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Math Help - Probability Questions

  1. #1
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    Probability Questions

    A bag contains 4 Red and 3 Green balls. Helen and Tony play a game where, starting with Helen, they alternately draw a ball at random and do not replace it. The game …finishes when a Red ball is drawn.

    It is decided that each player will receive, from the other, n units if they draw the Red ball on the nth draw. Construct a table for the random variable X representing Helen’s winnings: thus X = 1 if Helen draws a Red on the fi…rst draw, -2 if Tony draws it on the next, etc.
    Calculate E(X). How much, approximately, would Helen win in a series of 100 games?
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  2. #2
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    Hello, mr_motivator!

    A bag contains 4 Red and 3 Green balls.
    Helen and Tony play a game where, starting with Helen,
    they alternately draw a ball at random and do not replace it.
    The game finishes when a Red ball is drawn.

    It is decided that each player will receive, from the other n units
    if they draw the Red ball on the n^{th} draw.

    Construct a table for the random variable X representing Helenís winnings:
    thus X = 1 if Helen draws a Red on the first draw,
    \text{-}2 if Tony draws it on the next, etc.

    Calculate E(X). How much, approximately, would Helen win in a series of 100 games?
    The tree diagram looks like this . . .

    . . . \begin{array}{ccccc}<br />
& * \\<br />
& \frac{4}{7}\swarrow\searrow\frac{3}{7} \\<br />
& R \qquad G \\<br />
& \qquad\;\; \frac{4}{6}\swarrow\searrow\frac{2}{6} \\<br />
& \qquad\;\;\; R \qquad G \\<br />
& \qquad\qquad\quad\;\; \frac{4}{5}\swarrow\searrow\frac{1}{5} \\<br />
& \qquad\qquad\quad\;\; R \qquad G \\<br />
& \qquad\qquad\qquad\qquad\quad \downarrow\frac{4}{4} \\<br />
& \qquad\qquad\qquad\qquad R<br />
\end{array}


    Helen's winnings look like this:

    . . \begin{array}{|c|ccc|}\hline<br />
X & \text{Prob} & &\\ \hline \hline<br />
+1 & \frac{4}{7} & = & \frac{4}{7}\\ \\[-4mm]<br />
-2 & \frac{3}{7}\cdot\frac{4}{6} &=& \frac{2}{7} \\ \\[-4mm]<br />
+3 & \frac{3}{7}\cdot\frac{2}{6}\cdot\frac{4}{5} &=& \frac{4}{35} \\ \\[-4mm] <br />
-4 & \frac{3}{7}\cdot\frac{2}{6}\cdot\frac{1}{5}\cdot\f  rac{4}{4} &=&\frac{1}{35}\\ \end{array}


    E(X) \;=\;\tfrac{4}{7}(+1) + \tfrac{2}{7}(-2) + \tfrac{4}{35}(+3) + \tfrac{1}{35}(-4) \;=\;\frac{8}{35}

    She can expect to win an average of \tfrac{8}{35} units per game.


    In 100 games, she would win about: . 100\cdot\tfrac{8}{35} \:=\:22\tfrac{6}{7} units.

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