 # A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. A ball is drawn at random from the bag.

Question:

A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is

(i) white or blue,

(ii) neither white nor black.

Solution:

Number of white balls in the bag = 5

Number of red balls in the bag = 7

Number of black balls in the bag = 4

Number of blue balls in the bag = 2

Total number of balls in the bag = 5 + 7 + 4 + 2 = 18

∴ Total number of outcomes = 18

(i) There are 7 balls (5 white and 2 blue) in the bag which are either white or blue. So, there are 7 ways to draw a ball from the bag which is white or blue.

Favourable number of outcomes = 7

$\therefore \mathrm{P}($ Drawn ball is white or blue $)=\frac{\text { Favourable number of outcomes }}{\text { Total number of outcomes }}=\frac{7}{18}$

(ii) There are 9 balls (7 red and 2 blue) in the bag which are neither white nor black. So, there are 9 ways to draw a ball from the bag which is neither white nor black.

Favourable number of outcomes = 9

$\therefore \mathrm{P}($ Drawn ball is neither white nor black $)=\frac{\text { Favourable number of outcomes }}{\text { Total number of outcomes }}=\frac{9}{18}=\frac{1}{2}$