Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Jul 2016, 09:20

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 02 May 2012
Posts: 11
GMAT 1: 530 Q48 V16
Followers: 0

Kudos [?]: 17 [2] , given: 1

### Show Tags

16 May 2012, 08:03
2
KUDOS
19
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

60% (02:53) correct 40% (02:34) wrong based on 317 sessions

### HideShow timer Statistics

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, 08:17, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 34043
Followers: 6081

Kudos [?]: 76408 [14] , given: 9974

### Show Tags

16 May 2012, 08:28
14
KUDOS
Expert's post
7
This post was
BOOKMARKED
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.

_________________
Senior Manager
Joined: 10 Jul 2013
Posts: 335
Followers: 3

Kudos [?]: 262 [3] , given: 102

### Show Tags

18 Aug 2013, 12:22
3
KUDOS
1
This post was
BOOKMARKED
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)
Attachments

triplets.png [ 35.25 KiB | Viewed 9167 times ]

_________________

Asif vai.....

Manager
Joined: 13 Dec 2012
Posts: 60
Followers: 9

Kudos [?]: 23 [1] , given: 4

### Show Tags

07 May 2013, 20:50
1
KUDOS
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
_________________

Got questions?
gmatbydavid.com

Math Expert
Joined: 02 Sep 2009
Posts: 34043
Followers: 6081

Kudos [?]: 76408 [1] , given: 9974

### Show Tags

21 Aug 2013, 08:04
1
KUDOS
Expert's post
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.
_________________
Intern
Joined: 11 Jul 2013
Posts: 43
Followers: 0

Kudos [?]: 10 [1] , given: 92

### Show Tags

22 Aug 2013, 04:32
1
KUDOS
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.....
Intern
Joined: 09 Jul 2012
Posts: 11
Followers: 0

Kudos [?]: 3 [1] , given: 6

### Show Tags

15 Sep 2013, 23:20
1
KUDOS
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.
Intern
Joined: 02 May 2012
Posts: 11
GMAT 1: 530 Q48 V16
Followers: 0

Kudos [?]: 17 [0], given: 1

### Show Tags

16 May 2012, 16:52
Thank you so much,i will take into account.
Intern
Joined: 29 Aug 2012
Posts: 26
Schools: Babson '14
GMAT Date: 02-28-2013
Followers: 0

Kudos [?]: 22 [0], given: 56

### Show Tags

25 Nov 2012, 15:42
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.
Math Expert
Joined: 02 Sep 2009
Posts: 34043
Followers: 6081

Kudos [?]: 76408 [0], given: 9974

### Show Tags

26 Nov 2012, 01:55
Expert's post
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.
_________________
Manager
Joined: 20 Aug 2012
Posts: 73
Schools: Jones '15
Followers: 1

Kudos [?]: 10 [0], given: 5

### Show Tags

26 Nov 2012, 07:21
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 ???
Intern
Joined: 22 Jul 2012
Posts: 1
Followers: 0

Kudos [?]: 5 [0], given: 0

### Show Tags

30 Nov 2012, 13:55
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
Manager
Joined: 26 Jul 2011
Posts: 125
Location: India
WE: Marketing (Manufacturing)
Followers: 1

Kudos [?]: 88 [0], given: 15

### Show Tags

30 Nov 2012, 22:49
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??
Senior Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 287
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)
Followers: 3

Kudos [?]: 243 [0], given: 85

### Show Tags

07 Feb 2013, 16:47
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.
_________________

KUDOS is the good manner to help the entire community.

Intern
Joined: 18 Nov 2011
Posts: 37
Concentration: Strategy, Marketing
GMAT Date: 06-18-2013
GPA: 3.98
Followers: 0

Kudos [?]: 11 [0], given: 0

### Show Tags

07 Feb 2013, 18:16
1
This post was
BOOKMARKED
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.
Senior Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 287
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)
Followers: 3

Kudos [?]: 243 [0], given: 85

### Show Tags

07 Feb 2013, 18:37
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
_________________

KUDOS is the good manner to help the entire community.

Intern
Joined: 18 Nov 2011
Posts: 37
Concentration: Strategy, Marketing
GMAT Date: 06-18-2013
GPA: 3.98
Followers: 0

Kudos [?]: 11 [0], given: 0

### Show Tags

07 Feb 2013, 20:11
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.
Senior Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 287
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)
Followers: 3

Kudos [?]: 243 [0], given: 85

### Show Tags

09 Feb 2013, 13:33
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
_________________

KUDOS is the good manner to help the entire community.

Intern
Joined: 20 Nov 2012
Posts: 2
Concentration: Technology, Strategy
GMAT 1: 760 Q49 V46
GPA: 3.5
Followers: 0

Kudos [?]: 5 [0], given: 0

### Show Tags

07 May 2013, 20:17
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, 05:35, edited 1 time in total.
Intern
Joined: 20 Nov 2012
Posts: 2
Concentration: Technology, Strategy
GMAT 1: 760 Q49 V46
GPA: 3.5
Followers: 0

Kudos [?]: 5 [0], given: 0

### Show Tags

07 May 2013, 21:11
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, 21:11

Go to page    1   2    Next  [ 32 posts ]

Similar topics Replies Last post
Similar
Topics:
The ratio of ages of Aman, Bren, and Charlie 3 25 Feb 2016, 05:00
3 Alice and Bruce each bought a refrigerator, and the sum of their purch 5 12 Mar 2015, 06:47
18 Adam and Brianna plan to install a new tile floor in a classroom. Adam 6 06 Oct 2014, 04:00
5 If Charlie decides to cut the salary per employee at his 5 09 Jul 2013, 01:39
2 triplet combinations puzzle 1 11 Nov 2010, 04:02
Display posts from previous: Sort by