Chance that sample contains more individuals from one group than from another

Question

Question:

18 men and 12 women have agreed to participate in a randomized controlled experiment. A simple random sample of 15 of these 30 people is chosen as the treatment group; the others form the control group.

a) Find a decimal value for the chance that the treatment group contains more women than men.

b) Find the chance that all the women are in the same group.

For a) I chose the hypergeometric distribution as it is sampling without replacement. I set it up in the following way:

$\frac{\binom{12}{8} \cdot \binom{15}{7}}{\binom{30}{15}} + \frac{\binom{12}{9} \cdot \binom{15}{6}}{\binom{30}{15}} + ... + \frac{\binom{12}{12} \cdot \binom{15}{3}}{\binom{30}{15}}$

Is this a correct approach to the problem?

for b) I am confused why it can't be $\frac{\binom{12}{12} \cdot \binom{15}{3}}{\binom{30}{15}}$. I think the $\frac{1}{2}$ should also be taken into account for choosing between either the Treatment Group or Control group, but I don't know how to use this information.

Answer 1

Your approach for a) is correct.

As for part b),you must multiply by 2 instead of $\frac{1}{2}$. Another way to calculate this is $2\cdot\frac{\binom{18}{15}}{\binom{30}{15}}$, i.e. by looking at the 15 men in the other group.