I don't know how to approach this other than using Bayes' rule. Let $A$ be the event that the coin is fair and let $B$ be the event of getting two heads. We get
$$P(A \mid B) = \frac{P(A\cap B)}{P(B)} = \frac{1/4}{P(B)}.$$
I don't know how to get $P(B)$. Any suggestions?
Also, how would I generalize this — say, if we get $n$ heads rather than $2$ heads?
Best Answer
In any realistic model, the distribution of the probability of heads for a physical coin (given a particular flipping method) should be continuous. Therefore the probability that the coin is exactly fair should be 0.