Your line of thought is a little wrong.
You are choosing $3$ people out of $5$. This means, the order of choice doen't matter. More specifically, choosing $1$ then $2$ then $3$ is the same as $2$ then $1$ then $3$ and so on....
Therefore, the total number of ways of selecting $3$ people out of $5$, is the binomial coefficient $\binom{5}{3}=^5C_3$
The answer to (a) is, since the first guy is always chosen, you have a handle only on the choice of the other two. So, the number of possible ways of selecting $2$ people out $4$ is $^4C_2$
So, the probability is $\dfrac{\binom{4}{2}}{\binom{5}{2}}=\frac{6}{10}=\frac{3}{5}$ which completes (a).
Along the same lines, (c) is also solved this way: You have handle only on one guy which means, you have to choose $1$ out of $3$ guys, So, the probability is $\dfrac{\binom 3 1}{\binom 5 3}=\frac{3}{10}$
For (b), the condition imposed forbids you from choosing the candidate ranked (1). So, You need to choose $3$ out of $4$ people. So, the answer is $\dfrac{\binom 4 3}{\binom 5 3}=\frac{4}{10}=\frac{2}{5}$
It doesn't matter how many trainees there are, so let's suppose there are $100$. Then $80$ are female, and $72$ of the $80$ went to college, so the probability the person selected is a female who attended college is ... ?
Best Answer
$\left( \begin{array}{c} 9 \\ 3 \end{array} \right) = \frac{9!}{6!\cdot3!}=\frac{9\cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} = 3\cdot 4 \cdot 7 = 84$
The probability that we get a male upon randomly selecting an employee: $\frac{5}{9}$. If we were to select another male, however, we'd have a $\frac{4}{8}$ chance. For a third male, it'd be $\frac{3}{7}$.
Multiply the three probabilities together.