Originally Posted by

**Soroban** Hello, alexandrabel90!

It took me several readings to understand the problem,

. . but I *think* I've got it now.

With no gender bias, we have these possible votes:

. . $\displaystyle \begin{array}{c|c||ccc}\text{For} & \text{Against} & \text{Ways} & & \\ \hline \\[-3mm]

\text{5W, 0M} & \text{0W, 3M} & {5\choose5}{3\choose0} &=& 1 \\ \\[-3mm]

\text{4W, 1M} & \text{1W, 2M} & {5\choose4}{3\choose0} &=& 15 \\ \\[-3mm]

\text{3W, 2M} & \text{2W, 1M} & {5\choose3}{3\choose2} &=& 30 \\ \\[-3mm]

\text{2W, 3M} & \text{3W, 0M} & {5\choose2}{3\choose3} &=& 10 \\ \\[-3mm] \hline \end{array}$

. . . . . . . . . . . . . . . $\displaystyle \text{Total: }\quad\;\;\; 56$

There are 56 ways that the vote could be split, all equally likely.

There is one way that is split along sex lines.

Therefore: .$\displaystyle P(\text{split along sex lines}) \;=\;\frac{1}{56}$