A teacher wants to randomly form two teams of 5 students from a group of 5 girls and 5 boys for a sports activity.

A teacher wants to randomly form two teams of 5 students from a group of 5 girls and 5 boys for a sports activity. Two of the girls, Ann and Alice, are selected as team leaders. Find the probability that one team has exactly 3 girls.

If i do this way $$\dfrac{{3 \choose 2}{5\choose 2}{1\choose 1}{3\choose 3} \times 2!}{{8 \choose 4}{4 \choose 4}} = \dfrac{6}{7}$$

Which turns out to tally with the answer. But i got one big question, is my way correct. Do i really need to times $2$ here? Please see my link below to my previous question cause the previous question make me doubt if i should times $2$.

Previous Question

• @ilovewet My result matches with your one, youshould explain how you derive your expression.
– user
Mar 18 '18 at 13:49

The team of e.g. Ann goes along with a selection of $4$ persons out $5$ boys and $3$ girls.

So the probability that her team will have exactly $3$ girls is: $$\frac{\binom32\binom52}{\binom84}$$

The same is true for the team of Alice, and the two events are mutually exclusive.

Then the probability that one of the teams has exactly $3$ girls is:$$\mathsf P(\text{team of Ann has }3\text{ girls})+\mathsf P(\text{team of Alice has }3\text{ girls})=2\cdot\frac{\binom32\binom52}{\binom84}=\frac67$$

Note that the the only case for which we don't have a team with $3$ girls are

• all $3$ girls with Ann with one boys among $5$ thus $5$ ways
• all $3$ girls with Alice with one boys among $5$ thus $5$ ways

then we have $10$ ways for which we don't have a team with $3$ girls.

The total numner of team is $\binom{8}{4}$ indeed once I've selected 4 people for Ann the others go with Alice.

Thus

$$p=1-\frac{10}{\binom{8}{4}}=\frac67$$

we have 2 groups headed by $G_1 and G_2$

How many ways we can arrange groups such that $G_1$ always get 2 girls

This can be done by choosing 2 girls out of three and choosing 2 boys out of 5 = $\binom{3}{2} \binom{5}{2}$

Now in your question this restriction is not there i.e. $G_2$ also has the possibility of getting 2 girls. Therefore we multiply by 2