A box contains a certain number of balls. Some of these balls are marked A, some are marked B and the remainings are marked C. When a ball is drawn at random from the box P(A) = $\dfrac{1}{3}$ and P(B) = $\dfrac{1}{4}$. If there are 40 balls in the box which are marked C, find the number of balls in the box.

Since, A, B and C are mutually exclusive events.

∴ P(A) + P(B) + P(C) = 1

$\Rightarrow \dfrac{1}{3} + \dfrac{1}{4} + P(C) = 1 \\[1em] \Rightarrow \dfrac{4 + 3}{12} + P(C) = 1 \\[1em] \Rightarrow \dfrac{7}{12} + P(C) = 1 \\[1em] \Rightarrow P(C) = 1 - \dfrac{7}{12} \\[1em] \Rightarrow P(C) = \dfrac{12- 7}{12} \\[1em] \Rightarrow P(C) = \dfrac{5}{12}.$

We know that,

Probability of drawing ball marked C = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of total balls}}$

$\Rightarrow \dfrac{5}{12} = \dfrac{40}{\text{No. of total balls}} \\[1em] \Rightarrow \text{No. of total balls} = \dfrac{40 \times 12}{5} \\[1em] \Rightarrow \text{No. of total balls} = 8 \times 12 \\[1em] \Rightarrow \text{No. of total balls} = 96.$

Hence, there are 96 balls in the box.