Question

Jones figures that the total number of thousands of miles that an auto can be driven before it would need to be junked is an exponential random variable with a parameter$\frac{1}{20}$. Smith has a used car that he claims has been driven only $10,000$miles. If Jones purchases the car, what is the

probability that she would get at least $20,000$additional miles out of it? Repeat under the assumption that the life-

time mileage of the car is not exponentially distributed, but rather is (in thousands of miles) uniformly distributed over$(0,40)$.

Short answer

For exponential distribution we have$0.368$and for uniform$\frac{1}{3}$.

Step-by-step solution

Given Information.

Let $X$be an exponential random variable that represents the number of thousands of miles that$a$used auto can be driven$X~exp\left(\frac{1}{20}\right)$.

Explanation.

So, what we want to calculate is the probability that it has already crossed$10$thousands of miles.

Given Information.

$P(X>30|X>10)=P(X>20+10|X>10)$

$$\begin{gathered} =P(X>20) \\ =e^{\frac{-1}{20}.20} \\ =0.368 \end{gathered}$$

Explanation.

Now, Let $X$be uniformly distributed role="math" localid="1646708965270" $X~U(0,40)$

Explanation.

We have conditional probability:

$P(X>30|X>10)=\frac{P(X>30)}{P(X>10)}$

$$\begin{gathered} =\frac{1-P(X\le 30)}{-P(X\le 10)} \\ =\frac{1-{\displaystyle \frac{30}{40}}}{1-{\displaystyle \frac{10}{40}}} \\ =\frac{1}{3} \end{gathered}$$