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Probability - rolling a dice

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Probability - rolling a dice [#permalink] New post 12 Aug 2012, 13:23
A gambler rolls three fair-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number?

A) 1/8
B) 5/18
C) 1/3
D) 5/12
E) 5/6



[Reveal] Spoiler:
D

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Joined: 14 May 2011
Posts: 277
Location: Costa Rica
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Re: Probability - rolling a dice [#permalink] New post 12 Aug 2012, 13:26
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?
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Re: Probability - rolling a dice [#permalink] New post 12 Aug 2012, 22:35
Aximili85 wrote:
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.
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Re: Probability - rolling a dice [#permalink] New post 13 Aug 2012, 04:33
EvaJager wrote:
Aximili85 wrote:
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
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Re: Probability - rolling a dice [#permalink] New post 13 Aug 2012, 17:35
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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)
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Re: Probability - rolling a dice [#permalink] New post 26 Aug 2012, 07:07
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My approach:

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