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Math Help - Fun probability question

  1. #1
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    Fun probability question

    A bowl contains w white balls and b black balls. One ball is selected at random from the bowl, its color noted, and is returned to the bowl along with n additonal balls of the same color. Another ball is randomly selected (now containing w + b + n balls) and is observed black. Show that the conditional probability is w/w+b+n.
    Last edited by vexiked; October 1st 2009 at 06:45 AM.
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  2. #2
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    Quote Originally Posted by vexiked View Post
    A bowl contains w white balls and b black balls. One ball is selected at random from the bowl, its color noted, and is returned to the bowl along with n additonal balls of the same color. Another ball is randomly selected (now containing w + b + n balls) and is observed black. Show that the conditional probability is w/w+b+n.
    Again, I suspect no one is quite sure what this question means.
    Read this. “Show that the conditional probability is w/w+b+n”
    Does that make any sense to you?
    “the conditional probability” of what?
    Last edited by mr fantastic; October 1st 2009 at 01:46 PM.
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  3. #3
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    The problem should state:

    A bowl contains w white balls and b black balls. One ball is selected at random from the bowl, its color noted, and is returned to the bowl along with n additonal balls of the same color. Another ball is randomly selected (now containing w + b + n balls) and is observed black. Show that the conditional probability that the first ball selected was white is w/w+b+n.
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  4. #4
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    Quote Originally Posted by vexiked View Post
    A bowl contains w white balls and b black balls. One ball is selected at random from the bowl, its color noted, and is returned to the bowl along with n additonal balls of the same color. Another ball is randomly selected (now containing w + b + n balls) and is observed black. Show that the conditional probability that the first ball selected was white is w/w+b+n.
    You are asked to find P(W_1|B_2) and that is equal to  \frac{{P(B_2 |W_1 )P(W_1 )}}{{P(B_2 |W_1 )P(W_1 ) + P(B_2 |B_1 )P(B_1 )}}.

    I will do part of it for you.  P(B_2 |B_1 )P(B_1 )=\frac{(n+b)}{w+(n+b)}\cdot\frac{b}{w+b}
    Last edited by Plato; October 2nd 2009 at 01:35 AM.
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  5. #5
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    Thank you! This makes things alot more clear for me
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