Question

Jones figures that the total number of thousands of miles that used auto can be driven before it would need to be junked

Probability
ANSWERED
asked 2021-08-17

Jones figures that the total number of thousands of miles that a used auto can be driven before it would need to be junked is an exponential random variable with parameter \(\displaystyle\frac{{1}}{{20}}\).
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).

Expert Answers (1)

2021-08-18
1. Let X exponential random variable that represents the number of thousands of miles that a used auto can be driven, \(\displaystyle{X}\sim{\exp{{\left(\frac{{1}}{{20}}\right)}}}\). So what we want to calculate is a probability that the car will cross 30 thousands miles if we have that it has already crossed 10 thousands miles: \(\displaystyle{P}{\left({X}{>}{30}{\mid}{X}{>}{10}\right)}={P}{\left({X}{>}{20}+{10}{\mid}{X}{>}{10}\right)}={P}{\left({X}{>}{20}\right)}={e}^{{-\frac{{1}}{{20}}\cdot{20}}}={0.368}\)
2. Now let X be uniformly distributed< X~U(0,40). Now we have conditional probability: \(\displaystyle{P}{\left({X}{>}{30}{\mid}{X}{>}{10}\right)}={\frac{{{P}{\left({X}{>}{30}\right)}}}{{{P}{\left({X}{>}{10}\right)}}}}={\frac{{{1}-{P}{\left({X}\le{30}\right)}}}{{-{P}{\left({X}\le{10}\right)}}}}={\frac{{{1}-\frac{{30}}{{40}}}}{{{1}-\frac{{10}}{{40}}}}}=\frac{{1}}{{3}}\)
Result: For exponential distribution we have 0.368, and for uniform 1/3.
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