Q.5.34 - A First Course in Probability

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$(0,40)$.

Step-by-Step Solution

Answer

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

Step 1 Given Information.

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

Step 2 Explanation.

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

Step 3 Given Information.

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

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

Step 4 Explanation.

Now, Let $X$be uniformly distributed $X~U(0,40)$

Step 5 Explanation.

We have conditional probability:

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

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