probability - triplets

agarwalva wrote:Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon, and medals are awarded for first, second, and third place, what is the probability that at least two of the triplets will win a medal?

(A) 3/14
(B) 19/84
(C) 11/42
(D) 15/28
(E) 3/4

OA B

Please give your explanations

P(good outcome) = 1 - P(bad outcome).

Bad outcome #1: Exactly 1 triplet wins.
P(triplet wins 1st place) = 3/9. (Of the 9 competitors, 3 are triplets)
P(non-triplet wins 2nd place) = 6/8. (Of the 8 remaining competitors, 6 are non-triplets)
P(non-triplet wins 3rd place) = 5/7. (Of the 7 remaining competitors, 5 are non-triplets).
Since we want all of these events to happen together, we multiply:
3/9 * 6/8 * 5/7 = 5/28.

Since the triplet who wins could be in 1st, 2nd or 3rd place, we multiply by 3:
3 * 5/28 = 15/28.

Bad outcome #2: None of the triplets wins.
P(non-triplet wins 1st place) = 6/9. (Of the 9 competitors, 6 are non-triplets)
P(non-triplet wins 2nd place) = 5/8. (Of the 8 remaining competitors, 5 are non-triplets)
P(non-triplet wins 3rd place) = 4/7. (Of the 7 remaining competitors, 4 are non-triplets).
Since we want all of these events to happen together, we multiply:
6/9 * 5/8 * 4/7 = 5/21.

P(at least 1 triplet wins) = 1 - 15/28 - 5/21 = 1 - 45/84 - 20/84 = 19/84.

The correct answer is B.