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June 4th, 2015, 09:21 PM   #1
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Question Probability problem

I am in high school and my Maths skills is worst.

Two cubical dice are thrown and their scores added together.

If X = "The sum of the scores on the two dice", what is P(X is divisible by 4)?

I know that 6*6= 36 because there were two dice.

My answer is this ?/36
I can't find the numerator for this one it is pretty hard for sure in my capability.
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June 4th, 2015, 09:46 PM   #2
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There are 6 * 6 = 36 possible permutations
The permutations that are divisible BY 4 are: (1,3), (3, 1), (2, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4)...a total of 8.

So P(sum divisible by 4) = 8/36 = 2/9
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June 4th, 2015, 10:05 PM   #3
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Quote:
Originally Posted by shunya View Post
There are 6 * 6 = 36 possible permutations
The permutations that are divisible BY 4 are: (1,3), (3, 1), (2, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4)...a total of 8.

So P(sum divisible by 4) = 8/36 = 2/9
Thanks I am beginning to understand now. But why do you call them permutations
and (1,3) and (3,1) aren't the same? I thought you should only take (1,3), (2,2), (2,6), (3,5), (4,4) anyway thanks again. Can I get a brief explanation why (1,3) and (3,1) are both counted or both pemutations.
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June 5th, 2015, 05:30 AM   #4
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Quote:
Originally Posted by shunya View Post
There are 6 * 6 = 36 possible permutations
The permutations that are divisible BY 4 are: (1,3), (3, 1), (2, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4)...a total of 8.

So P(sum divisible by 4) = 8/36 = 2/9
How about (6,6)?
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June 5th, 2015, 05:38 AM   #5
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Quote:
Originally Posted by sensational View Post
Thanks I am beginning to understand now. But why do you call them permutations
and (1,3) and (3,1) aren't the same? I thought you should only take (1,3), (2,2), (2,6), (3,5), (4,4) anyway thanks again. Can I get a brief explanation why (1,3) and (3,1) are both counted or both pemutations.
Pretend that one die is colored red and the other die is colored green. Then red 1, green 3 is different from red 3, green 1.
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June 5th, 2015, 07:00 AM   #6
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The main reason you have to count $\{1,3\}$ and $\{3,1\}$ is that your 36 possibilities which you identified in the denominator include both $\{1,3\}$ and $\{3,1\}$.
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June 7th, 2015, 08:02 PM   #7
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How about (6,6)?
You're right. That makes a total of 9 possibilities that define the event: sum divisible by 4

So the Probability = 9/36 = 1/4
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