Question

From $21$ tickets numbered $1,2,3,........,21$ one ticket is drawn at random. Find the probability that the ticket has a number divisible by $3$.

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{1}{3}$

$S$ be the set of possible outcomes

$n\left(S\right)=21$

Let $E$ be an event of getting a ticket that has a number divisible by $3$

$\therefore E=\{3,6,9,12,15,18,21\}\Rightarrow n\left(E\right)=7$

Hence required probability is $=\frac{7}{21}=\frac{1}{3}$