# Teams

Two brothers belong to a club with 10 members. Four are selected
for a match. Find the probability that both brothers are selected.

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A team of $4$ members is selected to play a match. Suppose that each member has an equal chance of being selected what is the probability that both brothers are in the team and what is the probability that neither brother is in the team?

What is the probability that there are at least two women in the team and what is the probability that there are at least two men in the team?

How many ways can you choose a team of $4$ from $10$?

Phil sent us this solution:

There are $^{10}C_4=210$ possible chess teams.

- To find the number of teams containing both brothers, we need the number of ways of choosing the other $2$ team members from the remaining $8$ people, which is $^8C_2=28$. So the probability is $28/210=2/15$.
- To find the number of teams containing neither brother, we need the number of ways of choosing all $4$ team members from the remaining $8$ people, which is $^8C_4=70$. So the probability is $70/210=1/3$.
- To find the number of teams containing at least $2$ women, we can take $210$ - the number of teams containing $0$ women - the number of teams containing $1$ woman. There are $^6C_4=15$ teams with $0$ women, and $4 \times$ $^6C_3=80$ with $1$ woman, so the required probability is $1-\frac{15}{210}-\frac{80}{210}=\frac{ 115}{210}=\frac{23}{42}$.
- Similar to the last one. There is only $1$ team containing $0$ men. There are $4\times$ $^6C_1=24$ teams containing $1$ man, so the probability we want is $1-\frac{1}{210}-\frac{24}{210}=\frac{185}{210}=\frac{37}{42}$.