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 ?
 A: 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, $\dfrac{\binom{45}1\binom{15}1}{\binom{60}2} = \dfrac{45}{118}$ 
A: Unloading one by one there are two mutually disjoint events that lead to the scenario: VS and SV. That gives a total probability of:$$\frac{15}{60}\frac{45}{59}+\frac{45}{60}\frac{15}{59}=\frac{45}{118}$$
There is no essential difference between unloading one by one and unloading both at the same moment. 

Something to think about in order to train your intuition on this matter:
If e.g. there is exactly one red ball and exactly one blue ball and two balls are randomly picked at the same time then it is for certain that a red ball and a blue ball are picked. This is also the case if you pick the balls one by one.
The answer is $\frac12\frac11+\frac12\frac11=1$ and not $\frac12\frac11=\frac12$.
A: There are $_{15}C_2$ ways to pick two going to Vancouver, and $_{45}C_2$ ways to pick two going to Seattle.  What's left is one of each.  So the probability is
$$P = \frac{_{60}C_2 - _{45}C_2 - _{15}C_2}{_{60}C_2} = \frac{1770 - 990 - 105}{1770}=\frac{675}{1770} = \frac{45}{118}.$$
