Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co

Question

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

Manhattan Advanced Gmat Quant Work out set1 4th question

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

Manhattan Advanced Gmat Quant Work out set1 4th question

Bunuel · Accepted Answer

Welcome to GMAT Club. Below is a solution to the question.

The probability that  at least two  of the triplets will win a medal is the sum of the probability that  exactly  two of the triplets will win a medal and the probability that  all three  will win a medal.

The probability that  exactly  two of the triplets will win a medal is  \frac{C^2_3*C^1_6}{C^3_9}=\frac{18}{84} , where  C^2_3  is ways to select which two of the triplets will win a medal,  C^1_6  is ways to select third medal winner out of the remaining 6 competitors and  C^3_9  is total ways to select 3 winners out of 9;

The probability that  all three  will win a medal is  \frac{C^3_3}{C^3_9}=\frac{1}{84} ;

P=\frac{18}{84}+\frac{1}{84}=\frac{19}{84} .

Answer: B.

Hope it's clear.

P.S. Please post answer choices for PS problems.

Asifpirlo · Answer

My explanation with solution:(Diagram)

himanshuhpr · Answer

From what I understand in the question is that there are 3 teams (triplets) of 3 people each competing together , so how is it possible that only 2 triplets/ teams win medals as there would be a definite 1,2,3 position in case of 3 teams. ?? ---. because the word triathlon implies to me that the teams win not the individuals (teams with cumulative individual lowest score times win) ....

Now if we consider this otherwise that individuals win then :

The probability that at least two of the triplets will win a medal is the sum of the probability that exactly two of the individuals from same team/triplet wins the medal and third medal is won by a individual from one of the remaining 2 teams/triplets + 1 Individual from each team/triplet wins the medal...

P(at least two of the triplets will win a medal is the sum of the probability that exactly two of the individuals from same team/triplet wins the medal and third medal is won by a individual from one of the remaining 2 teams/triplets)=3P2 * 3/9 * 2/8 * 6/7 = 3/7

P(1 Individual from each team/triplet wins the medal)= 3P3 * 3/9 * 3/8 * 3/7 = 9/28

3/7 + 9/28 = 3/4 ---- (E)

Pl. help me know where I am going wrong.
Thanks.

Bunuel · Answer

Your understanding of the question is not correct. Adam, Bruce, and Charlie are not in one team, they compete between each other and 6 other competitors.

wfa207 · Answer

It's been quite some time since this was posted, but I have a quick question about how to solve this problem using fractional probability instead of combinatorics. I tried to solve the problem using the following:

P(\geq{	ext{2 Triplets}})=P(	ext{2 Triplets})+P(	ext{3 Triplets})

P(\geq{ 	ext{2 Triplets}})=(\frac{3}{9} 	imes \frac{2}{8} 	imes \frac{6}{7})+(\frac{3}{9} 	imes \frac{2}{8} 	imes \frac{1}{7})

P(\geq{ 	ext{2 Triplets}})=\frac{6}{84}+\frac{1}{84}=\frac{7}{84}

I realize that in order to get to the original answer, I would have had to multiply the initial  \frac{6}{84}  by 3, and proceed to calculate  \frac{19}{84} , but I do not understand why this is the case, since the order of the triplets should not matter.

In another problem from MGMAT's CAT test, I was able to solve for the desired probability using fractional probabilities only:

"Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?"

P(\geq 	ext{1 pair})=1- 	ext{ P}( 	ext{No Pairs})

P(\geq 	ext{1 pair})=1-(\frac{12}{12} 	imes \frac{10}{11} 	imes \frac{8}{10} 	imes \frac{6}{9})=1-\frac{16}{33}=\frac{17}{33}

Of course, this problem could have also been solved using combinatorics, but I feel the fractional approach would probably be quicker in a live scenario. However, over the course of my review, I've found the fractional method to be unreliable, whereas combinatorics are typically universally applicable (though they can take significantly longer in some instances).

I am curious as to why the solution for the second problem involving the playing cards worked whereas my solution to the original triplet question did not. I am not really sure why I have to multiply that initial  \frac{6}{84}  by 3 to get the right answer, and I would like to have a concrete understanding of this issue to help me solve similar problems.

DDB1981 · Answer

Hey wfa,

You're right about order not mattering but your math suggests you're missing a fundamental counting concept.

