A box contains $4$ white and $5$ black balls. A ball is drawn at random and its colour is noted. A ball is then put back in the box along with two additional balls of its opposite colour. If a ball is drawn again from the box, then the probability that the ball drawn now is black, is

Consider the following events:

    $E_1 :$Ball drawn from the box is white

    $E_2 :$Ball drawn from the box is black

   $A =$Getting a black ball from the box after replacing the drawn ball with two balls of opposite colour

  $P\left(E_1\right)=\frac{4}{9},P\left(E_2\right)=\frac{5}{9},P\left(A/E_1\right)=\frac{7}{11},P\left(A/E_2\right)=\frac{5}{11}$

$\begin{array}{r}\therefore P\left(A\right)=P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)\\ =\frac{4}{9}\times \frac{7}{11}+\frac{5}{9}\times \frac{5}{11}=\frac{53}{99}\end{array}$