Question

The number of years a radio functions is exponentially distributed with parameter λ = 1 8 . If Jones buys a used radio, what is the probability that it will be working after an additional 8 years?

Answer 1

`P(X>8)=e^(-1)`

Explanation:

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

`P(X=x)=\lambda e^(-\lambda x), x>0`

And 0 for other case. Let X the random variable that represent "The number of years a radio functions" and we know that the distribution is given by:

`X \sim Exp(\lambda=(1)/(8))`

We can assume that the random variable t represent the number of years that the radio is already here. So the interest is find this probability:

`P(X>8|X>t)`

We have an important property on the exponential distribution called "Memoryless" property and says this:

`P(X>a+t| X>t)=P(X>a)`

Where a represent a shift and t the time of interest.

On this case then
`P(X>8|X>t)=P(X>8+t|X>t)=P(X>8)`

We can use the definition of the density function and find this probability:

`P(X>8)=\int_(8)^(\infty) (1)/(8)e^{-(1)/(8)x}dx`

`=(1)/(8) \int_(8)^(\infty) e^{-(1)/(8)x}dx`

`=[lim_(x\to\infty) (-e^{-(1)/(8)x})+e^(-1)]=0+e^(-1)=e^(-1)`