probability of a board containing a certain number of accountants A corporate board contains twelve members. The board decides to create a five-person Committee
to Hide Corporation Debt. Suppose four members of the board are accountants. What is the probability that the Committee will contain two accountants and three nonaccountants?

cant figure this one out
There are $\binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.

But let's hold our noses and go on.  Call a choice good if it consists of $2$ accountants and $3$ others. There are $\binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $\binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?
Using the Hypergeometric distribution formular,
$$p(x) = {{k \choose x} {N-k \choose n-x} \over {N \choose n} }$$

$N=12$  , $n=5$ ,
$k=4$  ,
$x=2$  ,
$n-x=3$ (non-accountants) , $p(x) = 14/33$