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 isOriginally Posted by whatsanihar
The probability of choosing a biased coin and obtaining HT from 2 throws is
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{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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