Simple Conditional Probability Among 60 automobile repair parts loaded on a truck in San Fransico , 45 are destined for seattle and 15 for vancouver. If two of the parts are unloaded in portland by mistake and the selection is random , what are the probabilities that one should have gone to seattle and one to Vancouver. My attempt : So I first took out the probability of one part going to seattle and the other one going to vancouver which for me turned out to be $\frac{45}{60} * \frac{15}{59} $ But the correct answer that is listed is 1/243 What am I doing wrong , or is the listed answer wrong ?

The listed answer is wrong, and so is your attempt, where you have forgotten to multiply by $2$ to take into account that $2$ sequences are possible for the choices.

Alternatively, $\dfrac{\binom{45}1\binom{15}1}{\binom{60}2} = \dfrac{45}{118}$