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

A
$\frac{7}{11}$
B
$\frac{5}{11}$
C
$\frac{53}{99}$
D
$\frac{48}{99}$

Solution:

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

$P (E_{1}) = \frac{4}{9}, P (E_{2}) = \frac{5}{9}, P (A / E_{1}) = \frac{7}{11}, P (A / E_{2}) = \frac{5}{11}$

$\begin{aligned} ∴ P (A) = P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) \\ = \frac{4}{9} \times \frac{7}{11} + \frac{5}{9} \times \frac{5}{11} = \frac{53}{99} \end{aligned}$

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

Solution:

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