# Math Help - easy question

1. ## easy question

the board of 8 mbrs, 5 women and 3men,received a complaint of a woman about sex discrimination and they voted 5-3 in favour of the plaintiff. the attorney representing the company appealed the board/s decision by claiming sex bias on the part of the board members. if there was no sex bias among them, it might be reasonable to say that any group of 5 mbrs would be as likely to vote for the complainant. if this were the case, what is the probability that the vote would split along sex lines?

thanks!

2. Hello, alexandrabel90!

It took me several readings to understand the problem,
. . but I think I've got it now.

The board of 8 members, 5 women and 3men,
and they voted 5-3 in favour of the plaintiff.

The attorney representing the company appealed the board's decision
by claiming sex bias on the part of the board members.

If there were no sex bias among them, it might be reasonable to say that
any group of 5 members would be as likely to vote for the complainant.

If this were the case, what is the probability that the vote would split along sex lines?
With no gender bias, we have these possible votes:

. . $\begin{array}{c|c||ccc}\text{For} & \text{Against} & \text{Ways} & & \\ \hline \\[-3mm]
\text{5W, 0M} & \text{0W, 3M} & {5\choose5}{3\choose0} &=& 1 \\ \\[-3mm]
\text{4W, 1M} & \text{1W, 2M} & {5\choose4}{3\choose0} &=& 15 \\ \\[-3mm]
\text{3W, 2M} & \text{2W, 1M} & {5\choose3}{3\choose2} &=& 30 \\ \\[-3mm]
\text{2W, 3M} & \text{3W, 0M} & {5\choose2}{3\choose3} &=& 10 \\ \\[-3mm] \hline \end{array}$

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

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

There is one way that is split along sex lines.

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

3. Originally Posted by alexandrabel90
the board of 8 mbrs, 5 women and 3men,received a complaint of a woman about sex discrimination and they voted 5-3 in favour of the plaintiff. the attorney representing the company appealed the board/s decision by claiming sex bias on the part of the board members. if there was no sex bias among them, it might be reasonable to say that any group of 5 mbrs would be as likely to vote for the complainant. if this were the case, what is the probability that the vote would split along sex lines?

thanks!
If there was no sex discrimination of any kind,
then strictly speaking, as there is no bias, the probability is zero.

However, it may "appear" that there is bias if....

4 or 5 of the 5 women side with the plaintiff,
or if 2 or 3 of the men side with the defendant.

The probability of 4 or 5 women siding with the plaintiff is

$\frac{\binom{5}{4}+\binom{5}{5}}{\binom{5}{0}+\bin om{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\b inom{5}{5}}$

The probability of 2 or 3 men voting for the defendant is

$\frac{\binom{3}{2}+\binom{3}{3}}{\binom{3}{0}+\bin om{3}{1}+\binom{3}{2}+\binom{3}{3}}$

4. My interpretation of the question is different.
Soroban's reply may be the result.

5. 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:

. . $\begin{array}{c|c||ccc}\text{For} & \text{Against} & \text{Ways} & & \\ \hline \\[-3mm]
\text{5W, 0M} & \text{0W, 3M} & {5\choose5}{3\choose0} &=& 1 \\ \\[-3mm]
\text{4W, 1M} & \text{1W, 2M} & {5\choose4}{3\choose0} &=& 15 \\ \\[-3mm]
\text{3W, 2M} & \text{2W, 1M} & {5\choose3}{3\choose2} &=& 30 \\ \\[-3mm]
\text{2W, 3M} & \text{3W, 0M} & {5\choose2}{3\choose3} &=& 10 \\ \\[-3mm] \hline \end{array}$

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

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

There is one way that is split along sex lines.

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

why is there one way that it is split along sex lines?