Phil sent us this solution:

There are ${}^{10}{C}_4=210$ possible chess teams.

(i) 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 ${}^8{C}_2=28$. So the probability is 28/210=2/15.
(ii) 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 ${}^8{C}_4=70$. So the probability is 70/210=1/3.
(iii) 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 ${}^6{C}_4=15$ teams with 0 women, and $4{\times }^6{C}_3=80$ with 1 woman, so the required probability is $1-\frac{15}{210}-\frac{80}{210}=\frac{115}{210}=\frac{23}{42}$.
(iv) Similar to the last one. There is only 1 team containing 0 men. There are $4{\times }^6{C}_1=24$ teams containing 1 man, so the probability we want is $1-\frac{1}{210}-\frac{24}{210}=\frac{185}{210}=\frac{37}{42}$.