A bag contains 6 white balls and 4 black balls. A ball is drawn and is put back in the bag with 5 balls of the same colour as that of the ball drawn. A ball is drawn again at random. What is the probability that the ball drawn now is white.

Let A, B be the events of drawing white ball and black ball from the bag in 1st
draw respectively.
Here A, B are exclusive and exhaustive

$P\left(A\right)=\frac{6}{10}=\frac{3}{5};P\left(B\right)=\frac{4}{10}=\frac{2}{5}$

Let E be the event of getting white ball in 2nd draw.

Now $P\left(E\right)=P(E\cap S)=P[E\cap (A\cup B\left)\right]$  $(\because A\cup B=S)$

$=P\left[\right(E\cap A)\cup (E\cap B\left)\right]=P(E\cap A)+(E\cap B)$

$=P\left(A\right)P(E/A)+P\left(B\right)P(E/B)$         .... (1)

$P(E/A)=P$(getting white ball when 1st ball drawn from the bag is white)

$=\frac{11}{15}[\because \text{ white ball is put back with }5\text{ white balls }]$

similarly $P(E/B)=\frac{6}{15}$

[since black ball is put back along with 5 black balls]

from (1), $P\left(E\right)=\frac{11}{15}\times \frac{3}{5}+\frac{6}{15}\times \frac{2}{5}=\frac{3}{5}$