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Mathematics 14 Online
OpenStudy (anonymous):

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OpenStudy (anonymous):

There are 36 possible outcomes. (2:1, 3:2, 4:3, 5:4, 6:5, 7:6, 8:5, 9:4, 10:3, 11:2, 12:1) So you just find out how many numbers fit into the requirements, add together the number of outcomes each number has, and divide it by 36.

OpenStudy (rajee_sam):

rolling doubles mean (1,1), (2,2) (3,3), (4,4), (5,5), (6,6) A no. that is evenly divisible by 3 is only 6 (1,6) (2,6) (3,6) (4,6) (5,6) ...... (6,6) is already under the first one. ( 6,1) (6,2) 96,3) (6,4 )) (6,5) ---- again (6,6) is already under one

OpenStudy (rajee_sam):

so there are 16 possible outcomes. total possible outcomes are 36 P = 16/36 = 4/9

OpenStudy (anonymous):

OH. Okay, do I put P(a) = 4/9 then?

OpenStudy (rajee_sam):

Yes

OpenStudy (anonymous):

Okay, thank you! :) I appreciate it!

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