Question:
A bag contains 5 white, 6 red and 4 green balls. One ball is drawn at random. What is the probability that the ball drawn is
(i) green?
(ii) white?
(iii) non-red?
Solution:
Total number of balls $=5+6+4=15$
(i) Number of green balls $=4$
$\therefore \mathrm{P}_{(\text {green ball })}=\frac{4}{15}$
(ii) Number of white balls $=5$
$\therefore \mathrm{P}_{(\text {white }}$ ball $)=\frac{5}{15}=\frac{1}{3}$
(iii) Number of balls that are not red (i.e., 5 white and 4 green) $=9$
$\therefore \mathrm{P}_{(\text {non-red balls })}=\frac{9}{15}=\frac{3}{5}$