Question

A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is
(i) Black (ii) Red (iii) Not green.

Answer 1

Total number of balls in the bag = 5 + 8 + 4 + 7 = 24
$\therefore$ Total number of elementary events = 24
(i) There are 7 black balls in the bag.
$\therefore$ Favourable number of elementary events = 7
Hence, P (Getting a black ball) = $\frac{7}{24}$ .
(ii) There are 5 red balls in the bag.
$\therefore$ Favourable number of elementary events = 5
Hence, P (Getting a red ball) =$\frac{5}{24}$ .
(iii) There are 5 + 8 + 7 = 20 balls which are not green. $\therefore$ Favourable number of elementary events = 20 Hence, P (No getting a green ball) = $\frac{20}{24}=\frac{5}{6}$