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Math Help - Probability

  1. #1
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    Probability

    A and B are two independent events in the sampling space S. Given that P(A U B) = 5/8 and P(A ∩ B')= 7/24, Find

    (a) P(B)
    (b) P(A ∩ B)
    (c) P(A)
    (d) P(A' U B')


    The answer for a,b,c,d are 1/3, 7/48, 7/16, 41/48 respectively.....I'm not sure how to get those probabilities....
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  2. #2
    MHF Contributor harish21's Avatar
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    Do you know what independent events mean? Look at the definition of Independent Events and then show your attempt at the problem.

    and  P(A \cup B) = P(A)+P(B)-P(A\cap B)
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  3. #3
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    Quote Originally Posted by nav92 View Post
    A and B are two independent events in the sampling space S. Given that P(A U B) = 5/8 and P(A ∩ B')= 7/24, Find

    (a) P(B)
    (b) P(A ∩ B)
    (c) P(A)
    (d) P(A' U B')


    The answer for a,b,c,d are 1/3, 7/48, 7/16, 41/48 respectively.....I'm not sure how to get those probabilities....
    \Pr(A \cup B) = 1 - \Pr(A' \cap B') = \frac{5}{8} \Rightarrow \Pr(A' \cap B') = \frac{3}{8}.

    Let \Pr(A \cap B) = x.

    Now try to draw a Karnaugh Table.

    Note: \Pr(A \cap B) = \Pr(A) \cdot \Pr(B) since A and B are independent and so x = \left(x + \frac{7}{24}\right) \cdot \frac{8}{24} \Rightarrow x = ....
    Last edited by mr fantastic; November 1st 2010 at 01:52 PM.
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  4. #4
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    \frac{5}{8} = P(A) + P(B) - P(A)P(B) = P(B) + P(A)P(B') = P(B)+\frac{7}{{24}}

    Solve for P(B)
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