# Math Help - Coins Probability

1. ## 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..

2. 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

3. 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.

4. Originally Posted by whatsanihar
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

$\frac{\frac{f}{4(f+b)}}{\frac{f}{4(f+b)}+\frac{2b} {9(f+b)}}=\frac{\frac{f}{4(f+b)}}{\frac{9f(f+b)+4( f+b)2b}{36(f+b)^2}}=\frac{f}{4(f+b)}\frac{36(f+b)} {9f+8b}=\frac{36f}{36f+32b}=\frac{9f}{9f+8b}$

5. thanks a lot Archie Meade makes sense .