Solution
= Solution
Use the <law of total probability> according to the first draw. Without replacement, the second-draw red <probability> is $1/6$ after a red and $2/6$ after a green. Thus
$$
\boxed{\mathbb P(E_2)=\frac27\frac16+\frac57\frac26=\frac27.}
$$
Equivalently, each counter has the same <probability> of occupying the second position as the first. Equal marginal <probabilities> do not make the two draw events <independent>.