A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) white?
(ii) red?
(iii) black?
(iv) not red?
Number of red balls $=3$
Number of black balls $=5$
Number of white balls $=4$
Total number of balls $=3+5+4=12$
Therefore, the total number of cases is 12 .
(i) Since there are 4 white balls, the number of favourable outcomes is 4 .
$\mathrm{P}($ a white ball $)=\frac{\text { Number of favourable cases }}{\text { Total number of cases }}=\frac{4}{12}=\frac{1}{3}$
(ii) Since there are 3 red balls, the number of favourable outcomes is 3 .
$\mathrm{P}($ a red ball $)=\frac{\text { Number of favourable cases }}{\text { Total number of cases }}=\frac{3}{12}=\frac{1}{4}$
(iii) Since there are 5 black balls, the number of favourable outcomes is 5 .
$\mathrm{P}($ a black ball $)=\frac{\text { Number of favourable cases }}{\text { Total number of cases }}=\frac{5}{12}$
(iv) $\mathrm{P}($ not a red ball $)=1-\mathrm{P}($ a red ball $)=1-\frac{1}{4}=\frac{3}{4}$