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Math Help - Probability and Stats

  1. #1
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    Probability and Stats

    I have two questions that I need help on.
    1. Bag M contains 5 white balls and 2 red balls and bag N contains 3 white balls and 4 red balls and you randomly select a ball from Bag M and place it in Bag N. If a white ball is selected from Bag N, what is the probability that a red ball was transferred from Bag M to Bag N.

    2. There are 10 horses in a race. A particular bet requires a customer to choose the first three horses in the correct finishing order. If all 10 horses have an equal chance of finishing in any position, determine the probability that a single bet wins.

    Thanks Bye
    ~G
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  2. #2
    Member SengNee's Avatar
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    Quote Originally Posted by the-G View Post
    I have two questions that I need help on.
    1. Bag M contains 5 white balls and 2 red balls and bag N contains 3 white balls and 4 red balls and you randomly select a ball from Bag M and place it in Bag N. If a white ball is selected from Bag N, what is the probability that a red ball was transferred from Bag M to Bag N.

    2. There are 10 horses in a race. A particular bet requires a customer to choose the first three horses in the correct finishing order. If all 10 horses have an equal chance of finishing in any position, determine the probability that a single bet wins.

    Thanks Bye
    ~G
    1) P(Red ^M)
     =\frac {n(Red)}{n(All)}
     =\frac {2}{7}

    2) P(Win)
     =\frac {n(Win\quad order)}{n(All\quad order)}
     =\frac {1}{10P3}
     =\frac {1}{\frac {10!}{7!}}
     =\frac {1}{720}

    I'm not sure... Just try...
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  3. #3
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    1) P(Red^M)
    =\frac {n(Red)}{n(All)}
    =\frac {2}{7}
    We have an additional piece of information that we can use: a white ball was selected from bag N.

    Before we take the ball out, there is Sengnee's 2/7 chance of there being 5 red balls and 3 white balls (case one), and a 1-2/7=5/7 chance of there being 4 red balls and 4 white balls (case 2).
    So taking a white ball out of case one has probability \frac {3}{5+3} * \frac {2}{7} = \frac {3}{28} and the probability of taking a white ball out of case 2 is \frac {5}{7} * \frac {1}{2} = \frac {5}{14}
    So knowing we took a white ball out, the probability that a red ball was transferred is the probability of taking a white ball out of case one divided by the total probability of taking out a white ball ie. \frac {3/28} {3/28+5/14} = \frac {3}{13}
    Last edited by badgerigar; January 10th 2008 at 04:21 PM. Reason: fixed latex syntax error
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  4. #4
    Member SengNee's Avatar
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    Quote Originally Posted by badgerigar View Post
    We have an additional piece of information that we can use: a white ball was selected from bag N.

    Before we take the ball out, there is Sengnee's 2/7 chance of there being 5 red balls and 3 white balls (case one), and a 1-2/7=5/7 chance of there being 4 red balls and 4 white balls (case 2).
    So taking a white ball out of case one has probability \frac {3}{5+3} * \frac {2}{7} = \frac {3}{28} and the probability of taking a white ball out of case 2 is \frac {5}{7} * \frac {1}{2} = \frac {5}{14}
    So knowing we took a white ball out, the probability that a red ball was transferred is the probability of taking a white ball out of case one divided by the total probability of taking out a white ball ie. \frac {\frac {3}{28}}{\frac{3}{28}}+\frac {5}{14}} = \frac {3}{13}

    I don't understand...
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  5. #5
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    sorry I have fixed the latex error now.
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