival, so each component remains with probability 0.49. A binomial calculation then gives 0.3674 for exactly two survivors and choice C.
Setup
Setup
Let p be the conditional probability that one currently operating component lasts another ten hours.
Pr(T>5)=e−5λ=0.70
Model
Model
The exponential survival function multiplies over two successive five-hour intervals.
p=e−10λ=(e−5λ)2=0.702=0.49
Compute
Compute
Independence makes the number of components still operating binomial with size three and success probability 0.49.
Pr(K=2)=(23)(0.49)2(0.51)
=3(0.2401)(0.51)=0.367353
Answer
Answer
The probability of exactly two surviving components is about 0.3674.
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