Tickets numbered 3, 5, 7, 9, …., 29 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on ticket is

(i) a prime number

(ii) a number less than 16

(iii) a number divisible by 3.

Tickets are mixed thoroughly and a ticket is drawn at random from the box means that all the outcomes are equally likely.

Sample space = {3, 5, 7, 9, …., 29}, which has 14 equally likely outcomes.

(i) Let E₁ be the event of choosing ticket with prime number.

E₁ = {3, 5, 7, 11, 13, 17, 19, 23, 29}.

∴ The number of favourable outcomes to the event E₁ = 9.

$\therefore P(E$1) = \dfrac{\text{No. of favourable outcomes to } E1}{\text{Total no. of possible outcomes}} = \dfrac{9}{14}.

Hence, the probability of choosing a ticket with prime number is $\dfrac{9}{14}$.

(ii) Let E₂ be the event of choosing ticket with number less than 16.

E₂ = {3, 5, 7, 9, 11, 13, 15}.

∴ The number of favourable outcomes to the event E₂ = 7.

$\therefore P(E$2) = \dfrac{\text{No. of favourable outcomes to } E2}{\text{Total no. of possible outcomes}} = \dfrac{7}{14} = \dfrac{1}{2}.

Hence, the probability of choosing a ticket with number less than 16 is $\dfrac{1}{2}$.

(iii) Let E₃ be the event of choosing ticket with number divisible by 3.

E₃ = {3, 9, 15, 21, 27}.

∴ The number of favourable outcomes to the event E₃ = 5.

$\therefore P(E$3) = \dfrac{\text{No. of favourable outcomes to } E3}{\text{Total no. of possible outcomes}} = \dfrac{5}{14}.

Hence, the probability of choosing a ticket with number divisible by 3 is $\dfrac{5}{14}$.