Question

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number, which is a multiple of 7, is [CBSE 2013C]

(a) $\frac{1}{7}$
(b) $\frac{1}{8}$
(c) $\frac{1}{5}$
(d) $\frac{7}{40}$

Answer 1

​Total number of tickets = 40.

Out of the given numbers, multiples of 7 are 7, 14, 21, 28 and 35.

Numbers of favourable outcomes = 5.

∴ P(getting a number which is a multiple of 7) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{5}{40}=\frac{1}{8}$

Thus, the probability that the selected ticket has a number, which is a multiple of 7, is $\frac{1}{8}$.

Hence, the correct answer is option (b).