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 r.v. with parameter 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 uniformly distributed over (0,40)? I did first question as below, is that correct? Let X denotes the no. of thousands of miles that an auto can be driven P(X>30) = 1 – F(30) = 1 – (1 - e-µx) = e-µx = e –(1/20)(30) = 0.223
but i have no idea for the 2nd question, any hints?
For the uniform, we want $$\Pr(X\ge 30|X\ge 10).$$ By the definition of conditional probability, this is $$\frac{\Pr(X\ge 30\cap X\ge 10)}{\Pr(X\ge 10}.$$ Now calculate. We get $$\frac{10/40}{30/40}.$$
The exponential distribution case was dealt with in a comment. By the memorylessness of the exponential, we have $$\Pr(X\ge 30|X\ge 10)=\Pr(X\ge 20)=\int_{20}^\infty \frac{1}{20}e^{-t/20}\,dt=e^{-20/20}.$$
Alternately, we can do a conditional probability calculation as in the uniform case. We we do that, we will get $\frac{e^{-30/20}}{e^{-10/20}}$, which is the same as the number we got using memorylessness.