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