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.

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.