Compare sequential draws with unordered counter pairs

Let $G_i,W_i$ mean green or white on draw i. First-draw probabilities are 3/5 and 2/5. After green, second-draw probabilities are 2/4 green and 2/4 white; after white, they are 3/4 green and 1/4 white. Each outgoing pair sums to 1.

$P(G_1\cap W_2)+P(W_1\cap G_2)=(3/5)(2/4)+(2/5)(3/4)=3/5$. These are disjoint paths.

$P(\text{at least one green})=1-P(W_1\cap W_2)=1-(2/5)(1/4)=9/10$.

With two draws, at least one white and at least one green means exactly one of each. Thus $P(\text{at least one white}\mid\text{at least one green})=(3/5)/(9/10)=2/3$. This integrates the Teacher annotations pointing to part (i) for the numerator and part (ii) for the denominator.

Label the individual counters G1,G2,G3,W1,W2. These are physical identities, distinct from draw-index event notation. The 10 unordered pairs are equally likely, so the common denominator is $\binom52=10$.

There are $\binom31\binom21=6$ favourable pairs: choose one of the 3 green counters and one of the 2 white counters. The probability is 6/10=3/5.

There are $10-\binom22=9$ pairs with at least one green, giving 9/10.

The favourable overlap is the same one-green-one-white event, with 6 of the 9 eligible pairs. Equivalently reuse the printed ratio (3/5)/(9/10)=2/3. Both methods count the same physical experiment with consistent outcome spaces.