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

(Sorry to repost)
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?

zarkon · Answer

I like your answer better

anonymous · Answer

Thanks for the response.  Hopefully my prof will too.

zarkon · Answer

obviously we have the 350 as being fixed.. 

The expected loss for the warranty is

$$E(L)=-40\cdot P(W\leq 12)+0\cdot P(W>12)$$
$$=-40(1-e^{-2/3})+0\approx19.46$$

Thus the expected profit is 350-19.46=330.54

anonymous · Answer

Super.  That's what I got as well.  Couldn't talk myself into 2/3, just wanted to make sure I wasn't missing something.  Appreciate the effort.