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when 2 fair dice are rolled what is the probability of having 6 as sum of the resulting numbers? A 1/12 B 1/6 C 5/6 D 5/36 E 1/2

can you solve it.... because i got an answer but i cannot understand why it is wrong.

There are 36 possible outcomes when a pair of dice is rolled (6 for the first die X 6 for the second one). From this 36 outcomes five have a total of 6, {(1,5), (5,1), (2,4), (4,2), (3,3)}, hence the probability of the two numbers adding up to 6 is \(\frac{5}{36}\).

Re: When 2 fair dice are rolled what is the probability of having 6 as sum [#permalink]

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16 Jan 2010, 03:41

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i have choosen 1/6 cuz I have thought that there are 6 possible outcomes 5,1 1,5 4,2 2,4 3,3 and again 3,3 because 3,3 can happen 2 times... why dont you count 3,3 2 times?

i have choosen 1/6 cuz I have thought that there are 6 possible outcomes 5,1 1,5 4,2 2,4 3,3 and again 3,3 because 3,3 can happen 2 times... why dont you count 3,3 2 times?

When we count (4,2) and (2,4), it means that we get: 4 on die #1 and 2 on die #2 in first case and 2 on dies #1 and 4 on die #2 in the second case.

With (3,3) we have only one case: 3 on #1 die and 3 on #2 die, there is no case two.

Re: When 2 fair dice are rolled what is the probability of having 6 as sum [#permalink]

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16 Jan 2010, 06:35

thank you... my approach was wrong! KUDOS FOR YOU!

Bunuel wrote:

lucalelli88 wrote:

i have choosen 1/6 cuz I have thought that there are 6 possible outcomes 5,1 1,5 4,2 2,4 3,3 and again 3,3 because 3,3 can happen 2 times... why dont you count 3,3 2 times?

When we count (4,2) and (2,4), it means that we get: 4 on die #1 and 2 on die #2 in first case and 2 on dies #1 and 4 on die #2 in the second case.

With (3,3) we have only one case: 3 on #1 die and 3 on #2 die, there is no case two.

Re: When 2 fair dice are rolled what is the probability of having 6 as sum [#permalink]

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12 Aug 2010, 19:43

Bunuel wrote:

With (3,3) we have only one case: 3 on #1 die and 3 on #2 die, there is no case two.

Hope it's clear.

well bit confused.... i think it means it is not times ie 1 time and 2nd time rather it is in 1st dice and in second dice... but suppose if die we colored green and blue would it be like 3 on G ,3 on B and 3 on B ,3 on G?

With (3,3) we have only one case: 3 on #1 die and 3 on #2 die, there is no case two.

Hope it's clear.

well bit confused.... i think it means it is not times ie 1 time and 2nd time rather it is in 1st dice and in second dice... but suppose if die we colored green and blue would it be like 3 on G ,3 on B and 3 on B ,3 on G?

Can you please clarify?

Not sure I understood your question...

There are only following 5 cases possible to have sum of 6:

#1|#2 1---5 2---4 3---3 4---2 5---1

Do we have any other case? It doesn't matter whether dice are colored, they are already numbered. (3,3) means 3 on die #1 and 3 on die #2 (3 on die #2 and 3 on die #1 is basically the same case).
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14 Aug 2017, 02:06

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Re: When 2 fair dice are rolled what is the probability of having 6 as sum [#permalink]

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14 Aug 2017, 03:34

Bunuel wrote:

frank1 wrote:

Bunuel wrote:

With (3,3) we have only one case: 3 on #1 die and 3 on #2 die, there is no case two.

Hope it's clear.

well bit confused.... i think it means it is not times ie 1 time and 2nd time rather it is in 1st dice and in second dice... but suppose if die we colored green and blue would it be like 3 on G ,3 on B and 3 on B ,3 on G?

Can you please clarify?

Not sure I understood your question...

There are only following 5 cases possible to have sum of 6:

#1|#2 1---5 2---4 3---3 4---2 5---1

Do we have any other case? It doesn't matter whether dice are colored, they are already numbered. (3,3) means 3 on die #1 and 3 on die #2 (3 on die #2 and 3 on die #1 is basically the same case).

In such a case, 1,1 2,2 3,3 4,4 5,5 and 6,6 will be the same.

Re: When 2 fair dice are rolled what is the probability of having 6 as sum [#permalink]

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15 Aug 2017, 06:36

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nkimidi7y wrote:

In such a case, 1,1 2,2 3,3 4,4 5,5 and 6,6 will be the same.

So our possible outcomes are just 30.

Our answer would then be 1/6.

Am i missing something?

Don't think of it like combination choices where you can flip and come up with the same thing. Think of 36 possibilities in 6 groups where first roll is 1 for all pairs in row 1, 2 for all pairs in row 2 and so on. You have 5 bold faced choices that have sum 6. Every row starting 1-5 has one matching pair where sum is 6, but the last row starting with 6 cannot come up with any second dice roll number to get to a sum of 6. So we only have 5 of 36 choices that have a sum of 6.

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