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Math Help - Probability and Bayes' formula

  1. #1
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    [SOLVED] Probability and Bayes' formula

    I need these solutions in 3 hours.

    I've 2 boxes. THe first box has 2 new and 3 old balls. The second box has 3 new and 4 old balls. I took one ball from the first box and put it into the second box. Now I take the ball from the second box.
    What's the probability that:
    1) The ball taken from the second box is new.
    2) The ball taken from the first box was new if I know that the ball taken from the second box was old.

    Answers
    1) 17/40
    2) 8/23
    Last edited by totalnewbie; September 28th 2008 at 06:09 AM. Reason: solved
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  2. #2
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    1)
    2 cases where the ball is a new one
    4 new 4 old -> 1/2
    3 cases where it's an old one
    3 new 5 old -> 3/8
    5 cases total
    2/5*1/2 + 3/5*3/8 = 17/40

    too tired for the second one sorry
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  3. #3
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    Quote Originally Posted by totalnewbie View Post
    I need these solutions in 3 hours.

    I've 2 boxes. THe first box has 2 new and 3 old balls. The second box has 3 new and 4 old balls. I took one ball from the first box and put it into the second box. Now I take the ball from the second box.

    [snip]

    2) The ball taken from the first box was new if I know that the ball taken from the second box was old.

    Answers
    1) 17/40
    2) 8/23
    Using the result from 1), Pr(Old ball taken from box 2) = 1 - (17/40) = 23/40.

    Using Bayes Theorem: \Pr \left( \, \text{New ball taken from box 1} ~ | ~ \text{old ball taken from box 2} \, \right) = \frac{ \left( \frac{2}{5}\right) \, \left( \frac{1}{2}\right)}{\frac{23}{40}} = \, ....
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  4. #4
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    Quote Originally Posted by jbpellerin View Post
    1)
    2 cases where the ball is a new one
    4 new 4 old -> 1/2
    3 cases where it's an old one
    3 new 5 old -> 3/8
    5 cases total
    2/5*1/2 + 3/5*3/8 = 17/40

    too tired for the second one sorry
    I am solving different exercises and I understand how 2/5*1/2 + 3/5*3/8 comes but it's unclear how it equals to 17/40. It actually equals to 23/40. 17/40+23/40=1 Therefore these numbers are related, but how?
    It seems to me that one of these is the old ball (23/40) taken and the second one of these is the new ball (17/40) taken? How to make difference?
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  5. #5
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    Quote Originally Posted by totalnewbie View Post
    I am solving different exercises and I understand how 2/5*1/2 + 3/5*3/8 comes but it's unclear how it equals to 17/40. It actually equals to 23/40. 17/40+23/40=1 Therefore these numbers are related, but how?
    It seems to me that one of these is the old ball (23/40) taken and the second one of these is the new ball (17/40) taken? How to make difference?
    Pr(New ball taken from box 2) = 2/5*1/2 + 3/5*3/8 = 2/10 + 9/40 = 8/40 + 9/40 = 17/40 so I'm not sure what your problem is here .....
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Pr(New ball taken from box 2) = 2/5*1/2 + 3/5*3/8 = 2/10 + 9/40 = 8/40 + 9/40 = 17/40 so I'm not sure what your problem is here .....
    Sorry, my bad!
    I got that point now! I just wrongly substituted 3/8 by 5/8 and didn't notice it was wrong substitution.
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