A bag contains 3 white balls, 3 black balls & 2 red balls.

My attempt:

I did not get the OA, but here is my understanding...

For the third ball to be a red one, here are the combination.

NR -- Non red ball.
R -- Red ball.

NR -- NR -- R (Non red ball drawn followed by a non-red ball followed by a red ball.)
R -- NR -- R
NR -- R -- R

Case 1: NR -- NR -- R

(1/7) * (1/6) * (1) -- (1/42)

(In the above calculation, I have included one red ball since we need only one red ball to be drawn as the third one.)

Case 2: R -- NR -- R

(1/2)*(1/6)*(1) = (1/12)

Case 3: NR -- R -- R

(1/6) * (1/2) * (1) -- (1/12)

Adding all the probabilities -- (1/42) + (1/12) + (1/12)

Gives me (4/21) --- which is not an option at all


Any thoughts on what went wrong in my approach.