<s>[INST] From a group of three biologists, two physicists and one mathematician, a committee of two people is to be randomly selected. From a group of three biologists, two physicists and one mathematician, a committee of two people is to be randomly selected. Denote by $X$ the random variable representing the number of biologists and by $Y$ the random variable representing the number of physicists on the committee. Calculate: $f_{X, Y}$, $f_{X}$, $f_{Y}$.
Attempt
$$f_{X}=\frac{\binom{3}{1} \binom{2}{1}}{\binom{5}{2}}+\frac{\binom{3}{1} \cdot 1}{\binom{5}{2}}$$
$$f_{Y}=\frac{\binom{2}{1} \binom{3}{1}}{\binom{5}{2}}+\frac{\binom{2}{1} \cdot 1}{\binom{5}{2}}$$
Then, $$f_{X, Y}=f_{X}+f_{Y}$$
Am I understanding the problem?
 [/INST] No that is not correct. You are supposed to find probability mass function $ \small f_X(x), f_Y(y), f_{XY} (x, y)$. Below is a tabular representation of what you need to come up with,
$ \small \begin{array}{|c|c|c|}
  \hline
   & Y=0 & Y=1 & Y=2 & f_X\\
 \hline
  X=0 &  &  \\
  \hline
  X=1 &  &  \\
  \hline
X=2 &  &  \\
  \hline 
f_Y &  & & & - \\
  \hline 
\end{array}$
We have $3$ biologists, $2$ physicists and $1$ mathematician. We can first find pmf $ \small f_{XY}(x, y)$. For example,
$\small f_{XY}(X = 0, Y = 0) = 0 ~ $ as a two member committee cannot be formed with just mathematician.
$ \displaystyle \small f_{XY}(X = 1, Y = 0) = {3 \choose 1} {1 \choose 1} / {6 \choose 2} ~ $ as we choose one of the biologists and the other member is mathematician.
$ \displaystyle \small f_{XY}(X = 2, Y = 0) = {3 \choose 2}  / {6 \choose 2}$
Similarly find
$\small f_{XY}(X = 0, Y = 1), f_{XY}(X = 1, Y = 1), f_{XY}(X = 0, Y = 2)$
Then,
$ \small f_Y(Y = 0) = f_{XY}(X = 0, Y = 0) + f_{XY}(X = 1, Y = 0) + f_{XY}(X = 2, Y = 0)$
You can similarly find $ \small f_Y(Y = 1), f_Y(Y = 2), f_X(X = 0), f_X(X = 1)$ and $ \small f_X(X = 2)$
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