This Exam P sample reference tests Exponential Distribution. The four-year survival probability gives exp(-4 λ)=0.30, so λ=-ln(0.30)/4. Substitution into λ exp(-λ x) yields the expression in choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis expression has the form one minus a power with a negative exponent; it is neither the required decreasing density nor a valid exponential CDF.
BThis is a CDF-shaped expression and also substitutes 0.70 for the stated four-year survival probability.
CThe expression 1-(0.30)⁽ˣ⁄⁴⁾ is the calibrated cumulative distribution function, not its derivative.
DThis has the density form but calibrates the survival at four years to 0.70 rather than 0.30.
Original practice · fully worked
Original variant: median from conditional exponential survival
The operating time T of a sealed sensor follows an exponential distribution. A sensor that has already operated for five hours has probability 0.64 of continuing for at least four more hours. Determine the unconditional median operating time.
A 2.714 hours
B 4.000 hours
C 6.213 hours
D 8.963 hours
E 12.425 hours
Variant answer in brief
Memorylessness turns the conditional statement into exp(-4 λ)=0.64. The median is ln(2)/λ=6.213 hours, choice C.
Setup
Setup
Use memorylessness to remove the already-completed operating time.
Pr(T>9∣T>5)=Pr(T>4)=0.64
Model
Model
Recover the exponential rate from the four-hour survival probability.
e−4λ=0.64
λ=−4ln(0.64)=0.1115718
Compute
Compute
Set the survival probability at the median equal to one-half.
e−λm=0.5
m=λln2=6.2125674
Answer
Answer
The median operating time is approximately 6.213 hours.
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