Drawing a Ball!

Relevant wiki: Bayes' Theorem and Conditional Probability

$\large \displaystyle E_1 = \text{First ball drawn is red}\\ \large \displaystyle E_2 = \text{First Ball drawn is black}\\ \large \displaystyle A = \text{Second Ball drawn is red}\\ \large \displaystyle \therefore P(E_1) = \frac{3}{10}, \, P(E_2) = \frac{7}{10}\\ \large \displaystyle P(A/E_1) = \frac{2}{9}, \, P(A/E_2) = \frac{3}{9}$

$\large \displaystyle \implies P(E_1/A) = \frac{P(E_1) \times P(A/E_1)}{P(E_1) \times P(A/E_1) + P(E_2) \times P(A/E_2)}\\ \large \displaystyle = \frac{\frac{3}{10} \times \frac{2}{9}}{\frac{3}{10} \times \frac{2}{9} + \frac{7}{10} \times \frac{3}{9}} = \frac{6}{6+21} = \frac{6}{27} = \color{#3D99F6}{\frac{2}{9}}\\ \large \displaystyle \therefore \frac{a}{b} = \frac{2}{9}\\ \large \displaystyle \implies \color{#D61F06}{a} + \color{#20A900}{b} = \color{#D61F06}{2} + \color{#20A900}{9} = \color{#BA33D6}{\boxed{11}}$

I did it this way

If the second ball is red first can be red or black.

i.e., (RR,RB) these are the events that may happen.

And we can even choose different red and black balls as we have 3 red and 7 black balls.

Now, total no. of combinations for RR = 3 * 2 = 6 {combination : no. of choices in first position * no. of choices in second position}

And total no. of combination for BR = 7 * 3 = 21

Total no. of combinations are = 21 + 6 = 27

Probability of First ball red = {RR}/{RR + BR} = 6/27 = 2/9 ==> a = 2 and b = 9

a + b = 11

Is this way of doing this question valid (OK).