Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 02 May 2012
Posts: 10

Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
Updated on: 23 Dec 2019, 08:47
Question Stats:
56% (02:29) correct 44% (02:48) wrong based on 516 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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by cnon on 16 May 2012, 07:03.
Last edited by Bunuel on 23 Dec 2019, 08:47, edited 2 times in total.
Edited the question, added the answer choices and OA




Math Expert
Joined: 02 Sep 2009
Posts: 62396

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
16 May 2012, 07:28
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 \(\frac{C^3_3}{C^3_9}=\frac{1}{84}\); \(P=\frac{18}{84}+\frac{1}{84}=\frac{19}{84}\). Answer: B. Hope it's clear. P.S. Please post answer choices for PS problems.
_________________




Senior Manager
Joined: 10 Jul 2013
Posts: 276

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
18 Aug 2013, 11:22
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 28339 times ]




Intern
Joined: 29 Aug 2012
Posts: 25
GMAT Date: 02282013

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
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.



Math Expert
Joined: 02 Sep 2009
Posts: 62396

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
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.
_________________



Manager
Joined: 26 Jul 2011
Posts: 76
Location: India
WE: Marketing (Manufacturing)

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
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??



Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 223
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
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. Can someone shed some light about this please ?



Intern
Joined: 18 Nov 2011
Posts: 29
Concentration: Strategy, Marketing
GMAT Date: 06182013
GPA: 3.98

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
07 Feb 2013, 17:16
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. Can someone shed some light about this please ? 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.



Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 223
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
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 \(3P2 = \frac{3!}{(32)!}\) 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



Intern
Joined: 18 Nov 2011
Posts: 29
Concentration: Strategy, Marketing
GMAT Date: 06182013
GPA: 3.98

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
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 \(3P2 = \frac{3!}{(32)!}\) 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 \(\frac{n!}{k!(nk)!}\) 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.



Intern
Joined: 20 Nov 2012
Posts: 1
Concentration: Technology, Strategy
GPA: 3.5

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
Updated on: 08 May 2013, 04:35
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.
Originally posted by wfa207 on 07 May 2013, 19:17.
Last edited by wfa207 on 08 May 2013, 04:35, edited 1 time in total.



Manager
Joined: 13 Dec 2012
Posts: 60

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
07 May 2013, 19:50
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 nontriplet 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 noorder issue, but really, you were attempting sequential counting, but should have been using combinatorials Hope this clarifies things
_________________
Got questions? gmatbydavid.com



Manager
Joined: 12 Mar 2010
Posts: 242
Concentration: Marketing, Entrepreneurship

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
18 Aug 2013, 07:57
Are Adam, Bruce and Charlie in the same triathlon team or are they in separate teams?



Math Expert
Joined: 02 Sep 2009
Posts: 62396

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
19 Aug 2013, 00:58
gmatter0913 wrote: Are Adam, Bruce and Charlie in the same triathlon team or are they in separate teams? No, they are competing each other, they do not form a team.
_________________



Manager
Joined: 24 Nov 2012
Posts: 143
Concentration: Sustainability, Entrepreneurship
WE: Business Development (Internet and New Media)

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
19 Aug 2013, 01:55
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 \(3P2 = \frac{3!}{(32)!}\) 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 In case the order does matter we would still get the same answer i believe.. Experts please correct me if i am wrong... No of ways of choosing 3 ppl out of 9 = 3C9=84 No of ways those three ppl can be arranged on the podium = 6 So hence total number of options = 504 No of ways A,B and C can be arranged on the podium = 6 No of ways you can choose two ppl from A, B, C = 3 No of ways you can chose the 3rd person from 6 = 6 No of ways you can arrange 3 ppl on the podium = 6 So total number of ways = 6 x 6 x3 = 108 So probability of at least two on the podium = 108+6/504 = 19/84
_________________
You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper!  Rumi
http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprepcom/  This is worth its weight in gold
Economist GMAT Test  730, Q50, V41 Aug 9th, 2013 Manhattan GMAT Test  670, Q45, V36 Aug 11th, 2013 Manhattan GMAT Test  680, Q47, V36 Aug 17th, 2013 GmatPrep CAT 1  770, Q50, V44 Aug 24th, 2013 Manhattan GMAT Test  690, Q45, V39 Aug 30th, 2013 Manhattan GMAT Test  710, Q48, V39 Sep 13th, 2013 GmatPrep CAT 2  740, Q49, V41 Oct 6th, 2013
GMAT  770, Q50, V44, Oct 7th, 2013 My Debrief  http://gmatclub.com/forum/fromtheashesthoushallrise770q50v44awa5ir162299.html#p1284542



Intern
Joined: 11 Jul 2013
Posts: 31

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
20 Aug 2013, 10:11
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?
Thanks in advance.



Math Expert
Joined: 02 Sep 2009
Posts: 62396

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
21 Aug 2013, 07:04
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?
Thanks in advance. 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 515 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. Hope this helps.
_________________



Intern
Joined: 09 Jul 2012
Posts: 8

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
15 Sep 2013, 22:20
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. The correct answer is B.



Intern
Joined: 23 Apr 2014
Posts: 9

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
24 Jul 2014, 23:03
# of participants is not clear. is it 12 (triplets + 9) or just 9(including triplets)
Language is confusing :/



Math Expert
Joined: 02 Sep 2009
Posts: 62396

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
Show Tags
24 Jul 2014, 23:20
Gyanendra wrote: # of participants is not clear. is it 12 (triplets + 9) or just 9(including triplets)
Language is confusing :/ I don't see anything confusing there. Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon...Would it be correct to write that there are 9 competitors in the triathlon if there were 12 competitors?
_________________




Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co
[#permalink]
24 Jul 2014, 23:20



Go to page
1 2
Next
[ 26 posts ]



