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

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Intern
Joined: 11 Jul 2013
Posts: 34

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20 Aug 2013, 11:11
1
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?

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Joined: 02 Sep 2009
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21 Aug 2013, 08:04
1
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, 04:32
1
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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Joined: 09 Jul 2012
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15 Sep 2013, 23:20
1
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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25 Jul 2014, 00:03
# of participants is not clear. is it 12 (triplets + 9) or just 9(including triplets)

Language is confusing :/
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25 Jul 2014, 00:20
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?
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01 Aug 2014, 00:34
Bunuel wrote:
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.

Hi Bunuel, I wonder how we can solve this question in 2.5 mins. I spent almost 3.5 mins to solve this question. Is there any shortcuts?
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28 Nov 2014, 05:34
Hi Bunuel, My take was probability of getting a winner but from not a single one from the triplets was 6/9*5/8*4/7.
so, atleast 2 of the winning the prizes will be 1-(6/9*5/8*4/7)=16/21.

What am I missing? pLz help
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Joined: 26 Jan 2016
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04 Oct 2016, 13:39
This took me a while but...

There are two different ways that we can have at least two twins win medals either all the twins win medals or only two.

Two twins winning medals

(3/9)*(2/8)*(6/7)=1/14

Three twins winning medals
(3/9)*(2/8)*(1/7)=1/84

For the way 2 twins can win medals (1/14) there are three different ways this can happen
1&2
1&3
2&3
3*(1/14)=3/14

So now we have 3/14 OR 1/84. Or in probability means add. so we get 6*(3/14)=18/14+1/84=19/84
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Posts: 7954
Re: Triplets Adam, Bruce, and Charlie enter a triathlon. There are nine  [#permalink]

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13 Feb 2017, 20:44
vikasp99 wrote:
Triplets Adam, Bruce, and Charlie enter a triathlon. There are nine competitors in the triathlon. If every competitor has an equal chance of winning, and three medals will be awarded, 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

Hi,

Note :- all medals are equal that is not different for 1,2 and 3 POSITION

Two cases
1) only two get medals..
Choose 2 out of three A,B, and C---3C2=3..
The third can be any of the remaining 6..
So total ways=3*6=18..
2) all three get medals..
Only 1 way..

Total ways ATLEAST 2 get medals= 18+1=19..

TOTAL ways=9C3=$$\frac{9!}{6!3!}=84$$..

Probability=$$\frac{19}{84}$$
B
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04 Aug 2017, 02:48
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

Total number of outcomes = 9C3 = 9!/3!/6! = 9*8*7/3/2 = 3*4*7 = 84

Number of outcomes of winning atleast 2 medals by triplet = Number of outcomes of winning 2 medals by triplets + Number of outcomes of winning 3 medals by triplets
= 3C2*6C1 + 3C3 = 18+1 = 19

So, probability that at least two of the triplets will win a medal = 19/84

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04 Nov 2018, 11:32
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

$$\left\{ \matrix{ \,A\,,B\,,\,{\kern 1pt} C\,\,\,\,\, \to \,\,\,3\,\,{\rm{competitors}} \hfill \cr \,{\rm{D,}}\,\, \ldots \,\,{\rm{,}}\,I\,\,\, \to \,\,\,{\rm{other}}\,\,6\,\,{\rm{competitors}}\, \hfill \cr} \right.$$

$$\left( {\rm{i}} \right)\,\,A,B,C\,\,{\rm{win}}\,\,{\rm{the}}\,\,3\,\,{\rm{medals}}$$

$$\left( {{\rm{ii}}} \right)\,\,A,B,C\,\,{\rm{win}}\,\,{\rm{exactly}}\,\,2\,\,{\rm{of}}\,\,{\rm{the}}\,\,3\,\,{\rm{medals}}$$

$$? = P\left( {{\rm{i}}\,\,{\rm{or}}\,\,{\rm{ii}}} \right) = P\left( {\rm{i}} \right) + P\left( {{\rm{ii}}} \right)\,\,\,\,\,\,\,\,\left[ {\,{\rm{i}}\,\,{\rm{and}}\,\,{\rm{ii}}\,\,{\rm{are}}\,\,{\rm{mutually}}\,\,{\rm{exclusive}}\,} \right]$$

$$\left( {\rm{i}} \right)\,\,\,\left\{ \matrix{ \,{\rm{favorable}} = \,\,3! \hfill \cr \,{\rm{total}}\,\, = \,\,9 \cdot 8 \cdot 7\,\,{\rm{equiprobables}}\,\,\, \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,P\left( {\rm{i}} \right) = {1 \over {84}}$$

$$\left( {{\rm{ii}}} \right)\,\,\,\left\{ \matrix{ \,{\rm{favorable}} = \,\,\underbrace {C\left( {3,2} \right)}_{{\rm{say}}\,\,A,B}\,\,\, \cdot \,\,\underbrace {C\left( {3,2} \right)}_{{\rm{say}}\,\,\,1{\rm{st}}\,\,,\,\,2{\rm{nd}}}\,\,\, \cdot \,\,\underbrace {2!}_{{\rm{say}}\,\,\,A\,\,1{\rm{st}}\,\,,\,B\,\,2{\rm{nd}}}\,\,\, \cdot \,\,\,\underbrace {C\left( {6,1} \right)}_{{\rm{say}}\,\,D\,\,\left( {3{\rm{rd}}} \right)}\, \hfill \cr \,{\rm{total}}\,\, = \,\,9 \cdot 8 \cdot 7\,\,{\rm{equiprobables}}\,\,\, \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,P\left( {{\rm{ii}}} \right) = {3 \over {14}}$$

$$? = {1 \over {84}} + {{3 \cdot 6} \over {14 \cdot 6}} = {{19} \over {84}}$$

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

Regards,
Fabio.
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07 Nov 2018, 19:04
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

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.

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16 Aug 2019, 07:19
Bunuel wrote:
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.

Plz explain Bunuel EgmatQuantExpert chetan2u
why this approach is wrong
select 2 out of 3 then selected 1 out of 6
3c2/9c2*6c1/7c1
why is this wrong
Triplets Adam, Bruce, and Charlie enter a triathlon. If   [#permalink] 16 Aug 2019, 07:19

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