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Math Help - Probability

  1. #1
    Senior Member Mukilab's Avatar
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    Probability

    Amy plays one game of snooker and one game of billards each friday. The probability that she wins every friday is as follows

    Snooker: Win=1/3

    Billards: Wind=3/4

    If she plays one game of each for a number of fridays, calculate how many games Amy did not win at either if she won at both on 21 fridays

    So I did

    1/4=21/x (x being the amount of games played)

    so x=84

    then since 1/6 is the probability that she loses both I did
    1/6=z/84 z being the number of fridays

    z=14

    correct?
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  2. #2
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    Hello, Mukilab!

    Your method is excellent . . . and your answer is correct!

    I'll show you a method I often use.


    Amy plays one game of snooker and one game of billards each Friday.

    The probability that she wins every Friday is as follows:
    . . Snooker: Win = 1/3
    . - Billards: Win = 3/4

    If she plays one game of each for a number of Fridays,
    calculate how many games Amy did not win at either
    if she won at both on 21 Fridays.

    The probability that Amy wins both games on a particular Friday is:
    . . P(\text{win both}) \:=\:\tfrac{1}{3}\cdot\tfrac{3}{4} \:=\:\tfrac{1}{4}

    Since she won both games 21 times, we assume she played on 84 Fridays.


    Place the information in a chart . . .

    . . \begin{array}{c||c|c||c|}<br />
& \text{win B} & \text{lose B} & \text{Total} \\ \hline \hline <br />
\text{win S} & 21 & & \\ \hline<br />
\text{lose S} & & & \\ \hline\hline<br />
\text{Total} & & & 84 \end{array}



    We know that she wins \tfrac{1}{3} of her snooker games.
    . . \text{win S} \:=\:\tfrac{1}{3}(84) \:=\:28

    We know that she wins \tfrac{3}{4} of her billiards games.
    . . \text{win B} \:=\:\tfrac{3}{4}(84) \:=\:63

    Enter that information in our chart:

    . . \begin{array}{c||c|c||c|}<br />
& \text{win B} & \text{lose B} & \text{Total} \\ \hline \hline <br />
\text{win S} & 21 & & 28\\ \hline<br />
\text{lose S} & & & \\ \hline\hline<br />
\text{Total} & 63 & & 84 \end{array}


    Now you can supply the missiing numbers and answer the question.

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  3. #3
    Senior Member Mukilab's Avatar
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    Thanks, that method is far clearer!
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