Question

Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. Then, the probability that the ticket has a number which is a multiple of 3 or 7 is _____.

• A. $\frac{5}{8}$
• B. $\frac{3}{8}$
• C. $\frac{2}{5}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{2}{5}$

Tickets numbered from 1 to 20,
hence total no. of outcomes = 20
Out of 20 tickets numbered from 1 to 20, tickets bearing numbers which are multiples of 3 or multiples of 7 are $3,6,7,9,12,14,15and18.$
$\therefore$ The number of favourable outcomes = 8
Hence, required probability $=\frac{Totalnumberoffavourableoutcomes}{Totalnumberofoutcomes}=\frac{8}{20}$
$=\frac{2}{5}$