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Probability all balls are white.

  1. Dec 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A bag contains 4 balls. Two balls are drawn at random, and are found to be white. What is the probability that all balls are white?

    2. Relevent equations
    At school, I'm currently learning Bayes' theorem, probability disribution and Bernoulli trials.

    3. The attempt at a solution
    The 4 balls can be
    1) all white
    2) 3 white, 1 non-white
    3) 2 white, 2 non-white
    4) 1 white, 3 non-white
    5) all non-white

    P(all white)=1/5

    But given, two balls taken at random are white.

    I can't figure out what to do next. When I ignore the 2 white balls, I get the answer as 1/3 which is wrong according to my text book.

    [Edit]: I got it!
    A-2 balls taken are white
    E1-4W
    E2-3W, 1Non-white
    E3-2W, 2N
    E4-1W, 3N
    E5-4N

    P(E1/A)= {P(E1)P(A/E1)} / {P(E1)P(A/E1)+...+P(E5)P(A/E5)}

    ={1/5*1} / {1/5*1 + 1/5*3C2/4C2 + 1/5*2C2/4C2 + 1/5*0 + 1/5*0}
    =3/5
     
    Last edited: Dec 18, 2012
  2. jcsd
  3. Dec 18, 2012 #2

    mfb

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    Does the problem statement say that all 5 cases have the same probability?
    Otherwise, you cannot know this.
     
  4. Dec 18, 2012 #3

    phinds

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    I'm not too good at this, so may have it wrong, but it seems to me that once you have pulled 2 balls and they are white, what you have is a situation where you have 2 balls (remaining) and you want to know what are the odds that they are both white.

    This assumes that the starting condition becomes irrelevant once you draw the first two, and that's where I may be going wrong, though I don't think so.
     
  5. Dec 18, 2012 #4

    haruspex

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    As Bayes showed, the problem is not solvable without plugging in some a priori probabilities. E.g. you could argue that, by and large, bags of balls tended to be, about equally often say, mixed colours or all the same colour. This would be for reasons related to why balls are put in bags in the first place.
    Another reasonable choice says each ball is independently white or non-white, but since there are many possible colours these might not be with prob 1/2. Let's say a ball is white with prob p. Now we find that drawing two white balls tells us nothing about the remaining two (as phinds surmised). The prob that they are both white is p2.
     
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