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Math Help - probability of dice throw

  1. #1
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    probability of dice throw

    Question: A die has 6 sides numbered 1-6. If 3 dices are thrown, what is the probability that at least one of the dices shows the number six?

    For this question, do I have to take into consideration that the other two dices are not sixes? This is what I get:

    1/6 for first die to be a 6
    (5/6)*(1/6) for second die to be a 6 if the first is not
    (25/36)*(1/6) for the third die to be a 6 if the first two are not

    adding them up I get 91/216

    Is my analysis correct?
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  2. #2
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    Hello, chemekydie!

    A die has 6 sides numbered 1-6.
    If 3 dices are thrown, what is the probability that
    at least one of the dices shows the number six?

    For this question, do I have to take into consideration that the other two dices are not sixes?


    This is what I get:

    \frac{1}{6} for first die to be a 6

    \frac{5}{6}\cdot\frac{1}{6} for second die to be a 6 if the first is not.

    \frac{25}{36}\cdot\frac{1}{6} for the third die to be a 6 if the first two are not.

    Adding them up I get: \frac{91}{216}

    Is my analysis correct?

    Yes! . . . Good work!


    There is a back-door approach to this problem.

    The opposite of "at least one 6" is "no 6's."

    The probability of no 6's is: . \left(\frac{5}{6}\right)^3 \:=\:\frac{125}{216}


    Therefore: . P(\text{at least one 6}) \;=\;1 - \frac{125}{216} \;=\;\frac{91}{216}

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