A committee of 3 is to be chosen from 15, 9 Gophers and 6 Badgers, at random. Bucky is one of the Badgers. What is the probability that Bucky and at least one Gopher are selected?

Answer 1

While normally when at least one appears, you would want to avoid adding all the at least one ways (1, 2, 3, ...), since we only have three committee members, that is the most straightforward way in this case.

The total number of committees is C(15, 3) = 15*14*13/3! = 455

Committees with Bucky, one gopher, and one other badger is 1 * 9 * (6-1) = 45 - the - 1 is because you don't count Bucky in the other badger.

Committees with Bucky and 2 gophers = 1 * C(9, 2) = 1 * 36 = 36

Committees with Bucky and at least 1 gopher = 36 + 45 = 81

P(Bucky and at least one gopher) = 81/455 = .1780 (rounded to four digits)

Thinking about this conditionally brings another way to solve this problem.

P(Bucky and at least one gopher) = P(Bucky)P(at least one gopher given Bucky)

P(Bucky), as there are 3 selections from 15 people, is 3/15 = 1/5

As there are 14 non-Buckys, with 9 gophers, P(picking at least one gopher in selection of 2 remaining) = 1 - C(5,2)/C(14,2) = 1 - 10/91 = 81/91

1/5 * 81/91 = 81/455

If you wanted to check the "reasonableness" of this answer, P(Bucky) = 3/15 = 1/5, and P(at least one gopher) = 1 - C(6,3)/C(15,3) = 1 - 20/455 = 87/91. As having Bucky, a badger, increases the likelihood of having only badgers, you would expect your answer to be a little less than 1/5 * 87/91 = 87/455, and it is.