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Math Help - Rolling dices

  1. #1
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    Rolling dices

    Six fair dice are tossed independently. Find the probability that the number of 1ísminus the number of 2ís will be 3?

    I know there will be two cases!
    Case 1: 3 ones and 0 twos
    Case 2: 4 ones, 1 two, and one (any 3-6)

    I dont know how to calculate the probability. I keep getting the wrong answer.
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  2. #2
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    Quote Originally Posted by bigb View Post
    Six fair dice are tossed independently. Find the probability that the number of 1ísminus the number of 2ís will be 3?
    Case 1: 3 ones and 0 twos
    Case 2: 4 ones, 1 two, and one (any 3-6)
    Case 1: 111XXX in any order. X \notin \left\{ {1,2} \right\}
    \frac{{6!}}{{\left( {3!} \right)^2 }}\left( {\frac{1}{6}} \right)^3 \left( {\frac{4}{6}} \right)^3

    Case 2: 11112X in any order:
    \frac{{6!}}{{\left( {4!} \right)}}\left( {\frac{1}{6}} \right)^5 \left( {\frac{4}{6}} \right)
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  3. #3
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    Quote Originally Posted by Plato View Post
    Case 1: 111XXX in any order. X \notin \left\{ {1,2} \right\}
    \frac{{6!}}{{\left( {3!} \right)^2 }}\left( {\frac{1}{6}} \right)^3 \left( {\frac{4}{6}} \right)^3

    Case 2: 11112X in any order:
    \frac{{6!}}{{\left( {4!} \right)}}\left( {\frac{1}{6}} \right)^5 \left( {\frac{4}{6}} \right)
    I was being really really dumb...its should be 4/6 not 5/6, which is what i kept saying.You dont want 1 or 2 to come up again. Thanks for the help man!
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