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 the ticket is
(i) a prime number
(ii) a number less than 16
(iii) a number divisible by 3.
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 the ticket is
(i) a prime number
(ii) a number less than 16
(iii) a number divisible by 3.

Solution:

The possible outcomes are {3,5,7,9..…29}

Number of possible outcomes = 14

(i) Let E be the event of getting the number on the ticket is a prime number.

Outcomes favourable to E are {3,5,7,11,13,17,19,23,29}

Number of favourable outcomes = 9

P(E) = 9/14

Hence the probability of getting the number on the ticket is a prime number is 9/14.

(ii) Let E be the event of getting the number on the ticket is less than 16.

Outcomes favourable to E are {3,5,7,9,11,13,15}

Number of favourable outcomes = 7

P(E) = 7/14 = 1/2

Hence the probability of getting the number on the ticket is less than 16 is 1/2.

(iii) Let E be the event of getting the number on the ticket is a number divisible by 3.

Outcomes favourable to E are {3,9,15,21,27}

Number of favourable outcomes = 5

P(E) = 5/14

Hence the probability of getting the number on the ticket is a number divisible by 3 is 5/14.