My school's Future Mathematicians of America club has 16 members, 7 boys and 9 girls. A president and a 3-person executive committee are chosen (where the president cannot serve on the committee). What is the probability that the president is the same gender as the majority of the committee?
I ended up getting:$${{7\left(9\binom{6}{2} + \binom{6}{3}\right) + 9\left(7\binom{8}{2} + \binom{8}{3}\right)}\over{\binom{16}{4}}} = {{479}\over{260}},$$which I know can't be right since that's greater than $1$. What did I do wrong?
EDIT: Okay, so with the helpful comment by JMoravitz in the comments section, maybe it should be the following instead?
$${{7\left(9\binom{6}{2} + \binom{6}{3}\right) + 9\left(7\binom{8}{2} + \binom{8}{3}\right)}\over{16 \binom{15}{3}}} = {{479}\over{1040}}$$
Best Answer
Here's a different approach that yields the same result. First choose the four boys and girls, and then choose a president from the majority: $$\frac{\binom{7}{0}\binom{9}{4}4+\binom{7}{1}\binom{9}{3}3+\binom{7}{3}\binom{9}{1}3+\binom{7}{4}\binom{9}{0}4}{\binom{16}{4}4}=\frac{479}{1040}$$
Note the absence of $\binom{7}{2}\binom{9}{2}$ in the numerator.