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Give that two six sided dice are rolled once

  1. Jun 8, 2012 #1
    1. Given that two six sided dice are rolled once, what are the odds against rolling a sum that is a multiple of 3



    3. The answer I got was 2:3

    Was just wondering if this is correct.
     
  2. jcsd
  3. Jun 8, 2012 #2

    vela

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    Show your work.
     
  4. Jun 8, 2012 #3
    The highest possible number with a single roll of two dice is 12. Multiples of three going up to 12 are 3,6,9, and 12. Looking at the grid, there are 12 out of 36 combinations that give a sum that is a multiple of three ((1,2), (1,5), (2,1), (2,4), (3,3), (3,6), (4,2), (4,5), (5,1), (5,4), (6, 3),(6,6)). This means that the probability against rolling a sum that is a multiple of three is 24/36 because 36-12=24.

    24/36 can be simplified to 2/3 , thus, the odds against rolling a sum that is a multiple of 3 is 2:3.
     
  5. Jun 8, 2012 #4

    vela

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    Your calculation of the probability is fine. However, the odds would be 2 to 1 because there are twice as many ways to not roll a multiple of 3 as there are ways to roll a multiple of 3. The odds is the ratio of the relative probabilities of an event and its complement, so in this case you have (2/3) to (1/3) or 2 to 1.
     
  6. Jun 8, 2012 #5
    Thank you!
     
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