Question

A bag contains 3 red, 3 white and 3 green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is –
(1) red.
(2) not red
(3) either red or white.

Answer 1

Total number of balls = 3 + 3 + 3
= 9
(1) Number of red balls = 3
Therefore, required probability = $\frac{\mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{balls}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{balls}}$
$$\begin{gathered} =\frac{3}{9} \\ =\frac{1}{3} \end{gathered}$$
Hence, the probability that a red ball is drawn is $\frac{1}{3}$.
(2) Number of balls which are not red = 3 + 3
Therefore, required probability = $\frac{\mathrm{Number}\mathrm{of}\mathrm{not}\mathrm{red}\mathrm{balls}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{balls}}$
$$\begin{gathered} =\frac{6}{9} \\ =\frac{2}{3} \end{gathered}$$
Hence, the probability that a ball drawn is not red is $\frac{2}{3}$.
(3) Number of red and white balls = 3 + 3
Therefore, required probability = $\frac{\mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{and}\mathrm{white}\mathrm{balls}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{balls}}$
$$\begin{gathered} =\frac{6}{9} \\ =\frac{2}{3} \end{gathered}$$
Hence, the probability that a ball drawn is either red or white is $\frac{2}{3}$.