A bag contains 5 white, 6 red and 4 green balls. One ball is drawn at random.

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) $=9$

$\therefore \mathrm{P}_{(\text {non-red balls })}=\frac{9}{15}=\frac{3}{5}$

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