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Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If [#permalink]

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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?

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 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.

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If [#permalink]

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22 Aug 2013, 03:32

1

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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?

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 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.

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If [#permalink]

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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. The correct answer is B.

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If [#permalink]

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31 Jul 2014, 23:34

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

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

Answer: B.

Hope it's clear.

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

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?
_________________

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If [#permalink]

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28 Nov 2014, 04:34

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.

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..

Concentration: General Management, Entrepreneurship

GPA: 3.8

WE: Engineering (Energy and Utilities)

Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If [#permalink]

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04 Aug 2017, 01:48

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