 Solution

Each position is equally likely to receive any coin, so the law of total probability gives $\mathbb{P}(A_i)=(1/3+1/2+2/3)/3=1/2$. For two specified positions, the ordered pair of distinct coin biases is uniform over six possibilities. Assuming independent tosses conditional on the order,Thus the events are not independent, even pairwise. The covariance of draws without replacement explains the difference: the head indicator random variables have covariance $13/54-1/4=-1/108$. Independent physical tosses given the assigned coins do not imply independence after that hidden assignment is averaged out.