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
Source: Manhattan Prep
Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon, and medals are awarded
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P(at least two triplets win a metal) = P(exactly two triplets win a middle OR exactly three triplets win a medal)BTGmoderatorDC wrote: ↑Wed Oct 20, 2021 6:19 pmTriplets 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
Source: Manhattan Prep
= P(exactly two win a middle) + P(exactly three win a medal)
Let's examine each probability separately....
P(exactly two triplets win a middle)
There are three ways for this to happen:
case i) a triplet places 1st, another triplet places 2nd, a nontriplet places 3rd
case ii) a triplet places 1st, a nontriplet places 2nd, a triplet places 3rd
case iii) a nontriplet places 1st, a triplet places 2nd, a nontriplet places 3rd
P(case i) = (3/9)(2/8)(6/7) = 1/14
P(case ii) = (3/9)(6/8)(2/7) = 1/14
P(case iii) = (6/9)(3/8)(2/7) = 1/14
P(exactly two triplets win a middle) = 1/14 + 1/14 + 1/14 = 3/14
P(exactly three triplets win a middle)
P(exactly three triplets win a middle) = P(a triplet places 1st, another triplet places 2nd, another triplet places 3rd)
= (3/9)(2/8)(1/7) = 1/84
We get:
P(at least two triplets win a metal) = P(exactly two win a middle) + P(exactly three win a medal)
= 3/14 + 1/84
= 18/84 + 1/84
= 19/84
Answer: B
Cheers,
Brent