# Jones figures that the total number of thousands of miles that an auto

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 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 lifetime mileage of the car is not exponentially distributed, but rather is (in thousands of miles) uniformly distributed over (0,40).
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Dawn Neal
Step 1
Let X be the number of miles the car can be driven before being junked.
a) First, let $X\sim Exp\left(\frac{1}{2}\right)$.
We know that the car has been driven 10,000 miles and we want to find the probability that it will last an additional 20,000 miles. We use the memoryless property of exponential distribution, where .
Then, $P\left(X\ge 10000+20000\mid X>10000\right)=P\left(X\ge 20000\right)$.
$P\left(X\ge 30000\mid X>10000\right)=P\left(X\ge 20000\right)$
$={\int }_{20}^{\mathrm{\infty }}{e}^{-\frac{1}{20}x}dx$
$={e}^{-1}$
Step 2
b) Let $X\sim U\left(0,40\right)$. We want to find $P\left(X>30\mid X>10\right)$. By Bayes
###### Not exactly what you’re looking for?
Bertha Jordan
Let X denote the exponential random variable with parameter 1.20, U denote the uniform random variable with parameters 0 and 40. The probability function of an exponential random variable with parameter 1/20 is
$F\left(x\right)=1-{e}^{-\frac{x}{20}},x\ge 0$
The desired probability is $P\left(X>30\mid X>10\right)$. Since exponential random variable have memoryless property, we have
$P\left(X>30\mid X>10\right)=P\left(X>20\right)=1-P\left(X\le 20\right)={e}^{-1}\approx 0.3679$
The probability function of an uniform random variable with parameter 0 and 40 is
$F\left(x\right)=\frac{1}{40-0}x=\frac{x}{40},0\le x\le 40$
Hence $P\left(U>30\mid U>10\right)=\frac{P\left(U>30\right)}{P\left(U>10\right)}=\frac{1-P\left(U\le 30\right)}{1-P\left(U\le 10\right)}=\frac{1-\frac{30}{40}}{1-\frac{10}{40}}=\frac{1}{3}$ is the desired probability.
###### Not exactly what you’re looking for?
nick1337

Let X denote the total number of thousands of miles that the auto is driven before it needs to be junked. We want to compute $P\left(\left\{X\ge 30\right\}|\left\{X\ge 10\right\}\right)$. Assuming X is an exponential random variable with parameter $1/20$, we have
$P\left\{X\ge 30\right\}={\int }_{30}^{\mathrm{\infty }}\frac{1}{20}{e}^{-x/20}dx={\int }_{3/2}^{\mathrm{\infty }}{e}^{-t}dt={e}^{-3/2},$
$P\left\{X\ge 10\right\}={\int }_{10}^{\mathrm{\infty }}\frac{1}{20}{e}^{-x/20}dx={\int }_{1/2}^{\mathrm{\infty }}{e}^{-t}dt={e}^{-1/2},$
$P\left(\left\{X\ge 30\right\}|\left\{X\ge 10\right\}\right)=P\left\{X\ge 30\right\}/P\left\{X\ge 10\right\}={e}^{-3/2}/{e}^{-1/2}={e}^{-1}\approx .3678.$
Note that we could have saved some work here by using the memoryless property of the exponential random variable.
Assuming X is uniformly distributed over (0,40), we have
$P\left\{X\ge 30\right\}={\int }_{30}^{40}\frac{1}{40}dx=1/4,$
$P\left\{X\ge 10\right\}={\int }_{10}^{40}\frac{1}{40}dx=3/4,$
$P\left(\left\{X\ge 30\right\}|\left\{X\ge 10\right\}\right)=P\left\{X\ge 30\right\}/P\left\{X\ge 10\right\}=\left(1/4\right)/\left(3/4\right)=1/3.$