Hello we had a group divided into 4 teams of 7 each and there is a girl on each team. One of the girls thinks there was foul play in choosing the teams so I thought I'd calculate the chance of a girl on each team. This proved to be harder than I expected so I'm asking for help.
I think that since there are 28 total and 7 girls that we would have $\binom{28}{7}\binom{21}{7}\binom{14}{7}$ ways to assign the people to labelled teams.
I think now I would need to count the number of ways girls could be on teams and then divide?
I am unsure if I am doing anything correctly at this point, been years since I took math.
Best Answer
Label the teams as $1,2,3,4$.
The total number of ways to form the $4$ teams is $$ n = \binom{28}{7} \binom{21}{7} \binom{14}{7} \binom{7}{7} $$ Let $v$ be the list of counts, sorted in ascending order, for the number of girls on each of the $4$ teams.
Consider $3$ cases . . .
Case $(1)$:$\;v=[1,1,1,4]$.
For case $(1)$, the number of ways to form the $4$ teams is $$ x_1 = \left(\binom{4}{1}\binom{3}{3}\right) {\cdot} \left(\binom{7}{4}\binom{21}{3}\right) {\cdot} \left(\binom{3}{1}\binom{18}{6}\right) {\cdot} \left(\binom{2}{1}\binom{12}{6}\right) {\cdot} \left(\binom{1}{1}\binom{6}{6}\right) $$ hence the probability for case $(1)$ is \begin{align*} p_1&=\frac{x_1}{n}\\[4pt] &={\Large{\frac { \left(\binom{4}{1}\binom{3}{3}\right) {\cdot} \left(\binom{7}{4}\binom{21}{3}\right) {\cdot} \left(\binom{3}{1}\binom{18}{6}\right) {\cdot} \left(\binom{2}{1}\binom{12}{6}\right) {\cdot} \left(\binom{1}{1}\binom{6}{6}\right) } { \binom{28}{7} \binom{21}{7} \binom{14}{7} \binom{7}{7} }}} \\[4pt] &=\frac{2401}{59202}\approx .04055606230\\[4pt] \end{align*}
Case $(2)$:$\;v=[1,1,2,3]$.
For case $(2)$, the number of ways to form the $4$ teams is $$ x_2 = \left(\binom{4}{1}\binom{3}{1}\binom{2}{2}\right) {\cdot} \left(\binom{7}{3}\binom{21}{4}\right) {\cdot} \left(\binom{4}{2}\binom{17}{5}\right) {\cdot} \left(\binom{2}{1}\binom{12}{6}\right) {\cdot} \left(\binom{1}{1}\binom{6}{6}\right) $$ hence the probability for case $(2)$ is \begin{align*} p_2&=\frac{x_2}{n}\\[4pt] &={\Large{\frac { \left(\binom{4}{1}\binom{3}{1}\binom{2}{2}\right) {\cdot} \left(\binom{7}{3}\binom{21}{4}\right) {\cdot} \left(\binom{4}{2}\binom{17}{5}\right) {\cdot} \left(\binom{2}{1}\binom{12}{6}\right) {\cdot} \left(\binom{1}{1}\binom{6}{6}\right) } { \binom{28}{7} \binom{21}{7} \binom{14}{7} \binom{7}{7} }}} \\[4pt] &=\frac{2401}{6578}\approx .3650045607\\[4pt] \end{align*}
Case $(3)$:$\;v=[1,2,2,2]$.
For case $(3)$, the number of ways to form the $4$ teams is $$ x_3 = \left(\binom{4}{3}\binom{1}{1}\right) {\cdot} \left(\binom{7}{2}\binom{21}{5}\right) {\cdot} \left(\binom{5}{2}\binom{16}{5}\right) {\cdot} \left(\binom{3}{2}\binom{11}{5}\right) {\cdot} \left(\binom{1}{1}\binom{6}{6}\right) $$ hence the probability for case $(3)$ is \begin{align*} p_3&=\frac{x_3}{n}\\[4pt] &={\Large{\frac { \left(\binom{4}{3}\binom{1}{1}\right) {\cdot} \left(\binom{7}{2}\binom{21}{5}\right) {\cdot} \left(\binom{5}{2}\binom{16}{5}\right) {\cdot} \left(\binom{3}{2}\binom{11}{5}\right) {\cdot} \left(\binom{1}{1}\binom{6}{6}\right) } { \binom{28}{7} \binom{21}{7} \binom{14}{7} \binom{7}{7} }}} \\[4pt] &=\frac{7203}{32890}\approx .2190027364\\[4pt] \end{align*} Hence the probability that each team has at least one girl is $$ p = p_1+p_2+p_3 = \frac{2401}{59202} + \frac{2401}{6578} + \frac{7203}{32890} = \frac{16807}{26910}\approx .6245633593 $$