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Math Help - Coins Probability

  1. #1
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    Coins Probability

    A hat contains n=f+b coins, f of which are fair, and b of which are biased to land heads with probability 2/3. A coin is drawn from the hat and tossed twice. The …first time it lands heads and the second time it lands tails. What is the probability that it is a fair coin ?

    I am not really sure where to start..
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  2. #2
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    it depends on how many coins are in the hat. if there's 2 coins in a hat then the probability of picking either of them at random is 50% regardless of if they are fair or bias.

    P(f) = f/n
    P(b) = b/n
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  3. #3
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    ya thats what i thought. But i think its more complicated than that since they give the probabilities of the coin picked 1/2 and the probability of the biased coin.
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  4. #4
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    Quote Originally Posted by whatsanihar View Post
    A hat contains n=f+b coins, f of which are fair, and b of which are biased to land heads with probability 2/3. A coin is drawn from the hat and tossed twice. The …first time it lands heads and the second time it lands tails. What is the probability that it is a fair coin ?

    I am not really sure where to start..
    The probability of choosing a fair coin and having it land HT from 2 tosses is

    \frac{f}{f+b}\frac{1}{2}\frac{1}{2}=\frac{f}{4(f+b  )}

    The probability of choosing a biased coin and obtaining HT from 2 throws is

    \frac{b}{f+b}\frac{2}{3}\frac{1}{3}=\frac{2b}{9(f+  b)}

    the probability the coin was fair is the fair coins probability divided by
    the total probability for this scenario.

    Therefore the probability the coin was fair is

    \frac{\frac{f}{4(f+b)}}{\frac{f}{4(f+b)}+\frac{2b}  {9(f+b)}}=\frac{\frac{f}{4(f+b)}}{\frac{9f(f+b)+4(  f+b)2b}{36(f+b)^2}}=\frac{f}{4(f+b)}\frac{36(f+b)}  {9f+8b}=\frac{36f}{36f+32b}=\frac{9f}{9f+8b}
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  5. #5
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    thanks a lot Archie Meade makes sense .
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