Person A has a fair coin (only: head/tail) and an unfair coin (head/head)
He chooses one coin randomly, throws it and it lands on head.
He throws the same coin again and it lands on head again!
What is the probability he chose the fair coin
If he was to throw the coin only once, then according to bayes law it is: $\frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$ However I do not know how to consider the second throw, what do I do?
Thanks!
Use Bayes' Law again. Only the values will differ.
You seek the probability that the coin is fair, event $F$, given evidence of throwing two consecutive heads, event $E$. Because the selection of the coin is presumably unbiased ($\mathsf P(F)=\mathsf P(F^\complement)$) we then have.
$$\begin{align}\mathsf P(F\mid E)&=\dfrac{\mathsf P(E\mid F)\,\mathsf P(F)}{\mathsf P(E\mid F)\,\mathsf P(F)+\mathsf P(E\mid F^\complement)\,\mathsf P(F^\complement)}\\[2ex]&=\dfrac{\mathsf P(E\mid F)}{\mathsf P(E\mid F)+\mathsf P(E\mid F^\complement)}\end{align}$$
$\mathsf P(E\mid F)$ is the probability for obtaining two heads when twice tossing a fair coin.
$\mathsf P(E\mid F^\complement)$ is the probability for obtaining two heads when twice tossing a double-headed coin. …