melodykap

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 $\frac{1}{20}$.
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).

Alannej

1. Let X be a random variable that represents the number of thousands of miles that a used auto can be driven, $X\sim \mathrm{exp}\left(\frac{1}{20}\right)$. What we want to calculate is the probability that the car will cross 30,000 miles if we know it has already crossed 10,000 miles.$P\left(X>30\mid X>10\right)=P\left(X>20+10\mid X>10\right)=P\left(X>20\right)={e}^{-\frac{1}{20}\cdot 20}=0.368$
2. Now let X be uniformly distributed< X~U(0,40). Now we have conditional probability: $P\left(X>30\mid X>10\right)=\frac{P\left(X>30\right)}{P\left(X>10\right)}=\frac{1-P\left(X\le 30\right)}{-P\left(X\le 10\right)}=\frac{1-\frac{30}{40}}{1-\frac{10}{40}}=\frac{1}{3}$
Result: For exponential distribution we have 0.368, and for uniform 1/3.

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