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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
$1-\frac{1}{210}-\frac{24}{210}=\frac{185}{210}=\frac{37}{42}$.