**Question:**

A bag contains 24 balls of which x are red, 2x are white and 3x am are. A ball is selected at random. What is the probability that it

(i) not red?

(ii) white

**Solution:**

Given that, A bag contains total number of balls = 24 A bag contains number of red bails = 24

A bag contains number of white balls = 2x and a bag contains number of blue balls = x

By condition, x + 2x + 3x = 24

⇒ 6x = 24

∴ x = 4

∴Number of red balls = x = 4

Number of white balls = 2x = 2 x 4 = 8

and number of blue balls = 3x = 3 x 4 = 12

So, total number of outcomes for a ball is selected at random in a bag contains 24 balls.

⇒ n(S) = 24

(i) Let $E_{1}=$ Event of selecting a ball which is not red $i . e .$, can be white or blue.

$\therefore n\left(E_{1}\right)=$ Number of white balls + Number of blue balls

$\Rightarrow \quad n\left(E_{1}\right)=8+12=20$

$\therefore \quad$ Required probability $=\frac{n\left(E_{1}\right)}{n(S)}=\frac{20}{24}=\frac{5}{6}$

(ii) Let $E_{2}=$ Event of selecting a ball which is white

$\therefore n\left(E_{2}\right)=$ Number of white balls $=8$

So, required probability $=\frac{n\left(E_{2}\right)}{n(S)}=\frac{8}{24}=\frac{1}{3}$