Painting golf balls

There are $3^6$ ways the balls could have been chosen for painting (taking order into account)

Of these there are $\binom{6}{2}$ ways that one of them could have received 2 coats of paint. And the remaining two could have received 2 coats in $\binom{4}{2}$ ways.

So, the total probability of all three receiving 2 coats of paint is:

$P = \dfrac{\binom{6}{2}\cdot\binom{4}{2}}{3^6} = \dfrac{10}{81}$

$10+81 = \boxed{91}$

There are $3^{6}=729$ ways to choose the balls.

Of these, there are $\frac{6!}{2!\cdot 2!\cdot 2!}=90$ arrangements of AABBCC.

$\frac{90}{729}=\frac{10}{81}$ , so the solution is $10+81=\boxed{91}$

Number the balls 1, 2, and 3. The probability of choosing ball 1 twice is ${6\choose 2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^4 = 15\times\frac{1}{9}\times\frac{16}{81} = \frac{80}{243}$

Of the remaining two balls, the probability of choosing ball 2 twice is then ${4\choose 2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^2 = 6\times\frac{1}{4}\times\frac{1}{4} = \frac{3}{8}$

The probability of both these events happening is $\frac{80}{243}\times\frac{3}{8} = \frac{10}{81}$

$10+81 = \boxed{91}$