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-   -   Probability for sum = 3 with two dice (a 4-sided & a 6-sided) (https://www.askmehelpdesk.com/showthread.php?t=733826)

  • Feb 12, 2013, 01:47 PM
    Umu
    Probability for sum = 3 with two dice (a 4-sided & a 6-sided)
    I have two dice - one is 6-sided and the other is 4-sided.
    I throw the two dice. What is the probability of getting the sum 3 (I add the two results)

    I figure that the probability for 1 or 2 on the 4-sided dice is 50% = 1/2
    and
    that the probability for 1 or 2 on the 6-sided dice is 33 1/3% = 1/3

    but what do I do with that information? Do I add them: 3/6 + 2/6 = 5/6
    or do I * 1/2 * 1/3 = 1/6

    I don't think that any of the two answers is correct. The correct answer must be smaller - but I cannot figure out what to do
  • Feb 12, 2013, 02:38 PM
    ebaines
    There are two possble combinations that add to 3:

    a) Roll a 1 on the 4-sided die and 2 on the 6-sided.
    b) Roll a 2 on the 4-sided and 1 on the 6-sided.

    Calculate the probability of each of these and add then tgether. Post back with what you get for an answer and we'll check it for you.
  • Feb 13, 2013, 04:08 AM
    Umu
    Probability of 1 on the 4-sided die = 1/4
    Probability of 2 on the 6-sided die = 1/6

    or the other way around - also 1/4 and 1/6

    1/4 + 1/6 = 3/12 + 2/12 = 5/12

    No that doesn't seem right either??
  • Feb 13, 2013, 06:26 AM
    ebaines
    If the probability of event 'A' occurring is 1/4, and the probability of event 'B' is 1/6, and the two events are independent, then the probability of both 'A' and 'B' occurring is 1/4 x 1/6 = 1/24. So that's the probability of rolling a 1 with the first die and a 2 with the second. Can you take it from here?
  • Feb 14, 2013, 12:58 AM
    Umu
    Probability 1: Roll a 1 with the first die and a 2 with the second is 1/4 x 1/6 = 1/24
    Probability 2: Roll a 2 with the first die and a 1 with the second is 1/4 x 1/6 = 1/24

    Probability of getting sum 3 = 1/24 + 1/24 = 2/24 = 1/12


    That sounds right - I believe.
    Thanks for the help

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