In a box there are 4 white balls and 6 black balls. A ball is drawn at random. If it is white, it is put back along with two more white balls in the box. If it is black, it is put back in the box and then two black balls are thrown out of the box. Now a ball is drawn again at random from the box. Then, what is the probability that it is black?

Correct option is A

Given:
Initial: W = 4, B = 6, total = 10.
First draw white with probability 4/10; first draw black with probability 6/10. After the first draw the composition changes depending on colour, and then a second draw is made.

Formula / Logic Used:
Use total probability:
P(final black) = P(first white)·P(black | first white) + P(first black)·P(black | first black).

Solution:
P(first white) = 4/10 = 2/5.
If white is drawn: return it and add 2 whites → W = 4 + 2 = 6, B = 6, total = 12.
So P(black | first white) = 6/12 = 1/2.

P(first black) = 6/10 = 3/5.
If black is drawn: return it, then remove two blacks → W = 4, B = 6 − 2 = 4, total = 8.
So P(black | first black) = 4/8 = 1/2.

Therefore
P(final black) = (2/5)·(1/2) + (3/5)·(1/2) = (1/5) + (3/10) = (2/10 + 3/10) = 5/10 = 1/2.

Correct Answer: (a) 1/2