Phil sent us this solution:

There are ¹⁰C₄=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 ⁸C₂=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 ⁸C₄=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 ⁶C₄=15 teams with 0 women, and 4×⁶C₃=80 with 1 woman, so the required probability is


1-

15 210

-

80 210

=

115 210

=

23 42


.
(iv) Similar to the last one. There is only 1 team containing 0 men. There are 4×⁶C₁=24 teams containing 1 man, so the probability we want is


1-

1 210

-

24 210

=

185 210

=

37 42


.