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The question asks about the probability that two of the dice show the same number but the third dice shows a different number.

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=Favorable/Total=90/6^3=15/36.

Answer: B.

Or: P(XXY)=1*1/6*5/6*3!/2! --> (any, the same one, different one).

Re: Three dice are thrown. What is the probability that the firs [#permalink]

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10 Sep 2013, 06:48

did I read this questions wrong? How it was written made me think that all three dice were thrown individually, and the number on the first two had to match, with the number on the third die being thrown had to be the different number. So I interpreted that the pattern had to be xxy.

To do it this way it would be 1/1 * 1/6 * 5/6 = 5/36. Then multiply by 3 to get 15/36. It is 1/1 because the first die can be any number, not a specific number.

Re: Three dice are thrown. What is the probability that the firs [#permalink]

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11 Sep 2013, 00:22

The question talks about the "first two" rolls and the "last" roll, so I don't think that Bunuel's answer is correct. He's solving for the probability that ANY TWO are the same and the other is different.

Re: Three dice are thrown. What is the probability that the firs [#permalink]

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11 Sep 2013, 00:35

tchang wrote:

The question talks about the "first two" rolls and the "last" roll, so I don't think that Bunuel's answer is correct. He's solving for the probability that ANY TWO are the same and the other is different.

The correct answer is 1/1 x 1/6 x 5/6 = 5/36

Where did this question come from?

t1000

I approached it this way too....what are we doing wrong Bunuel??

To do it this way it would be 1/1 * 1/6 * 5/6 = 5/36. Then multiply by 3 to get 15/36. It is 1/1 because the first die can be any number, not a specific number.

ok but still can be any number out of the 6 on the die. it cant be just any number, for eg ten. Also why did you multiply 5/36 with three? first two had to be the same number, last had to be different. also, its manhattan

Total cases = 216. Favorable cases: 112,121,211.... If you fix one two numbers as above , their can be 15 such cases. You can write it down if needed for clarity and there are 6 numbers. So fav cases= 15x 6 = 90. probability = 15/36

The question asks about the probability that two of the dice show the same number but the third dice shows a different number.

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=Favorable/Total=90/6^3=15/36.

Answer: B.

Or: P(XXY)=1*1/6*5/6*3!/2! --> (any, the same one, different one).

Hope it's clear.

Bunuel it specifically says FIRST two show the same number and LAST one shows different. What you did here would be an answer to 'ANY TWO showing same and the REMAINING showing different'. The questions states the order too by mentioning FIRST/LAST. Please clear me if i'm wrong.

Re: Three dice are thrown. What is the probability that the firs [#permalink]

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20 Jun 2014, 19:48

this is clearly a problem where the sequence matters hance the favourable outcomes are 30 for XXY excluding XyX and YXX. But a great problem to sort out basic understanding and generate debate. Kudos

gauravkaushik8591 wrote:

Bunuel wrote:

sandra123 wrote:

Three dice are thrown. What is the probability that the first two show the same number, and the last one a different number?

A. 3/6 B. 15/36 C.1/216 D. 15/216 E. 1/6

So I did 1/6 * 1/6 * 5/6 (first two show the same number, the last one another number) = 5/216. then i multiply by 3 and get 15/216.

The question asks about the probability that two of the dice show the same number but the third dice shows a different number.

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=Favorable/Total=90/6^3=15/36.

Answer: B.

Or: P(XXY)=1*1/6*5/6*3!/2! --> (any, the same one, different one).

Hope it's clear.

Bunuel it specifically says FIRST two show the same number and LAST one shows different. What you did here would be an answer to 'ANY TWO showing same and the REMAINING showing different'. The questions states the order too by mentioning FIRST/LAST. Please clear me if i'm wrong.

I approached this question the same way as many others:

First dice (can be any number) = 6/6 Second dice (must be the same as first dice) = 1/6 Third dice (any number other than the first 2 dice) = 5/6

Therefore the probability is: 6/6 * 1/6 * 5/6 = 30/216 = 5/36

I'm not understanding why this number needs to be multiplied by 3 either.

This is explained in my solution above.

It should be P(XXY)=1*1/6*5/6*3!/2! --> (any, the same one, different one). XXY can occur in three different ways XXY, XYX, or YXX: \(\frac{3!}{2!}=3\) is # of permutation of 3 letters XXY out of which 2 X's are identical.
_________________

Re: Three dice are thrown. What is the probability that the firs [#permalink]

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24 Jun 2014, 12:06

Another way to look at this question: total # of outcomes are 216 (6^3...6 for each dice)...now consider 6 throws in which you have sequences such as 11(2,3,4,5,6), 22(1,3,4,5,6), ..., 66(1,2,3,4,5)...so these are 6*5=30 outcomes....so your result is 30/216 which is 5/36..now multiple it by 3 for 3 possible sequences on 3 dice for the above 30 outcomes.

Re: Three dice are thrown. What is the probability that the firs [#permalink]

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09 Jul 2014, 11:41

Bunuel wrote:

beef001 wrote:

I approached this question the same way as many others:

First dice (can be any number) = 6/6 Second dice (must be the same as first dice) = 1/6 Third dice (any number other than the first 2 dice) = 5/6

Therefore the probability is: 6/6 * 1/6 * 5/6 = 30/216 = 5/36

I'm not understanding why this number needs to be multiplied by 3 either.

This is explained in my solution above.

It should be P(XXY)=1*1/6*5/6*3!/2! --> (any, the same one, different one). XXY can occur in three different ways XXY, XYX, or YXX: \(\frac{3!}{2!}=3\) is # of permutation of 3 letters XXY out of which 2 X's are identical.

Hi Bunuel,

The confusion seems to be stemming from the question text. The question mentions that 'first two' dice should be same and the last dice should be different. So only XXY should be a success and not XYX or YXX.. So should the correct answer be 6x1x5/216 (Any number x same number x remaining numbers)/216 = 5/36?

Three dice are thrown. What is the probability that the firs [#permalink]

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15 Jul 2014, 21:26

The question asks about the probability that two of the dice show the same number but the third dice shows a different number.

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=Favorable/Total=90/6^3=15/36.

Answer: B.

Or: P(XXY)=1*1/6*5/6*3!/2! --> (any, the same one, different one).

Hope it's clear.[/quote]

Hello Bunuel,

Why are we calculating the permutation of the three letters which have been selected ? Shouldn't it be just 6.5.6 = 180 ? In case of permutation we are also considering the arrangement XXY, XYX, YXX. However, the question clearly states that the first two need to be identical while the last one is different ?

The question asks about the probability that two of the dice show the same number but the third dice shows a different number.

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=Favorable/Total=90/6^3=15/36.

Answer: B.

Or: P(XXY)=1*1/6*5/6*3!/2! --> (any, the same one, different one).

Hope it's clear.

Hello Bunuel,

Why are we calculating the permutation of the three letters which have been selected ? Shouldn't it be just 6.5.6 = 180 ? In case of permutation we are also considering the arrangement XXY, XYX, YXX. However, the question clearly states that the first two need to be identical while the last one is different ?

Any help will be appreciated. Thank you[/quote]

There are no first or second or third die there. Any die can be called first or second when thrown.
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