A bag contains 18 balls out of which x balls are red.

Solution:

Total number of balls = 18.
Number of red balls = x.

(i) Number of balls which are not red = 18 − x

$\therefore \mathrm{P}($ getting a ball which is not red) $)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$

$=\frac{18-x}{18}$

Thus, the probability of drawing a ball which is not red is $\frac{18-x}{18}$.

(ii) Now, total number of balls = 18 + 2 = 20.
Number of red balls now = x + 2.

$P($ getting a red ball now $)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$

$=\frac{x+2}{20}$

and $P$ (getting a red ball in first case) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$

$=\frac{x}{18}$

Since, it is given that probability of drawing a red ball now will be $\frac{9}{8}$ times the probability of drawing a red ball in the first case.

Thus, $\frac{x+2}{20}=\frac{9}{8} \times \frac{x}{18}$

$\Rightarrow 144(x+2)=180 x$

$\Rightarrow 144 x+288=180 x$

$\Rightarrow 36 x=288$

 

$\Rightarrow x=\frac{288}{36}=8$

Thus, the value of x is 8.