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Math Help - probabilities

  1. #1
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    probabilities

    13 . show that:
    i. P(A∩B∩C)=P(A|B∩C)P(B|C)P(C)
    ------------------------------------------------------------------------------------------------------
    14.two boxes contain the number of the ball as indicated. a box is chosen randomly and then two balls are
    drawn One after the other, from that box
    , without Replacement .what is the ....................W B W B
    probability that both Of the drawn balls ...............6 4 5 5
    have the same color?
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    15. 14.two boxes contain the number of the ball as indicated.aball is drawn from box 1 and put in box 2 .a ball
    Is then drawn randomly from box 2 .what
    Is the probability that the ball drawn is ...............................W B W B
    White ? .................................................. .....................m1 m2 m1 m2
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  2. #2
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    Quote Originally Posted by compufatwa View Post
    13 . show that:
    i. P(A∩B∩C)=P(A|B∩C)P(B|C)P(C)
    P(B|C) = \frac{P(B\cap C)}{P(C)}
    P(A|B\cap C) = \frac{P(A\cap B\cap C)}{P(B\cap C)}
    Thus,
    P(A|B\cap C)P(B|C)P(C) = \frac{P(A\cap B\cap C)}{P(B\cap C)} \cdot \frac{P(B\cap C)}{P(C)} \cdot P(C) = P(A\cap B\cap C)


    I do not understand 14 and 15.
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  3. #3
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    In #14, let’s name one box X containing six white balls and four black balls.
    Then box Y contains five of each color.
    P(X) = P(Y) = \frac{1}{2} and let W_1 W_2 denote the first and second balls drawn are white.

    Here is what the question asks: P\left( {W_1 W_2  \cup B_1 B_2 } \right).

    Because they come from either X or Y we have
    \begin{array}{l} P\left( {W_1 W_2  \cup B_1 B_2 } \right)  =  P\left( {W_1 W_2 X) + P(B_1 B_2 X} \right) + P\left( {W_1 W_2 Y) + P(B_1 B_2 Y} \right) \\ <br />
   =  P\left( {W_1 W_2 |X)P(X) + P(B_1 B_2 |X} \right)P(X) + P\left( {W_1 W_2 |Y)P(Y) + P(B_1 B_2 |Y} \right)P(Y) \\  \end{array}

    But since each box is equally likely the becomes
    P\left( {W_1 W_2  \cup B_1 B_2 } \right) = \frac{1}{2}\left[ {\frac{{6 \cdot 5}}{{10 \cdot 9}} + \frac{{4 \cdot 3}}{{10 \cdot 9}} + \frac{{5 \cdot 4}}{{10 \cdot 9}} + \frac{{5 \cdot 4}}{{10 \cdot 9}}} \right]

    You can do #15 the same way.
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