A box has 20 pots of plants.
There are 5 pink pots, 9 blue pots and the remaining pots are green.

1. Find the probability of drawing one pink and one blue in either order.

There are 5 pink pots and 9 blue pots out of 20 pots.

Since the pots are taken without replacement,
$P(\text{pink then blue})=\frac{5}{20}\times\frac{9}{19}$
$P(\text{blue then pink})=\frac{9}{20}\times\frac{5}{19}$

2. Add the two possible orders and simplify the fraction.

$P(\text{one pink and one blue})=\frac{5}{20}\times\frac{9}{19}+\frac{9}{20}\times\frac{5}{19}$
$=\frac{45}{380}+\frac{45}{380}$
$=\frac{90}{380}=\frac{9}{38}$

1. Find the numbers of green and non-green pots in the full box.

Since the two pots are returned, the box again has $20$ pots.

Number of green pots:
$20-5-9=6$

Number of non-green pots:
$5+9=14$

2. Find the probability of getting exactly one green pot in three draws without replacement.

Only one pot is green means: $1$ green and $2$ non-green.

$\frac{\binom{6}{1}\binom{14}{2}}{\binom{20}{3}}$

$=\frac{6\times 91}{1140}$

$=\frac{546}{1140}$

3. Simplify the probability fraction.

$\frac{546}{1140}=\frac{273}{570}=\frac{91}{190}$