Exponential distribution problem, solution clarification

Question

Problem:

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 $λ = 1/20$. 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?

Solution $$\begin{align}P(X \geq 10,000+ 20,000 \mid X >10,000) &= P(X \geq 20,000)\\[1ex] &= \int_{20}^{\infty} e^{-x/20} \end{align}$$

Why does the integral go from $20$ to infinity? My thinking is that is should go from $20,000$ to infinity.

Answer 1

Consider the integrand! But it's still wrong…

It's $$P(X \ge 20000) = \int_{20000}^\infty \lambda e^{-\lambda x} dx = 120 \int_{20000}^\infty 120 e^{-120 x} dx$$

now you could substitute $z=\frac{x}{1000}$ and get:

$$P(X \ge 20000) = 120000 \int_{20}^\infty e^{-0.12x} dx$$

But the given "solution" is wrong.

Answer 2

That is because exponential distributions have the memoryless property, which means that a car with 10,000 miles has the same probabillity of lasting $x$ more miles as a new car (that is what the first equality states in the solution). In this case the probability of the car lasting another 20,000 miles is therefore $$ \begin{align} P(X \geq 30,000 \mid X > 10,000) &= P(X \geq 20,000)\\ &= \int_{20}^{\infty} {1\over20}\, e^{-x/20}\, dx \\ &= e^{-20/20}\\ &= 36~\% \end{align} $$ --- there was a slight error in the stated solution; the density is ${1\over20}\, e^{-x/20}$.

(this is true of any exponential distribution : the probability of exceeding the mean is $1/e=0.36\approx1/3$).

Answer 3

The parameter $\lambda$ is written in thousands of miles. So to do a computation, you essentially need to either scale $\lambda$ to $1/20,000$ or scale the $20,000$ miles to $20$ miles. The integral you provided did the latter.