Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers?

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  • 1 month ago
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    This is conditional probability so we only have to consider the cases where we meet the given condition. Discounting doubles, there are 30 outcomes where the dice land on different numbers.

    Of these, there are 10 outcomes (1,6), (2,6), (3,6), (4,6), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5) where there is (at least) one six.

    So the conditional probability is 10/30 = 1/3

    UPDATE:

    If you want to be a little more mathematically rigorous, use the conditional probability formula:

    P(A|B) = P(A and B) / P(B)

    A : at least one 6

    B : dice land on different numbers

    P(A and B) = 10/36 = 5/18

    P(B) = 30/36 = 5/6

    P(A|B) = 5/18 / (5/6)

    = 5/18 * 6/5

    = 1/3 * 1/1

    = 1/3

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  • 1 month ago

    case 1: first is 6, prob 1/6, which means the other cannot be a 6, prob 5/6

    combination is 1/6 x 5/6 = 5/36

    case 2, first is not a 6, prob 5/6, second is a 6, prob 1/6, combo is 5/6 x 1/6 = 5/36

    sum of the two is 10/36 or 5/18

    alternative, list all 36 cases and pick the ones that match

    1 and 1,2,3,4,5,6

    2 and 1,2,3,4,5,6

    3 and 1,2,3,4,5,6

    4 and 1,2,3,4,5,6

    5 and 1,2,3,4,5,6

    6 and 1,2,3,4,5,6

    the ones that meet your criteria are

    1 6

    2 6

    3 6

    4 6

    5 6

    6 1, 6 2, 6 3, 6 4, 6 5

    total 10 out of 36

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