A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is

Let A₁, A₂ denote the events of drawing a red ball, a black ball respectively from the bag containing 4 red balls and 6 black balls$\therefore P\left(A_1\right)=\frac{4}{10}\text{ and }P\left(A_2\right)=\frac{6}{10}$

Let E be the event of drawing a red ball after adding two balls. 

Then $P\left(E\mid A_1\right)=\frac{6}{12}\text{ and }P\left(E\mid A_2\right)=\frac{4}{12}$

Required probability $=P\left(A_1\right)P\left(E\mid A_1\right)+P\left(A_2\right)P\left(E\mid A_2\right)=\frac{4}{10}\times \frac{6}{12}+\frac{6}{10}\times \frac{4}{12}=\frac{2}{5}$