Triplets Adam, Bruce, and Charlie enter a triathlon. If : GMAT Problem Solving (PS)
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# Triplets Adam, Bruce, and Charlie enter a triathlon. If

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16 May 2012, 07:03
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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

Manhattan Advanced Gmat Quant Work out set1 4th question
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 May 2012, 07:17, edited 1 time in total.
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16 May 2012, 07:28
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cnon 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

Manhattan Advanced Gmat Quant Work out set1 4th question

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 $$\frac{C^3_3}{C^3_9}=\frac{1}{84}$$;

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

Hope it's clear.

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18 Aug 2013, 11:22
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cnon 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

Manhattan Advanced Gmat Quant Work out set1 4th question

My explanation with solution:(Diagram)
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triplets.png [ 35.25 KiB | Viewed 11307 times ]

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07 May 2013, 19:50
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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
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21 Aug 2013, 07:04
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domfrancondumas wrote:
Questions like this: probability and combination take some time to solve.. In most cases you have to read the question again... Is there a shortcut for solving them?
If not, how many minutes should i give to the problem before quitting(or selecting some answer randomly) in the test?

I wouldn't spend more than 3 minutes on a question. When close to that and still don't know the answer spend the next 5-15 seconds for an educated guess.

As for combinatorics and probability questions: GMAT combination/probability questions are fairly straightforward and as practice shows you will encounter at max 3 questions from both fields combined.

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html

Hope this helps.
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22 Aug 2013, 03:32
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Bunuel wrote:
domfrancondumas wrote:
Questions like this: probability and combination take some time to solve.. In most cases you have to read the question again... Is there a shortcut for solving them?
If not, how many minutes should i give to the problem before quitting(or selecting some answer randomly) in the test?

I wouldn't spend more than 3 minutes on a question. When close to that and still don't know the answer spend the next 5-15 seconds for an educated guess.

As for combinatorics and probability questions: GMAT combination/probability questions are fairly straightforward and as practice shows you will encounter at max 3 questions from both fields combined.

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html

Hope this helps.

Thanks a ton Bunuel.....
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15 Sep 2013, 22:20
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9! / 3!6! = 84. So our total possible number of combinations is 84

We have 3! / 3! = 1. That is, there is only one instance when all three brothers win medals.

First, for the three who win medals, we have 3! / 2! = 3. For the six who don't win medals, we have 6! / 5! = 6. We multiply these two numbers to get our total number: 3 × 6 = 18.

The brothers win at least two medals in 18 + 1 = 19 circumstances. Our total number of circumstances is 84, so our probability is 19 / 84.
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16 May 2012, 15:52
Thank you so much,i will take into account.
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25 Nov 2012, 14:42
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.
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26 Nov 2012, 00:55
himanshuhpr wrote:
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.

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.
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26 Nov 2012, 06:21
I might be combinatorics challenged...
I got that for 2 of the triplets to get award and one other competitor -
probability = 3C2 * 6C1 / 9C3 = 3!*6!*3!*6! / 2!*5!*9! = 3*6*3 / 9*8*7 = 3/28

mCn = m!/(n!*(m-n)!)

How is it that you got 18/84 ???
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30 Nov 2012, 12:55
nanishora wrote:
I might be combinatorics challenged...
I got that for 2 of the triplets to get award and one other competitor -
probability = 3C2 * 6C1 / 9C3 = 3!*6!*3!*6! / 2!*5!*9! = 3*6*3 / 9*8*7 = 3/28

mCn = m!/(n!*(m-n)!)

How is it that you got 18/84 ???

This calculation is wrong. it will be 3*6*3*2/9*8*7=3/14. Bunuel multiplied it by 6, which comes to 18/84
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30 Nov 2012, 21:49
Hi Bunuel

Isnt this a permutation problem? I did in the same way as you did except that I used permutation since I thought that the rank and the order in which they are acquired does matter here. Obviously I am wrong since its not the official answer.. What am I missing here??
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07 Feb 2013, 15:47
ratinarace wrote:
Hi Bunuel

Isnt this a permutation problem? I did in the same way as you did except that I used permutation since I thought that the rank and the order in which they are acquired does matter here. Obviously I am wrong since its not the official answer.. What am I missing here??

