IMO A.

I used 1 - x probability formula to solve this.

First find total possibilities

8!/(6!*2!) = 28

Then find all possibilities of no white balls out of the two picked

Both Red = 1 possibility (there are only 2)

Both Black = 3 possibilities (3!/2!)

One Black and One Red

First Red = 3 possibilities
Second Red = 3 possibilities

Total Possibilities of no white = 1+3+3+3 = 10

28 - 10 = 18 Total Possibilities of at least one white

Total number of ways of choosing 2 balls is 8C2 = 8!/6!2! = 28

Prob of picking a white ball first time = 3/8

Prob of picking a white ball second time (having not picked it first time) = 5/8 * 3/7 = 15/56

Total prob of picking at least one white ball = 3/8 + 15/56 = (21+15)/56 = 36/56 = 18/28

Number of ways of picking a white ball = 28 * 18/28 = 18

Long winded but it gets you the right answer. My prob knowledge isnt that good.

Thanks, great explanation