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Thread: Conditional probability and independence

  1. #1
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    Conditional probability and independence

    A product can have three kinds of defects: A, B, C. It is rejected if$\displaystyle (A \cap B) \cup C$.
    It is known that $\displaystyle P(A) = 0.15 , P(B) = 0.25 , P(C) = 0.30$.
    It is also known that $\displaystyle P(A|C) = 1.3\cdot P(A)$ and $\displaystyle P(B|C) = 2\cdot P(B)$.
    Also A and B are independent.

    What is $\displaystyle P[(A \cap B) \cup C] $ ?

    Some one can help me with this?

    Thanks in advance
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  2. #2
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    Quote Originally Posted by johanS View Post
    A product can have three kinds of defects: A, B, C. It is rejected if$\displaystyle (A \cap B) \cup C$.
    It is known that $\displaystyle P(A) = 0.15 , P(B) = 0.25 , P(C) = 0.30$.
    It is also known that $\displaystyle P(A|C) = 1.3\cdot P(A)$ and $\displaystyle P(B|C) = 2\cdot P(B)$.
    Also A and B are independent.

    What is $\displaystyle P[(A \cap B) \cup C] $ ?

    Some one can help me with this?

    Thanks in advance
    I don't see it right now but to get you started, you can find $\displaystyle (A \cap B)$ easily because they are independent. So it is just $\displaystyle P(A)P(B)$. You know $\displaystyle P(C)$ So you are left to find $\displaystyle P(A\cap B\cap C)$

    Using the conditional probabilities you can find $\displaystyle P(A\cap C)$ and $\displaystyle P(B\cap C)$

    Try to mess with these and see what you can do.
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  3. #3
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    Quote Originally Posted by jass10816 View Post
    I don't see it right now but to get you started, you can find $\displaystyle (A \cap B)$ easily because they are independent. So it is just $\displaystyle P(A)P(B)$. You know $\displaystyle P(C)$ So you are left to find $\displaystyle P(A\cap B\cap C)$

    Using the conditional probabilities you can find $\displaystyle P(A\cap C)$ and $\displaystyle P(B\cap C)$

    Try to mess with these and see what you can do.
    $\displaystyle P(A\cap B\cap C)$ is exactly my problem
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