A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls. A ball is drawn at random from the bag. Find the probability that it is
(i) black,
(ii) not green,
(iii) red or white,
(iv) neither red nor green.
Total number of balls = 15
(i) Number of black balls = 2
$\therefore P($ getting a black ball $)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$
$=\frac{2}{15}$
Thus, the probability of getting a black ball is $\frac{2}{15}$.
(ii) Number of balls which are not green = 4 + 5 + 2 = 11
$\therefore \mathrm{P}$ (getting a ball which is not green) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$
$=\frac{11}{15}$
Thus, the probability of getting a ball which is not green is $\frac{11}{15}$.
(iii) Number of balls which are either red or white = 4 + 5 = 9
$\therefore P$ (getting a ball which is red or white) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$
$=\frac{9}{15}=\frac{3}{5}$
Thus, the probability of getting a ball which is red or white is $\frac{3}{5}$.
(iv) Number of balls which are neither red nor green = 4 + 2 = 6
$\therefore \mathrm{P}$ (getting a ball which is neither red nor green) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$
$=\frac{6}{15}=\frac{2}{5}$
Thus, the probability of getting a ball which is neither red nor green is $\frac{2}{5}$.