This Exam P sample reference tests Exponential Distribution. For an exponential lifetime, S(x)=exp(−λ x). Since S(1)=0.80, S(x)=(0.80)ˣ and the CDF is F(x)=1-(0.80)ˣ, selecting choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe expression (0.80)ˣ is the survival function S(x), not the cumulative distribution F(x). The CDF is its complement.
BThis density-like product is not a valid CDF: it does not approach 1 as x increases and uses 0.80 as though it were a rate.
DThis sets the exponential rate λ equal to 0.80. The supplied fact is exp(−λ)=0.80, so λ=-ln(0.80), not 0.80.
EThis both treats 0.80 as the rate and reports a decreasing survival expression instead of an increasing CDF.
Original practice · fully worked
Original variant: remaining decay time after inspection
A chemical cartridge has an exponential decay time T. By hour 6, 75% of cartridges have decayed. A cartridge is still active at hour 6. Calculate the probability that it remains active through hour 10.
A 0.0992
B 0.2500
C 0.3969
D 0.6300
E 0.7500
Variant answer in brief
The six-hour survival probability is 0.25. Memorylessness reduces the conditional event to four additional hours, giving S(4)=(0.25)⁽⁴⁄⁶⁾=(0.25)⁽²⁄³⁾=0.3969 and choice C.
Setup
Setup
Translate the reported decay percentage into a six-hour survival probability.
S(6)=1−0.75=0.25
e−6λ=0.25
Model
Model
Conditioning on survival to hour six leaves an exponential remaining lifetime, so only the next four hours matter.
Pr(T>10∣T>6)=Pr(T>4)=e−4λ
Compute
Compute
Express the four-hour survival in terms of the known six-hour survival.
e−4λ=(e−6λ)4/6
Pr(T>10∣T>6)=(0.25)2/3=0.3968502630…
Answer
Answer
The cartridge remains active through hour ten with conditional probability approximately 0.3969.
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