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 Probability and Statistics Basic Probability and Statistics Math Forum

 June 4th, 2015, 08:21 PM #1 Newbie   Joined: Jun 2015 From: Kiribati Posts: 5 Thanks: 0 Math Focus: Statistics 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. June 4th, 2015, 08:46 PM #2 Senior Member   Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18 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 from sensational June 4th, 2015, 09:05 PM   #3
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 Originally Posted by shunya 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. June 5th, 2015, 04:30 AM   #4
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 Originally Posted by shunya 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 June 5th, 2015, 04:38 AM   #5
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 Originally Posted by sensational 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. June 5th, 2015, 06:00 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,620 Thanks: 2609 Math Focus: Mainly analysis and algebra 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 $\{3,1\}$. Thanks from shunya June 7th, 2015, 07:02 PM   #7
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 Originally Posted by mrtwhs 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 Tags probability, problem ### two cubical dice are thrown ang their scores added together.

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