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Math Help - stuck on probability problem

  1. #1
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    Unhappy stuck on probability problem

    I'm stuck on this question. It is an expected value question involving marbles:

    "We have 20 marbles in a bag: 17 red and 3 blue. We randomly pull three marbles without replacement. If we pull three red marbles, we pay $10. In any other case we receive $10 for every blue marble pulled. (1 blue gets $10, 2 blue gets $20....) What is the expected value of this game?

    I'm thrown off by the third marble.
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  2. #2
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    Hey

    Quote Originally Posted by jlefholtz View Post
    I'm stuck on this question. It is an expected value question involving marbles:

    "We have 20 marbles in a bag: 17 red and 3 blue. We randomly pull three marbles without replacement. If we pull three red marbles, we pay $10. In any other case we receive $10 for every blue marble pulled. (1 blue gets $10, 2 blue gets $20....) What is the expected value of this game?

    I'm thrown off by the third marble.
    You need to find the probabilities first

    p("three red") =  \frac{17}{20}*\frac{16}{19}*\frac{15}{18}
    p("one blue") = prob(blue,red,red)+p(red,blue,red)+p(red,red,blue) = \frac{3}{20}*\frac{17}{19}*\frac{16}{18} + \frac{17}{20}*\frac{3}{19}*\frac{16}{18}+\frac{17}  {20}*\frac{16}{19}*\frac{3}{18}

    (I hope in this formula is no typo in it)

    p("two blue") = p(red,blue,blue)+p(blue,red,blue)+p(blue,blue,red)

    p("three blue") = p(blue,blue,blue)

    I think you can solve it yourself

    and then E(x) = -10*p(three red) + 10*p(one blue) + 20*p(two blue)+30*p(three blue)

    Rapha
    Last edited by Rapha; November 23rd 2008 at 07:06 PM.
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  3. #3
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    the answer i got.

    This is what I did, I'm not sure if I did it correctly though.

    P (3 red) + P (3 blue) + P(2 red, 1 blue) + P (2 blue, 1 red)=

    (17/20 x 16/19 x 15/18)(-$10) + (3/20 x 2/19 x 1/18)($30) + (17/20 x 16/19 x 3/18)($10) + (17/20 x 3/19 x 2/18)($20)= -$4.44
    Last edited by jlefholtz; November 23rd 2008 at 06:54 PM. Reason: typo
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  4. #4
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    thanks rapha! by the looks of the reply you gave me, it looks like i did it right. thank you for your reply.
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  5. #5
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    Quote Originally Posted by jlefholtz View Post
    This is what I did, I'm not sure if I did it correctly though.

    P (3 red) + P (3 blue) + P(2 red, 1 blue) + P (2 blue, 1 red)=

    (17/20 x 16/19 x 15/18)(-$10) + (3/20 x 2/19 x 1/18)($30) + (17/20 x 16/19 x 3/18)($10) + (17/20 x 3/19 x 2/18)($20)= -$4.44
    Im not sure about it, because i got


    P (3 red)*(-10) + P (3 blue)*(30) + P(2 red, 1 blue)*(10) + P (2 blue, 1 red)*(20)=

    17/20*16/19*15/18*(-10) + 3/20*2/19*1/18*(30) + 3*(17/20*16/19*3/18)*(10) + 3*(3/20*2/19*17/18)*20

    = - 1.47

    You forgot 3* P(2 red, 1 blue) ... 3* P(2blue, 1 red)

    because there are three different possibilities, (blue, red, red), (red,blue,red), (red,red,blue). Thus 3*p(blue,red,red) = p(1blue, 2 reds)
    Last edited by Rapha; November 23rd 2008 at 07:27 PM.
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  6. #6
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    oops

    oh, wow, thanks for catching that. i didn't check my work. thanks again.
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