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Math Help - Fair Dice Tossed Independently

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    Fair Dice Tossed Independently

    Seven fair dice are tossed independently. Find the probability that the number of 1's minus the number of 4's will be three.

    Thoughts?
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    Re: Fair Dice Tossed Independently

    What are all of the possible outcomes that satisfy that condition?

    # of 1's # of 4's # of (not 1's, not 4's)
    3 0 4
    4 1 2
    5 2 0

    Those are the only outcomes that you care about. So, what are the probabilities of each?

    I will help you out for the first case, see if you can figure out the other two. The probability that three out of the seven dice will roll a 1 and the other four will roll neither a 1 nor a 4: \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4
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    Re: Fair Dice Tossed Independently

    Quote Originally Posted by SlipEternal View Post
    I will help you out for the first case, see if you can figure out the other two. The probability that three out of the seven dice will roll a 1 and the other four will roll neither a 1 nor a 4: \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4

    So is it just: P = \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4 + \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{7}{1  }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2 + \binom{7}{5}\left(\dfrac{1}{6}\right)^5\binom{7}{2  }\left(\dfrac{1}{6}\right)^2?
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    Re: Fair Dice Tossed Independently

    Quote Originally Posted by vidomagru View Post
    So is it just: P = \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4 + \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{7}{1  }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2 + \binom{7}{5}\left(\dfrac{1}{6}\right)^5\binom{7}{2  }\left(\dfrac{1}{6}\right)^2?
    Not quite. You have \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{7}{1  }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2
    This means out of seven dice, you choose 4 to roll 1's, then you replace those four dice and choose 1 to roll a 4. You don't want to replace the dice. So, it would be:

    P = \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4 + \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{3}{1  }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2 + \binom{7}{5}\left(\dfrac{1}{6}\right)^5\binom{2}{2  }\left(\dfrac{1}{6}\right)^2
    Thanks from vidomagru and LimpSpider
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