That i was saying to my self tool : the rank and the order does matter here , so we must use permutations.
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07 Feb 2013, 17:16
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Rock750 wrote:
ratinarace wrote:
Hi Bunuel

Isnt this a permutation problem? I did in the same way as you did except that I used permutation since I thought that the rank and the order in which they are acquired does matter here. Obviously I am wrong since its not the official answer.. What am I missing here??

That i was saying to my self tool : the rank and the order does matter here , so we must use permutations.

It is a combinatorics question and not a permutation because it asks of two getting a medal (placing either 1st 2nd or 3rd) but does not consider the order.

As a result there are 3 ways in which the podium ( podium is a term used for the top three finishers in a race) can have 2 of the triplets: A and B ; A and C ; B and C.
There is only one way in which all three can occupy the podium: A, B, and C. Their finishing position does not count, simply that they made the podium.
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07 Feb 2013, 17:37
hitman5532 wrote:

As a result there are 3 ways in which the podium ( podium is a term used for the top three finishers in a race) can have 2 of the triplets: A and B ; A and C ; B and C.

Thanks hitman5532

I think there are more than 3 ways in which the podium can have 2 of the triplets. for instance , there are $$3P2 = \frac{3!}{(3-2)!}$$

which is equals to 6 ways :

A and B
A and C
B and C
B and A
C and A
C and B

Please correct if i am wrong
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07 Feb 2013, 19:11
Rock750 wrote:
hitman5532 wrote:

As a result there are 3 ways in which the podium ( podium is a term used for the top three finishers in a race) can have 2 of the triplets: A and B ; A and C ; B and C.

Thanks hitman5532

I think there are more than 3 ways in which the podium can have 2 of the triplets. for instance , there are $$3P2 = \frac{3!}{(3-2)!}$$

which is equals to 6 ways :

A and B
A and C
B and C
B and A
C and A
C and B

Please correct if i am wrong

It is not a Permutation, it is a combinatoric problem. So the formula is $$\frac{n!}{k!(n-k)!}$$

Even without the formula, if you think about what your list shows, you have doubled some selections. A and B is the same as B and A

Think of it this way: Treat the podium as a cup. If you put one A in the cup and one B in the cup, it is no different than one B and one A.
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09 Feb 2013, 12:33
hitman5532 wrote:
Rock750 wrote:
hitman5532 wrote:

As a result there are 3 ways in which the podium ( podium is a term used for the top three finishers in a race) can have 2 of the triplets: A and B ; A and C ; B and C.

Thanks hitman5532

I think there are more than 3 ways in which the podium can have 2 of the triplets. for instance , there are $$3P2 = \frac{3!}{(3-2)!}$$

which is equals to 6 ways :

A and B
A and C
B and C
B and A
C and A
C and B

Please correct if i am wrong

It is not a Permutation, it is a combinatoric problem. So the formula is $$\frac{n!}{k!(n-k)!}$$

Even without the formula, if you think about what your list shows, you have doubled some selections. A and B is the same as B and A

Think of it this way: Treat the podium as a cup. If you put one A in the cup and one B in the cup, it is no different than one B and one A.

Thanks

I get the point : the order doesn't matter here since all we need here is having at least two of the triplets in the podium ...

Regards
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07 May 2013, 19:17
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:

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

$$P(\geq{ \text{2 Triplets}})=(\frac{3}{9} \times \frac{2}{8} \times \frac{6}{7})+(\frac{3}{9} \times \frac{2}{8} \times \frac{1}{7})$$

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

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?"

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

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

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.

Last edited by wfa207 on 08 May 2013, 04:35, edited 1 time in total.
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07 May 2013, 20:11
DD, thanks for the quick response.

I'm still a little fuzzy on why $$\frac{3}{9} \times \frac{2}{8} \times \frac{6}{7}$$ from the triplets problem implies that order matters whereas the $$\frac{12}{12} \times \frac{10}{11} \times \frac{8}{10} \times \frac{6}{9}$$ from the card pairs problem does not.

Could you help explain this as well?
Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If   [#permalink] 07 May 2013, 20:11

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