- #1

IntegrateMe

- 217

- 1

[tex]p(t) = 1-e^{-λt}[/tex] for t ≥ 0, and 0 otherwise.

The cell phone company offers its clients a replacement phone after two years if they sign a new contract. What is the probability that the client will not have to replace his phone before the company will give him a new one?

I tried solving the problem as follows:

[tex]\int_{-\infty}^2 1-e^{-λt}dt[/tex] which would end up becoming [tex]\int_0^2 1-e^{-λt}dt[/tex] since the function is 0 for everything t < 0.

However, the solution says that the answer is actually [tex]\int_2^\infty 1-e^{-λt}dt[/tex]

I'm having trouble understanding why. If we're trying to find the probability that the client will not have to replace his phone before two years, why is the solution finding the probability that the phone will be defective after 2 years?