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.