Question

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

Answer 1

Total no. of possible outcomes = 20 {1, 2, 3, …. , 20}

E ⟶ event of drawing ticket with no multiple of 3 or 7

No. of favourable outcomes = 8 which are {3, 6, 9, 12, 15, 18, 7, 14}

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 8/20 = 2/5