A gambler has in his pocket a fair coin and a biased coin which will land heads with probability $\frac34$.
He selects one of the coins at random; when he tosses it, it lands heads.
What is the probability it is the fair coin?
(II)If he tosses the same coin a second time, and again it lands heads. What now
is the probability it is the fair coin?
(III) If he tosses the same coin for a third time, and this time it lands tails. What now is the probability it is the fair coin?
My solution using Bayes:
(I)$$P(\text{fair}|\text{heads})=\frac{P(\text{fair}\cap\text{heads})}{P(\text{heads})}=\frac{\frac12\cdot\frac12}{\frac12\cdot\frac12+\frac12\cdot\frac34}=\frac{\frac14}{\frac58}=\frac25$$
(II) $$P(\text{fair}|\text{heads,heads})=\frac{P(\text{fair}\cap\text{heads}\cap\text{heads})}{P(\text{heads}\cap\text{heads})}=\frac{\frac12\cdot\frac12\cdot\frac12}{\frac12\cdot\frac12\cdot\frac12+\frac12\cdot\frac34\cdot\frac34}=\frac{4}{13}$$
(III) $$P(\text{fair}|\text{heads,heads,Tails})=\frac{P(\text{fair}\cap\text{heads}\cap\text{heads}\cap\text{tails})}{P(\text{heads}\cap\text{heads}\cap\text{tails})}=\frac{\frac12\cdot\frac12\cdot\frac12\cdot\frac12}{\frac12\cdot\frac12\cdot\frac12\cdot\frac12+\frac12\cdot\frac34\cdot\frac34\cdot\frac14}=\frac{8}{17}$$ Please can someone help me if my understanding is correct.