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Math Help - Probability Q Again..

  1. #1
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    Probability Q Again..

    Hi >.<

    I got another Q:

    The number showing on the upper faces when 2 fair six-sided dice are tossed, is observed. Consider events A and B

    A- the total score is at most 5
    B- at least one die achieves a score of 3 or 4


    thanks
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  2. #2
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    Lexington, MA (USA)
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    Hello, procrastinator!

    With "dice" problems, there are no neat formulas.
    Usually, we must list the outcomes (or visualize them).


    The number showing on the upper faces when two fair six-sided dice are tossed, is observed.
    Consider events A and B

    A: the total score is at most 5
    B: at least one die achieves a score of 3 or 4
    I assume we are to find P(A) and P(B).

    (1) There are: . 6 \times 6 \:=\:36 possible outcomes.

    There are 10 outcomes with a sum of 5 or less:
    . . (1,1),\,(1,2),\,(1,3),\,(1,4),\,(2,1),\,(2,2),\,(2  ,3),\,(3,1),\,(3,2),\,(4,1)

    Therefore: . P(A) \;=\;\frac{10}{36} \;=\;\frac{5}{18}


    (2) I have a "back door" approach to the second part.

    The probability that a die does not show a 3 or 4 is: . \frac{4}{6} \,=\,\frac{2}{3}

    The probability that neither die shows a 3 or 4 is: . \frac{2}{3}\!\cdot\!\frac{2}{3}\:=\:\frac{4}{9}


    Therefore, the probability that at least one die shows a 3 or 4 is:

    . . P(B) \;=\;1 - \frac{4}{9}\;=\;\frac{5}{9}

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