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

Answer:

$P(X>8)=e^{-1}$

Step-by-step 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:

$X \sim Exp(\lambda=\frac{1}{8})$

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:

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

$=\frac{1}{8} \int_{8}^{\infty} e^{-\frac{1}{8}x}dx$

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