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.


1 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.

  • $\begingroup$ does the 2 only come from the fact that it could be in one group OR the other? $\endgroup$
    – Jayant.M
    Commented Feb 7, 2017 at 6:24
  • $\begingroup$ Right, this is the reason! $\endgroup$
    – sdd
    Commented Feb 7, 2017 at 7:19

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