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# HELP PROBABILITY QUESTION!!!!!

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Intern
Joined: 23 Dec 2009
Posts: 27
Followers: 0

Kudos [?]: 70 [0], given: 7

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16 Jan 2010, 02:42
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(N/A)

Question Stats:

75% (01:00) correct 25% (00:27) wrong based on 26 sessions

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 34496
Followers: 6297

Kudos [?]: 79871 [0], given: 10022

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16 Jan 2010, 03:02
lucalelli88 wrote:
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 $$\frac{5}{36}$$.

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Intern
Joined: 23 Dec 2009
Posts: 27
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Kudos [?]: 70 [0], given: 7

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16 Jan 2010, 03:41
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?
Math Expert
Joined: 02 Sep 2009
Posts: 34496
Followers: 6297

Kudos [?]: 79871 [1] , given: 10022

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16 Jan 2010, 04:36
1
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Expert's post
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.

Hope it's clear.
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Intern
Joined: 23 Dec 2009
Posts: 27
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Kudos [?]: 70 [0], given: 7

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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.

Hope it's clear.
Manager
Joined: 24 Apr 2010
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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?

Math Expert
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Posts: 34496
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Kudos [?]: 79871 [0], given: 10022

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13 Aug 2010, 03:03
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?

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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Re: HELP PROBABILITY QUESTION!!!!!   [#permalink] 13 Aug 2010, 03:03
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# HELP PROBABILITY QUESTION!!!!!

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