Question

Tickets numbered $1$ to $20$ are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of $3$?

• A. $\frac{3}{10}$
• B. $\frac{3}{20}$
• C. $\frac{2}{5}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{3}{10}$

No. of tickets$=20$

No. of tickets with number which is multiple of $3=6$

$\therefore$Probability$=\frac{{}^6{C}_1}{{}^{20}{C}_1}=\frac{6}{20}=\frac{3}{10}$