A bag contains 24 balls, out of which X are white. If one ball is drawn at random, the P(E) of drawing a white ball is Y. 12 more white balls are added to the bag. Now if a ball is drawn, from the bag, the P(E) of drawing a white ball is 5Y/3, find the value of X.

given , total number of balls = 24.

given, white balls = x

p(e) = x/24 = y (given)

ie, x = 24y ------------- (1)

now, given, 12 more white balls are added.

ie , white balls = x + 12

p(e) = x+12/24 = 5y/3 (given)

24y+12/24 = 5y/3 (from eqn - 1)

= y+12 = 5y/3 ( 24 and 24 gets cancelled)

bring 3 to left side.

3(y+12) = 5y

= 3y+36 = 5y

= 36 = 5y-3y

= 36 = 2y

= y = 36/2

y = 18.

now, w.k.t ,

x = 24y (from eqn -1 )

ie, x = 24 x 18

x = 432.