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Probability with and without replacement

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data

    An urn contains five balls, one marked WIN and four marked LOSE. You and another player take turns selecting a ball from the urn, one at a time. The first person to select the WIN ball is the winner. If you draw first, find the probability that you will win if the sampling is done

    (a) with replacement - answer = 5/9
    (b) without replacement - answer = 3/5

    3. The attempt at a solution

    (a) P(W) + P(LW) + P(LLW) + . . . . .

    (1/5) + (4/5)(1/5) + (4/5)(4/5)(1/5) . . . .

    so its is a series

    SUM (1/5)(4/5)^(n-1) starting at n=1 to infinity
    when I did this I got 1 and not 5/9

    (b) P(W) + P(LW) + P(LLW) . . .

    (1/5) + (4/5)(1/4) + (4/5)(3/4)(1/3) . . ..

    SUM (1/5)( . . . and I got stuck here

    Any Help would be appreciated

    Thanks
     
  2. jcsd
  3. Sep 13, 2007 #2

    learningphysics

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    You have not considered that the second person is also taking out a ball... you did the problem as if it was just one person...
     
  4. Sep 13, 2007 #3

    EnumaElish

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    For me (b) is easier: Let P(L|x) be the probability of "L" when there are x balls in the urn. P(W|x) = 1 - P(L|x).

    P(W|5) + P(L|5)P(L|4)P(W|3) + the 3rd term = 1/5 + 4/5 3/4 1/3 + the 3rd term.

    Can you guess what the 3rd term is?
     
  5. Sep 13, 2007 #4
    The third term should be:

    P(L|5)P(L|4)P(L|3)P(W|2) = 1/5
     
  6. Sep 13, 2007 #5

    learningphysics

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    Although the third term does come out to 1/5... just wanted to point out it should be:

    P(L|5)P(L|4)P(L|3)P(L|2)P(W|1) = 1/5

    because the expression P(L|5)P(L|4)P(L|3)P(W|2) means the second guy wins not you...
     
  7. Sep 13, 2007 #6

    EnumaElish

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    mutzy188, do you get 3/5 for b?
     
  8. Sep 13, 2007 #7
    Yes I did. Thank you very much for your help:smile:
     
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