A box contains 9 apples of variety A and 7 apples of variety B.

1. Find the probability of drawing two apples of the same variety by considering AA or BB.

Total number of apples $=9+7=16$.

$P(\text{same variety})=P(AA)+P(BB)$

$=\frac{9}{16}\times\frac{8}{15}+\frac{7}{16}\times\frac{6}{15}$

2. Simplify the probability to a fraction in simplest form.

$\frac{9}{16}\times\frac{8}{15}+\frac{7}{16}\times\frac{6}{15}$

$=\frac{72}{240}+\frac{42}{240}$

$=\frac{114}{240}=\frac{19}{40}$

1. Identify the new contents of the box after the apples are returned and variety C apples are added.

After returning the two apples, the box has the original $9+7=16$ apples.

After adding $4$ apples of variety C, total number of apples is
$16+4=20.$

There are $4$ variety C apples and $16$ non-C apples.

2. Find the probability of choosing exactly 2 variety C apples and 1 non-C apple from 3 apples.

$P(\text{exactly 2 C})=\frac{\binom{4}{2}\binom{16}{1}}{\binom{20}{3}}$

$=\frac{6\times16}{1140}$

3. Simplify the probability to a fraction in simplest form.

$\frac{6\times16}{1140}=\frac{96}{1140}=\frac{8}{95}$