Probability density function

[IntegrateMe]

The phones offered by a cell phone company have some chance of failure after they are activated. Suppose that the density function p(t) describing the time t in years that one of their phones will fail is

[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?

 

[LCKurtz]

Your density function is for the "T = time to failure". You want it to last at least two years so you want P(T>2).

 

[IntegrateMe]

OK, but if the function describes when the phones *will* fail, wouldn't the interval T>2 describe the probability the phone will fail after 2 years, so we actually need to subtract 1 from this value?

 

[LCKurtz]

OK, but if the function describes when the phones *will* fail, wouldn't the interval T>2 describe the probability the phone will fail after 2 years, so we actually need to subtract 1 from this value?

No. You want the phone to last two years so they don't have to replace it before then. That means you want it to fail sometime after 2 years. In other words, the probability of it lasting at least two years (which is what you want) is the same as the probability that the time of failure is greater than 2.