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Math Help - dice problem

  1. #1
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    dice problem

    Suppose that two fair dice are rolled. Let random variable M be given by the maximum of the values shown on the two dice.

    i) What is the probability mass function
    ii) What is the expected value and variance


    My answer:

    i) M can take on {1,2,3,4,5,6}. The probability that {M=1} is P(the first dice is 1 OR the second dice is 1)(the probability that 1 is a maximum) = (1/6 + 1/6)(1/6+1/6) = 1/9
    The probability that {M=2} is P(the first dice is 2 or second dice is 2)(the probability that 2 is a maximum) = (1/6+1/6)(2/3)
    The probability that {M=3} is P(the first dice is 3 or second dice is 3)(the probability that 3 is a maximum) = (1/6+1/6)(1)
    The probability that {M=4} is P(the first dice is 4 or second dice is 4)(the probability that 4 is a maximum) = (1/6+1/6)(8/6)
    The probability that {M=5} is P(the first dice is 5 or second dice is 5)(the probability that 5 is a maximum) = (1/6+1/6)(10/6)
    The probability that {M=6} is P(the first dice is 6 or second dice is 6)(the probability that 6 is a maximum) = (1/6+1/6)(12/6)

    What's wrong with my answer? The probabilities should add up to 1.
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  2. #2
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    Quote Originally Posted by lord12 View Post
    Suppose that two fair dice are rolled. Let random variable M be given by the maximum of the values shown on the two dice.

    i) What is the probability mass function
    ii) What is the expected value and variance


    My answer:

    i) M can take on {1,2,3,4,5,6}. The probability that {M=1} is P(the first dice is 1 OR the second dice is 1)(the probability that 1 is a maximum) = (1/6 + 1/6)(1/6+1/6) = 1/9
    The probability that {M=2} is P(the first dice is 2 or second dice is 2)(the probability that 2 is a maximum) = (1/6+1/6)(2/3)
    The probability that {M=3} is P(the first dice is 3 or second dice is 3)(the probability that 3 is a maximum) = (1/6+1/6)(1)
    The probability that {M=4} is P(the first dice is 4 or second dice is 4)(the probability that 4 is a maximum) = (1/6+1/6)(8/6)
    The probability that {M=5} is P(the first dice is 5 or second dice is 5)(the probability that 5 is a maximum) = (1/6+1/6)(10/6)
    The probability that {M=6} is P(the first dice is 6 or second dice is 6)(the probability that 6 is a maximum) = (1/6+1/6)(12/6)

    What's wrong with my answer? The probabilities should add up to 1.
    Draw a grid to get the probabilities:

    Pr(M = 1) = 1/36

    Pr(M = 2) = 3/36

    Pr(M = 3) = 5/36

    Pr(M = 4) = 7/36

    Pr(M = 5) = 9/36

    Pr(M = 6) = 11/36


    and in general Pr(M = x) = (2x-1)/36
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