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How to improve your RC score, pls Kudo if helpful! http://gmatclub.com/forum/how-to-improve-my-rc-accuracy-117195.html Work experience (as of June 2012) 2.5 yrs (Currently employed) - Mckinsey & Co. (US Healthcare Analyst) 2 yrs - Advertising industry (client servicing)

Hi, can somebody please explain a fundamental aspect of this question: I understand that the probability of getting 2 die with the same number AND one dice with a different number is = 6/6 x 1/6 x 5/6 = 5/36

BUT, beyond this, I get confused whether I should multiply this by 3 for the other 3 other combinations OR 3!?

Could somebody explain while accounting for different combinations, when we multiply by the # of combi VS the factorial of the # of combi? _________________

How to improve your RC score, pls Kudo if helpful! http://gmatclub.com/forum/how-to-improve-my-rc-accuracy-117195.html Work experience (as of June 2012) 2.5 yrs (Currently employed) - Mckinsey & Co. (US Healthcare Analyst) 2 yrs - Advertising industry (client servicing)

Hi, can somebody please explain a fundamental aspect of this question: I understand that the probability of getting 2 die with the same number AND one dice with a different number is = 6/6 x 1/6 x 5/6 = 5/36

BUT, beyond this, I get confused whether I should multiply this by 3 for the other 3 other combinations OR 3!?

Could somebody explain while accounting for different combinations, when we multiply by the # of combi VS the factorial of the # of combi?

In this case, you should multiply by 3 because you can get two identical results and the third different in three different scenarios: AAB, ABA and BAA. Each scenario has the same probability of 5/36. You can look at the 3 as 3C1, meaning how many choices we have to place the different result, or you can interpret 3 as 3!/2! because you have all the permutations of the triplet A,A,B, with A repeated twice. Obviously, as they should, they really give the same result. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Hi, can somebody please explain a fundamental aspect of this question: I understand that the probability of getting 2 die with the same number AND one dice with a different number is = 6/6 x 1/6 x 5/6 = 5/36

BUT, beyond this, I get confused whether I should multiply this by 3 for the other 3 other combinations OR 3!? ......................twice. Obviously, as they should, they really give the same result.

ok that helps, thanks _________________

How to improve your RC score, pls Kudo if helpful! http://gmatclub.com/forum/how-to-improve-my-rc-accuracy-117195.html Work experience (as of June 2012) 2.5 yrs (Currently employed) - Mckinsey & Co. (US Healthcare Analyst) 2 yrs - Advertising industry (client servicing)

Ways in which a given set can appear = 3 i.e. AAB ABA BAA For case AAB Probability of getting a number on first dice = 6/6 Probability of getting the same number on second dice = 1/6 Probability of getting a different number on third dice = 5/6

Thus probability = 6/6 * 1/6 * 5/6 * 3

5/12 (D) _________________

First say to yourself what you would be; and then do what you have to do.

Total number of possible way = 6*6*6 = 216 3 discs will appear in any one of the following arrangements: AAA, AAB, ABC where, AAA=All are same, AAB=Two are same, ABC=all three different

Now, total number of AAA = 6 Total number of ABC = 6*5*4 = 120 therefore, total number of AAA, ABC = 126 So, total number of AAB = 216-126=90 Probability of AAB= 90/216=5/12 _________________

My mantra for cracking GMAT: Everyone has inborn talent, however those who complement it with hard work we call them 'talented'.

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Re: Probability - rolling a dice
[#permalink]
26 Aug 2012, 08:07

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