**Question:**

A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:

(i) white

(ii) red

(iii) not black

(iv) red or white

**Solution:**

Number of red balls $=4$

Number of black balls $=5$

Number of white balls $=6$

Total number of balls in the bag $=4+5+6=15$

Therefore, the total number of cases is 15 .

(ii) Let A denote the event of getting a white ball.

Number of favourable outcomes, i.e. number of white balls $=6$

$\mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{6}{15}=\frac{2}{5}$

(ii) Let B denote the event of getting a red ball.

Number of favourable outcomes, i.e. number of red balls $=4$

$P(A)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{6}{15}=\frac{2}{5}$

(iii) Let $\mathrm{C}$ denote the event of getting a black ball.

Number of favourable outcomes, i.e. number of black balls $=5$

$P(C)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{5}{15}=\frac{1}{3}$

Therefore, the probabilty of not getting a black ball is as follows:

$\mathrm{P}(\overline{\mathrm{C}})=1-\mathrm{P}(\mathrm{C})=1-\frac{1}{3}=\frac{2}{3}$

(vi) Let $D$ denote the event of getting a red or a white ball.

$\mathrm{P}(\mathrm{D})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{4+6}{15}=\frac{10}{15}=\frac{2}{5}$