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Triplets Adam, Bruce, and Charlie enter a triathlon. If ther

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Triplets Adam, Bruce, and Charlie enter a triathlon. If ther

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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

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BTGmoderatorDC 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
One approach:
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.

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BTGmoderatorDC 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
Source: Manhattan Prep
$$\left\{ \matrix{
\,A\,,B\,,\,{\kern 1pt} C\,\,\,\,\, \to \,\,\,3\,\,{\rm{competitors}} \hfill \cr
\,{\rm{D,}}\,\, \ldots \,\,{\rm{,}}\,I\,\,\, \to \,\,\,{\rm{other}}\,\,6\,\,{\rm{competitors}}\, \hfill \cr} \right.$$
$$\left( {\rm{i}} \right)\,\,A,B,C\,\,{\rm{win}}\,\,{\rm{the}}\,\,3\,\,{\rm{medals}}$$
$$\left( {{\rm{ii}}} \right)\,\,A,B,C\,\,{\rm{win}}\,\,{\rm{exactly}}\,\,2\,\,{\rm{of}}\,\,{\rm{the}}\,\,3\,\,{\rm{medals}}$$
$$? = P\left( {{\rm{i}}\,\,{\rm{or}}\,\,{\rm{ii}}} \right) = P\left( {\rm{i}} \right) + P\left( {{\rm{ii}}} \right)\,\,\,\,\,\,\,\,\left[ {\,{\rm{i}}\,\,{\rm{and}}\,\,{\rm{ii}}\,\,{\rm{are}}\,\,{\rm{mutually}}\,\,{\rm{exclusive}}\,} \right]$$
$$\left( {\rm{i}} \right)\,\,\,\left\{ \matrix{
\,{\rm{favorable}} = \,\,3! \hfill \cr
\,{\rm{total}}\,\, = \,\,9 \cdot 8 \cdot 7\,\,{\rm{equiprobables}}\,\,\, \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,P\left( {\rm{i}} \right) = {1 \over {84}}$$
$$\left( {{\rm{ii}}} \right)\,\,\,\left\{ \matrix{
\,{\rm{favorable}} = \,\,\underbrace {C\left( {3,2} \right)}_{{\rm{say}}\,\,A,B}\,\,\, \cdot \,\,\underbrace {C\left( {3,2} \right)}_{{\rm{say}}\,\,\,1{\rm{st}}\,\,,\,\,2{\rm{nd}}}\,\,\, \cdot \,\,\underbrace {2!}_{{\rm{say}}\,\,\,A\,\,1{\rm{st}}\,\,,\,B\,\,2{\rm{nd}}}\,\,\, \cdot \,\,\,\underbrace {C\left( {6,1} \right)}_{{\rm{say}}\,\,D\,\,\left( {3{\rm{rd}}} \right)}\, \hfill \cr
\,{\rm{total}}\,\, = \,\,9 \cdot 8 \cdot 7\,\,{\rm{equiprobables}}\,\,\, \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,P\left( {{\rm{ii}}} \right) = {3 \over {14}}$$
$$? = {1 \over {84}} + {{3 \cdot 6} \over {14 \cdot 6}} = {{19} \over {84}}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or https://GMATH.com.br (Portuguese version)
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BTGmoderatorDC 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
The total number of ways to select 3 people from 9 is 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84.

The number of ways 2 of the triplets are among the 3 awarded is 3C2 x 6C1 = 3 x 6 = 18.

The number of ways all 3 triplets are the 3 awarded is 3C3 x 6C0 = 1 x 1 = 1.

Therefore, the probability is (18 + 1)/84 = 19/84.

Answer: B

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Scott Woodbury-Stewart Founder and CEO

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