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Math Help - probability

  1. #1
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    probability

    There are one blue, two black and three yellow balls. Two balls are randomly chosen without replacement. Define the following events: One of the balls is blue
    At least one ball is black
    Both balls are yellow
    Both balls are of the same color
    Find the following conditional probabilities:
    (a)
    (b)
    (c) =

    how can i solve this??
    Last edited by makaveli89; October 10th 2008 at 03:29 PM.
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  2. #2
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    (a) 0 (there are no black balls)
    (b) 0 (there are no black balls)
    (c) = 0 (if one is blue then both can not be yellow)

    Unless I misunderstand the question=)
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  3. #3
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    ive updated it, now my question is correct, so how can i solve this
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  4. #4
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    Quote Originally Posted by makaveli89 View Post
    ive updated it, now my question is correct, so how can i solve this
    Quote Originally Posted by makaveli89 View Post
    There are one blue, two black and three yellow balls. Two balls are randomly chosen without replacement. Define the following events: One of the balls is blue

    At least one ball is black

    Both balls are yellow

    Both balls are of the same color

    Find the following conditional probabilities:

    (a)

    (b)

    (c) =

    how can i solve this??
    The first thing you should do is draw a tree diagram. Have you done this?

    Then it's much easier to see that:



    \Pr(B | A) = \frac{\Pr(B \cap A)}{\Pr(A)} = \frac{\Pr(\text{Black ball and blue ball})}{\Pr(A)} = \frac{\left(\frac{1}{6}\right) \cdot \left(\frac{2}{5}\right) + \left(\frac{2}{6}\right) \cdot \left(\frac{1}{5}\right)}{\frac{1}{6} + \left(\frac{2}{6}\right) \cdot \left(\frac{1}{5}\right) + \left(\frac{3}{6}\right) \cdot \left(\frac{1}{5}\right)} = \, .....



    \Pr(D | B) = \frac{\Pr(D \cap B)}{\Pr(B)} = \frac{\Pr(\text{Both balls are black})}{\Pr(B)} = \frac{ \left(\frac{2}{6}\right) \cdot \left(\frac{1}{5}\right) }{\left(\frac{1}{6}\right) \cdot \left(\frac{2}{5}\right) + \frac{2}{6} + \left(\frac{3}{6}\right) \cdot \left(\frac{2}{5}\right) } = \, ....


    Pr(C | A) = 0.
    Last edited by mr fantastic; October 10th 2008 at 04:47 PM.
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  5. #5
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    wow thank you its correct. but where did u get all the numbers from?
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  6. #6
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    Quote Originally Posted by makaveli89 View Post
    wow thank you its correct. but where did u get all the numbers from?
    Draw the tree diagram like I suggested. It will help you see where the numbers came from.
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