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# A gambler rolls three fair six-sided dice. What is the probability

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A gambler rolls three fair six-sided dice. What is the probability [#permalink]

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31 Jan 2012, 16:50
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A gambler rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number?

OA:
[Reveal] Spoiler:
15/36

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Re: A gambler rolls three fair six-sided dice. What is the probability [#permalink]

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31 Jan 2012, 18:08
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This is a tricky question because there is a final twist that it is easy to overlook.

First off, we want to find the probability of rolling a pair of any number. There are six
pairs and a total of 36 different ways to roll a pair of dices: 6/36 = 1/6.

Next, we want to account for the third die - it cannot be the same as the first two,
so it can be any of five numbers out of 6: 5/6.

Therefore probability is 5/6 x 1/6 = 5/36.

It is tempting here to think that we have solved the problem. But if we read carefully, the
question does not say that the first two dice have to be the same and the third die has
to be different.

the question says a gambler rolls 3 dice. So the first two do not have to be pairs.
As long as 2 of the 3 dice are pairs and the last die is a different number. To find the
number of way this can occur, we use combinations formula: 3!/2!1! = 3.

We multiply this to the numerator to give us 3 x 5/36 = 15/36.

For some even trickier dice problems, check out our Magoosh GMAT blog post (these ones are really nasty!)

GMAT Probability: Difficult Dice Questions
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Re: A gambler rolls three fair six-sided dice. What is the probability [#permalink]

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01 Feb 2012, 00:02
Thanks chris for the solution.
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A gambler rolls three fair six-sided dice. What is the probability [#permalink]

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02 Feb 2012, 10:32
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abhi47 wrote:
A gambler rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number?

Responding to a pm.

Combinations approach:

Total # of outcomes is $$6^3$$;

Favorable outcomes are all possible scenarios of XXY:
$$C^1_6*C^1_5*\frac{3!}{2!}=6*5*3=90$$, where $$C^1_6$$ is # of ways to pick X (the number which shows twice), $$C^1_5$$ is # of ways to pick Y (out of 5 numbers left) and $$\frac{3!}{2!}$$ is # of permutation of 3 letters XXY out of which 2 X's are identical.

$$P=\frac{Favorable}{Total}=\frac{90}{6^3}=\frac{15}{36}$$.

Hope it's clear.
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Probability - rolling a dice [#permalink]

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12 Aug 2012, 13:23
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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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Re: Probability - rolling a dice [#permalink]

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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]

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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]

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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]

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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]

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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: A gambler rolls three fair six-sided dice. What is the probability [#permalink]

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08 Jul 2015, 02:00
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Re: A gambler rolls three fair six-sided dice. What is the probability [#permalink]

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24 Nov 2015, 09:35
Bunuel wrote:
abhi47 wrote:
A gambler rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number?

Responding to a pm.

Combinations approach:

Total # of outcomes is $$6^3$$;

Favorable outcomes are all possible scenarios of XXY:
$$C^1_6*C^1_5*\frac{3!}{2!}=6*5*3=90$$, where $$C^1_6$$ is # of ways to pick X (the number which shows twice), $$C^1_5$$ is # of ways to pick Y (out of 5 numbers left) and $$\frac{3!}{2!}$$ is # of permutation of 3 letters XXY out of which 2 X's are identical.

$$P=\frac{Favorable}{Total}=\frac{90}{6^3}=\frac{15}{36}$$.

Hope it's clear.

Hey Bunuel, you don't seem to agree with Chris on permutation/combination.

When you say $$\frac{3!}{2!}$$ is it a 3P1 or a 3C2 ?
And by the way, how would we know from this question, which to work with?

Thank you!
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Re: Probability - rolling a dice [#permalink]

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06 Jul 2016, 03:29
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Re: Probability - rolling a dice   [#permalink] 06 Jul 2016, 03:29
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