Here we are having a total of 1000 Balls, out of which we firstly draw 100 balls , and then 101st ball..
Firstly we have to find expected number of white and black balls in drawn 100 balls , as both can occur in 100 balls..
We have a situation like this:
Expected number of white balls = n*W/N = 100*(100/1000) = 10
Expected number of black balls = n*B/N = 100*(900/1000) = 90
So, we have drawn 100 balls(90 black, 10 white)
Left balls = (810 Black , 90 white) = 900 total
probability for 101st ball to be black = 810/900 = 9/10
So, option (A) is Correct