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

Question:

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

(i) If the ball is drawn at random from the bag, what is the probability that it is not red?

(ii) If two more red balls are put in the bag, the probability of drawing a red ball will be $\frac{9}{8}$ times the probability of drawing a red ball in the first case. Find the value of $x$.

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.