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Indentifying the ball

  1. Feb 21, 2009 #1
    we have 3 balls: 1 red, 1 green and 1 half red half green.

    1-the probability to pick a certain ball of these 3 is:1/3

    2-if after picking the ball, we were told that the ball contains the colour red (for exe.), then the probability of telling which ball it is becomes:1/2
    (same thing for the colour green).

    then why cant we by just picking the ball consider that it contains a certain colour and by that the probability of question 1 would become 1/2?
  2. jcsd
  3. Feb 21, 2009 #2


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    The trick is in the first part of the second statement: if ....
    You have some extra information, which changes the probabilities. We are talking about conditional probability here.

    In the first case, you are picking any ball with equal probability from a set of 3 (R, G and R/G). So for each one, the probability is 1/3.
    Now suppose that I tell you have the ball contains the colour red, and I ask you, what is the probability that it is completely red? Clearly, it is not 1/3 anymore. By giving you some information I have changed the experiment. Do you agree that a completely equivalent experiment would be if I took out the green ball and then asked you to compute the probability that you take the red ball? Well, there are two balls (R and R/G) with equal probability, so you have 1/2 probability of picking it.

    Another famous example of conditional probabilities is the three-door example. Suppose you are on a TV-show and the quiz master presents you with three closed doors. There is a grand prize behind only one of them. You pick a door. Then the quiz master opens one of the other two doors, and shows that there is no prize behind it. He asks you if you want to switch or stay with your original choice. Should you stick to your door, switch to the other one, or doesn't it matter?

    Surprisingly, the answer is not the latter. By opening a door and showing that there is no prize, you have been given extra information. So the probabilities are now conditional, and the chance of the prize being behind door 1 is not the same as the chance of the prize being behind door 1 given that it is not behind door 2.
  4. Feb 21, 2009 #3


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    It's not clear to me what you mean by "consider that it contains a certain colour". Do you mean pick a ball and then, without looking at it, say "it contains red". Obviously that does not make the probability that it is a specific ball equal to 1/2 because we might be wrong- we might have chosen the green ball.
    I by "consider that it contains a certain colour" you simply mean "it must contain some color and it doesn't matter which", that is also wrong. My guess as to which ball it is certainly does depend upon whether I know it "contains" the colour red or the colour green.

    If I pick one of the three balls and I am told it "contains" red, then I know i tis not the green ball- it must be either completely red or half red and half blue. Those are equally likely so the probability is 1/2.
    If I am told that it "contains" green, then I know that it must not be the red ball- it must be either the green ball or half red half green. Again, the probability that it is one of those is 1/2.
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