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
(a) $\frac{1}{7}$
(b) $\frac{1}{8}$
(c) $\frac{1}{5}$
(d) $\frac{7}{40}$
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.
$\therefore P($ getting a number which is a multiple of 7$)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$
$=\frac{5}{40}=\frac{1}{8}$
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).
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