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.
Best Answer
No, you have correctly employed Bayes' Rule and the Law of Total Probability to arrive at the correct answer, so there is nothing left to help you with.
Good work.