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.
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.
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 $)=\frac{\text { Favourable number of outcomes }}{\text { Total number of outcomes }}=\frac{9}{18}=\frac{1}{2}$