A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is $$p$$. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colours is $$q$$. If $$p : q = m : n$$, where $$m$$ and $$n$$ are co-prime, then $$m + n$$ is equal to

A bag contains 6 balls of different colours.

Now, finding $$p$$ (probability both balls same colour):

Two balls drawn with replacement. The probability that both are the same colour:

$$ p = 6 \times \left(\frac{1}{6}\right)^2 = \frac{6}{36} = \frac{1}{6} $$

Now, finding $$q$$ (probability exactly 3 balls same colour out of 4):

Four balls drawn with replacement. We need exactly 3 of the same colour.

Choose which colour appears 3 times: 6 ways.

Choose which 3 positions out of 4: $$\binom{4}{3} = 4$$ ways.

The 3 chosen positions have probability $$\left(\frac{1}{6}\right)^3$$ each for that colour.

The remaining position must be a different colour: probability $$\frac{5}{6}$$.

$$ q = 6 \times \binom{4}{3} \times \left(\frac{1}{6}\right)^3 \times \frac{5}{6} = 6 \times 4 \times \frac{5}{6^4} = \frac{120}{1296} = \frac{5}{54} $$

Now, finding the ratio $$p : q$$:

$$ \frac{p}{q} = \frac{1/6}{5/54} = \frac{1}{6} \times \frac{54}{5} = \frac{54}{30} = \frac{9}{5} $$

So $$p : q = 9 : 5$$, meaning $$m = 9$$ and $$n = 5$$ (which are co-prime).

$$ m + n = 9 + 5 = 14 $$