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Math Help - probability question

  1. #1
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    probability question

    A large bag contains 1200 red marbles and 2001 blue marbles. Joe and Lisa will play a game in which they draw a single marble from the bag with replacement. For every game they will bet $35. Lisa wins the money if the marble is red and Jow wins if it's blue. If they play this game for 7 straigth days at an average rate of one game per every 10 seconds, who should win money and exactly how much should he/she expect to win?

    I have a data management test tomorrow((

    I dunno if that's going to help but it's binomial distribution and we're suposed to use this formula

    (n)*p^k*q^n-k
    (k)

    I tried to solve it and the best solution I came up with is

    sorry it's not 2001 blue marbles, it's 1201. sorry about the typo

    (2001/2401)*604800*35=10592820

    I'm not sure if it's right thou
    Last edited by anna12345; December 17th 2008 at 03:10 AM.
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  2. #2
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    Quote Originally Posted by anna12345 View Post
    A large bag contains 1200 red marbles and 2001 blue marbles. Joe and Lisa will play a game in which they draw a single marble from the bag with replacement. For every game they will bet $35. Lisa wins the money if the marble is red and Jow wins if it's blue. If they play this game for 7 straigth days at an average rate of one game per every 10 seconds, who should win money and exactly how much should he/she expect to win?

    I have a data management test tomorrow((

    I dunno if that's going to help but it's binomial distribution and we're suposed to use this formula

    (n)*p^k*q^n-k
    (k)

    I tried to solve it and the best solution I came up with is

    (2001/2401)*604800*35=10592820

    I'm not sure if it's right thou
    n = (6)(60)(24)(7) = ....

    p = Pr(Red in a single trial) = 1200/(1200 + 2001).

    Now consider the expected number of wins (and losses) for red.
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  3. #3
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    Hello, anna12345!

    A large bag contains 1200 red marbles and 2001 blue marbles.
    Joe and Lisa will play a game in which they draw a single marble from the bag with replacement.
    For every game they will bet $35.
    Lisa wins the money if the marble is red, and Jow wine if it's blue.
    If they play this game for 7 days at a rate of one game every 10 seconds,
    who should win money and exactly how much should he/she expect to win?

    Just "eyeballing" the problem, we see that Joe has the advantage.
    . . (There are more blue marbles than red marbles.)


    There is a total of 3201 marbles: 1200 red and 2001 blue.

    P(\text{red}) \:=\:\frac{1200}{3201} \:\approx\:\frac{3}{8}

    P(\text{blue}) \:=\:\frac{2001}{3201} \:\approx\:\frac{5}{8}


    Joe wins \tfrac{5}{8} of the time and loses \tfrac{3}{8} of the time.
    He comes out ahead \tfrac{5}{8} - \tfrac{3}{8} \:=\:\tfrac{2}{8} \:=\:\tfrac{1}{4} of the time.

    In seven days, they will play 60,480 games.

    Joe comes out ahead in: \tfrac{1}{4} \times 60,\!480 \:=\:15,\!120 games.

    At $35 per game, Joe expects to win: . 15,\!120 \times \$35 \:=\:\$529,\!200



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