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Math Help - How to solve this probability question? Please help

  1. #1
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    How to solve this probability question? Please help

    A box contains 5 white balls and 3 red balls. 2 players, A and B, take turns at drawing one ball from the box at random, and balls drawn are not replaced. The play who first gets 2 red balls is the winner, and the drawing stops as soon as either player has drawn 2 balls. Player A draw first. Find the probability that
    i) play A is the winner on his second draw
    ii) play A is the winner, given that the winning player wins on his second draw.
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  2. #2
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    Quote Originally Posted by azuki View Post
    A box contains 5 white balls and 3 red balls. 2 players, A and B, take turns at drawing one ball from the box at random, and balls drawn are not replaced. The play who first gets 2 red balls is the winner, and the drawing stops as soon as either player has drawn 2 balls. Player A draw first. Find the probability that
    i) play A is the winner on his second draw
    ii) play A is the winner, given that the winning player wins on his second draw.
    For the first part, you just calculate in order the odds of what HAS to happen for player A to win on his second draw.
    So, first he has to draw a red ball, the odds of which are 3/8. Then, in one situation Player B draws a white, which is a 5/7 chance. Lastly A has to choose another Red, which would be a 2/6 chance. \frac{3}{8} * \frac{5}{7} * \frac{2}{6} = \frac{30}{336}

    However, B could have chosen a red ball on his draw, the odds of which were 2/7. Then, for B to get the last red ball would have been a 1/6 chance. \frac{3}{8} * \frac{2}{7} * \frac{1}{6} = \frac{6}{336} So the total chance is \frac{36}{336} = \frac{3}{28}
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  3. #3
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    The second part is somewhat similiar, but there are three different ways in which B could win on his second go, it'll take a little calculating.
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  4. #4
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    a

    Quote Originally Posted by JeWiSh View Post
    For the first part, you just calculate in order the odds of what HAS to happen for player A to win on his second draw.
    So, first he has to draw a red ball, the odds of which are 3/8. Then, in one situation Player B draws a white, which is a 5/7 chance. Lastly A has to choose another Red, which would be a 2/6 chance. \frac{3}{8} * \frac{5}{7} * \frac{2}{6} = \frac{30}{336}

    However, B could have chosen a red ball on his draw, the odds of which were 2/7. Then, for B to get the last red ball would have been a 1/6 chance. \frac{3}{8} * \frac{2}{7} * \frac{1}{6} = \frac{6}{336} So the total chance is \frac{36}{336} = \frac{3}{28}
    ii) Total chances to win in second draw:
    For A there is \frac{3}{28} chance as calculated by JeWiSh.
    Lets calculate for B:
    1) First A draws a white ball than B draws a red ball again A draws a white ball
    and B draws a red ball, prob. = \frac{5}{8}*\frac{3}{7}*\frac{4}{6}*\frac{2}{5} =  \frac{1}{14}
    2) First A draws a white ball than B draws a red ball again A draws a red ball now
    and B draws a red ball, prob. =  \frac{5}{8}*\frac{3}{7}*\frac{2}{6}*\frac{1}{5} = \frac{1}{56}
    3) First A draws a red ball than B draws a red ball, A draws a white ball
    and B draws a red ball, prob. =  \frac{3}{8}*\frac{2}{7}*\frac{5}{6}*\frac{1}{5} = \frac{1}{56}

    Now add these probabilities =  \frac{1}{14}+\frac{1}{56}+\frac{1}{56} = \frac{3}{28}

    prob. that A wins = \frac{\frac{3}{28}}{\frac{3}{28}+\frac{3}{28}} = \frac{1}{2}
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