P(2 triplets) is not simply equal to 3/9*2/8*6/7 as this would suggest a sequence where order matters. Rather, the math is (number of successful outcomes)/(total number of outcomes)

Total # of ways you could pick 2 of the triplets among the medallists is C(2,3), AND total # ways you could pick 1 non-triplet among the medallists is C(1,6), that's 3*6= 18

Total # outcomes is the # of ways you could pick ANYONE to be medallists, so C(6,9) = 84

P(2 triplets) = 18/84

I know u we're off by a factor of 3, and you think it's an order vs no-order issue, but really, you were attempting sequential counting, but should have been using combinatorials

Hope this clarifies things

gmatter0913 · Answer

Are Adam, Bruce and Charlie in the same triathlon team or are they in separate teams?

Bunuel · Answer

No, they are competing each other, they do not form a team.

ScottTargetTestPrep · Answer

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

Kritisood · Answer

Hi  Bunuel  isn't the probability that at least two of the triplets will win a medal equal to  none  of the triplets win the medal? But the answer coming through ths way isnt correct.. 1- 5/21 = 16/21

Bunuel · Answer

At least two of the triplets will win a medal, means that only two or all three of the triplets win a medal. The opposite event would be that only one or none of the triplets win a medal.

BrentGMATPrepNow · Answer

Since the other solutions in this thread use counting techniques, let's use  probability rules  to solve the question

P(at least two triplets win a medal) = P(exactly two triplets win a medal   OR   exactly three triplets win a medal)
= P(exactly two win a medal)   +   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 non-triplet places 3rd
case ii) a triplet places 1st, a non-triplet places 2nd, a triplet places 3rd
case iii) a non-triplet places 1st, a triplet places 2nd, a non-triplet 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 medal) = 1/14 + 1/14 + 1/14 = 3/14

P(exactly three triplets win a medal  ) 
P(exactly three triplets win a medal) = 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 medal) = P(exactly two win a medal)   +   P(exactly three win a medal)
= 3/14   +   1/84
= 18/84   +   1/84
= 19/84

Answer: B

Cheers,
Brent

GMATGuruNY · Answer

Alternate approach:

Case 1: Exactly two triplets win medals 
P(a triplet wins the first medal)  = \frac{3}{9}  (Of the 9 competitors, 3 are triplets)
P(a triplet wins the second medal)  = \frac{2}{8}  (Of the 8 remaining competitors, 2 are triplets)
P(a non-triplet wins the third medal)  = \frac{6}{7}  (Of the 7 remaining competitors, 6 are non-triplets)
To combine these probabilities, we multiply:
 \frac{3}{9} * \frac{2}{8} * \frac{6}{7} 
Since the non-triplet can be first, second, or third, we multiply by 3:
 \frac{3}{9} * \frac{2}{8} * \frac{6}{7} * 3 = \frac{3}{14}

Case 2: All 3 triplets win a medal 
P(a triplet wins the first medal)  = \frac{3}{9}  (Of the 9 competitors, 3 are triplets)
P(a triplet wins the second medal)  = \frac{2}{8}  (Of the 8 remaining competitors, 2 are triplets)
P(a triplet wins the third medal)  = \frac{1}{7}  (Of the 7 remaining competitors, 1 is a triplet)
To combine these probabilities, we multiply:
 \frac{3}{9 }* \frac{2}{8} *\frac{ 1}{7} = \frac{1}{84}

Since a favorable outcome will be yielded by Case 1 or Case 2, we ADD the fractions:
 \frac{3}{14} + \frac{1}{84} = \frac{19}{84}

B

skmaqeel51 · Answer

Why arent we considering the ordering of the medals won, should'nt it be permutation since they could have won first or second or thirds, order matters right?

Bunuel · Answer

We don’t care which medals they win, as long as they win any. Also, if we account for the order, we must do so in both the numerator and the denominator by multiplying both by 3!, which cancels each other out.

rak08 · Answer

Bunuel 
Where did i go wrong when i do this

3c1*2c1*1 / 9c1*8c1*7c1 + 3c1*2c1*6c1 / 9c1 * 8c1* 7c1 (3!)

why are we dividing with 2!? why are we considering the 2 of the triplets same?

Krunaal · Answer

Calculating it this way (selecting one by one), you are counting the order of arrangements (since you multiply your selections); whereas we only want to know ways that at least two of the three won medals (doesn't matter which medal) Thus, in one case we select 2 from the 3 in one go, using 3C2 and the remaining 1 using 6C1. In another case, 3C3 = 1. We add both and divide by 9C3 to get the probability. Very clearly explained here - https://gmatclub.com/forum/triplets-ada ... l#p1086111

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