A bag contains 10 balls: 4 white and 6 black.

A ball is removed at random and replaced; a second ball is then removed.

What is the probability that the balls removed were the same colour?

(Answer is in its simplest form.)

first case

probability of white ball in first replacement P (w) = 4/10 probability of black ball in first replacement P (b) = 6/10 probability of white ball in second replacement P' (w) = 3/9 probability of black ball in second replacement P' (b) = 5/9

Step-by-step explanation:

now for,

the probability that the balls removed were the same colour = P (w) * P' (w) + P' (b) * P (b)

=7/15