Because the number of even sides are equal to odd sides, you can simplify the problem to "What is the probability that a coin lands on heads three times out of five?
For which the probability is $5$ pick $3$ out of $32$ or $$\frac{1}{32} \cdot \frac{5 \cdot 4 \cdot 3}{3\cdot 2 \cdot 1}=\frac{10}{32}= \frac{5}{16}.$$
First it might be helpful to remember that the binomial coefficients are given by
$$
{n \choose k} \;\; =\;\; \frac{n!}{k!(n-k)!}.
$$
Assuming that in your notation $P(13,7) = {13 \choose 7}$ then you're correct for the first part of the question. For the second part, you need to pick $4$ women and $3$ men. You have the right idea in what (I believe) you're writing, but it needs to be refined a bit.
Note that we need to pick $4$ out of seven women and there are ${7\choose 4}$ ways to do this. Similarly there are ${6\choose 3}$ ways to choose the men. Since the choice of women can be done independently of the choice of men (every potential group of women can be combined with any of the possible groups of men) we simply multiply these quantities together. The answer should be ${7\choose 4}{6\choose 3}$.
When we get to questions of there being "at most" or "at least" a certain number of men and/or women, we need to consider each case separately and then add all of the possible numbers together. To illustrate, consider the problem of computing the number of groups containing at least one man. We can make a list of all of the different possible numbers simply by considering each case:
- The number of groups with zero men is ${7 \choose 7}{6\choose 0} = 1$, since all $7$ women just get chosen.
- The number of groups with one man is ${7\choose 6}{6\choose 1} = 42$.
- The number of groups with two men is ${7\choose 5}{6\choose 2}$.
- etc.
To find the number of cases with at least one man, we add up the numbers of groups with one man, two men, three men, all the way to six men. Another way of computing this number too is to realize that the total number of possible groups is just ${13\choose 7}$. The only possible case that we are eliminating from our analysis is when all $7$ women get chosen, thus we can equivalently come to the conclusion that the number of groups with at least one man is ${13\choose 7} - {7\choose 7}{6\choose 0}$.
Best Answer
The sample space would look like (T, 1), (H, 1), (T, 2), ... where (T, 1) is the event that you flip tails and a 1 is rolled on the die. The members of the event would be the events in the list that have an even roll.