= 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,
$$
\mathbb P(A_1\cap A_2)
=\frac13\left(\frac13\frac12+\frac13\frac23+\frac12\frac23\right)
=\frac{13}{54}\ne\frac14.
$$
Thus \b[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.
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