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Math Help - Conditional Probability #3

  1. #1
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    Conditional Probability #3

    Given two bags, where the first bag contains 3 gold coins and two silver coins and the second bag contains 4 gold coins and 6 coins, determine the probability that a coin chosen randomly between the two bags in gold.

    Let A be the event of choosing from the first bag
    B be the event of choosing from the second bag
    C be the event of choosing a gold coin.

    The probability of choosing a gold coin between the 2 bags is

    p(A\cap C) + p(B\cap C)=p(A)\cdot p(C|A) + p(B)\cdot p(C|B)

    Why does p(C|A)=\frac{3}{5}?, isn't it \frac {1}{2}? Or did they just write the answer reversed (which they shouldn't do)?
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  2. #2
    kin
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    the first bag contains totally 5 coins,
    if u take the a gold coin from it ,
    p(C|A)=3/5
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  3. #3
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    Why is p(A)=\frac{1}{2}?
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  4. #4
    kin
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    there are two bags ,
    one is bag A, one is bag B,
    so P(A) = 1/2
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  5. #5
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    p(A) is \frac{1}{2} because there are 2 bags???
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  6. #6
    kin
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    Quote Originally Posted by chengbin View Post
    p(A) is \frac{1}{2} because there are 2 bags???
    exactly
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