A Difficult Choice!

I solved it a bit differently.

This problem assumes that each of the four balls was sort of "magically generated"... i.e. Each has a 1/3 probability of being each of the 3 colors.

So, the fact that the first two were in the container is immaterial, and each has a 1/3 probability of being right.

Therefore the probability they are both red is $\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

$1+9 = \boxed{10}$

Since each ball can be of any of the three colors, there are total $3 \times 3 \times 3 \times 3 = 81$ possible combinations. The total number of combinations with at least two red balls can be calculated as given below.

The total number of combinations with exactly two red balls $N(2)$ can be calculated as this. The two balls out of four (which are to be colored red) can be chosen in $^4C_2$ ways. Then each of the remaining two balls can have two possible colors : Black or White. Thus, $N(2) = ^4C_2 \times 2 \times 2 = 24$ .

Similarly, we have,

$N(3) = 8$ and $N(4) = 1$ .

Suppose that the given combination contains $n_{r}$ red balls with $n_r > 1$ . Then the probability that from this combination the first drawn ball comes out red is $\frac{n_{r}}{4}$ and then the box contains 3 balls with $n_r - 1$ red balls. Thus, the probability that the second drawn ball is also red is: $\frac{n_r}{4} \times \frac{n_r-1}{3}$ . Combining all these, the total probability that two red balls are drawn from the original box is:

$P(\text{two reds}) = \frac{24}{81} \times \frac{2}{4} \times \frac{1}{3} + \frac{8}{81} \times \frac{3}{4} \times \frac{2}{3} + \frac{1}{81} = \frac{1}{9}$ .

Hence, $p + q = 1 + 9 = 10$

My answer turned out to be 7 so enlighten me:

Calculation of sample space :

Let there be x reds, y whites and z blacks so x+y+z=4 => ${6\choose2}{4\choose2}$ Ways of choosing 2 balls.

Favorable Cases:

Where the 3 ways of 2 Red are : The other 2 can be i) Both Black ii)Both White ii)1 of each

this gives total of 15 cases out of a total possible 90 cases i.e probability is $\frac{1}{6}$