**Question:**

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?

**Solution:**

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 $)=1-\frac{1}{4}=\frac{3}{4}$