Probability: exponential random variable.
Profit on a computer is $350. Suppose a warranty for one year, and the time between two failures of a computer is exponential with mean 18 months. If it costs $40 to repair a failed computer, what is the expected profit per computer?

(Set up to follow)

Question

Probability: exponential random variable.
Profit on a computer is $350.  Suppose a warranty for one year, and the time between two failures of a computer is exponential with mean 18 months.  If it costs $40 to repair a failed computer, what is the expected profit per computer?

(Set up to follow)

anonymous · Answer

Let X be an exponential random variable denoting the time until failure of a computer. 

	E[X] = 18 months, therefore λ = 1/E[X] = 1/18
	Warranty repairs must be made when X < 12
	P (X<12) = 1-e-λ*12  = 1 -  1/e^(2⁄3)  ≈ .48658


So, I think it should be Profit = $350 - $40 (Probability of warranty),
but others have said that it's just Profit = $350 - $40*2/3; where 2/3 is the expected failure.  Can anyone help clarify for me?