A probability question regarding combinatorics From a class of $300$ students, three are selected at random to receive three identical prizes. Of the students, $200$ are from department $A$, $60$ from department $B$ and $40$ from department $C$.


*

*Find the probability that the three winners come from different departments.

*Find the probability that all three winners come from the same department.

*Suppose that all three are from the same department. Compute the probability that they are all from department $A$.
Here is my attempt. I am not sure if this is correct.


*

*$P(\text{different departments})=3!\binom{200}{1}\binom{60}{1}\binom{40}{1}$ or $3!\frac{200}{300}\times \frac{60}{299} \times\frac{40}{298}$?

*$P(\text{same department})=\dfrac{\binom{200}{1}+\binom{60}{1}+\binom{40}{1}}{\binom{300}{3}}= \dfrac{3}{44551}$.

*$\begin{align} P(\text{from department A}|\text{same department}) & = \dfrac{P(\text{from department A}) P(\text{same department})}{P(\text{same department})} \\
& = \dfrac{\binom{200}{1} \frac{3}{4451}}{\frac{3}{4451}} \\
&=\binom{200}{1}. \end{align}$
Thank you in advance.
 A: Note: I am interpreting the problem to mean that the three winners must be different people.
Unfortunately all three proposed solutions are incorrect.
There are ${300 \choose 3}$ subsets of size 3.  This will be the denominator for the first two questions.  In the first question, the numerator is ${200\choose 1}{60\choose 1}{40 \choose 1}$.  We want one of our three winners to be from department $A$, one from $B$, and one from $C$; these three choices are made independently, so we use the multiplication principle.
The numerator of the second question is ${200\choose 3}+{60\choose 3}+{40\choose 3}$.  There are three types of subsets, those entirely from $A$, those entirely from $B$, and those entirely from $C$.  These are disjoint subsets of ${300\choose 3}$, so we use the addition principle.  To count the subsets entirely from $A$, we need to pick 3 winners.
The denominator in the third question is the same as the numerator of the second question, since now we are restricting to only those triplets that are from a single department, which the numerator from the second question was counting.  The numerator of the third question is ${200\choose 3}$.
