Bag A has 4 green counters, 6 red counters and 2 yellow counters.

1. Find the probability of taking a green counter on one draw and raise it to the power $n$ for repeated draws with replacement.

Total number of counters in Bag A:
$4+6+2=12$
Probability of taking a green counter in one draw:
$\frac{4}{12}=\frac{1}{3}$
Since the counter is replaced each time, the probability is the same for each of the $n$ draws.

Probability of taking only green counters in $n$ draws:
$\left(\frac{1}{3}\right)^n$

1. Find the probabilities of drawing a yellow counter and a non-yellow counter from Bag A.

Bag A has $4+6+2=12$ counters.

$P(Y)=\frac{2}{12}=\frac{1}{6}$

$P(\text{not }Y)=\frac{10}{12}=\frac{5}{6}$

2. Add the probabilities of the two mutually exclusive orders giving exactly one yellow.

Exactly one yellow can occur as:

$Y\text{ then not }Y \quad \text{or} \quad \text{not }Y\text{ then }Y$

With replacement, the probabilities on the two draws are unchanged.

$P(\text{exactly one yellow})=\frac{1}{6}\times\frac{5}{6}+\frac{5}{6}\times\frac{1}{6}$

$=\frac{5}{36}+\frac{5}{36}=\frac{10}{36}=\frac{5}{18}$

1. Identify the favourable cases that leave Bag A unchanged.

Bag A will have its original contents again only if the counter taken from Bag B and placed into Bag A is the same colour as the counter first taken from Bag A.

Possible favourable cases:

$G$ from A then $G$ from B, or $R$ from A then $R$ from B, or $Y$ from A then $Y$ from B.

2. Find the probability of each favourable case.

Initially, Bag A has $12$ counters and Bag B has $6$ counters.

For green:
$P(G\text{ from A then }G\text{ from B})=\frac{4}{12}\times\frac{4}{7}=\frac{4}{21}.$

For red:
$P(R\text{ from A then }R\text{ from B})=\frac{6}{12}\times\frac{4}{7}=\frac{2}{7}.$

For yellow:
$P(Y\text{ from A then }Y\text{ from B})=\frac{2}{12}\times\frac{1}{7}=\frac{1}{42}.$

3. Add the mutually exclusive favourable probabilities.