Tickets numbered from 1 to 20 are mixed up and a ticket is drawn at random.

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

Given: Tickets are marked from 1 to 20 are mixed up. One ticket is picked at random.

Required to find: Probability that the ticket bears a multiple of 3 or 7

Total number of cards is 20.

And, the cards marked which is multiple of 3 or 7 are 3, 6, 7, 9, 12, 14, 15 and 18.

So, the total number of cards marked multiple of 3 or 7 is 8.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a card that is a multiple of 3 or 7 is 8/20 = 2/5

Sample space, n(S) = 20 

Number of events of getting a multiples of 3 or 7 on the drawn ticket,

n(E) = 8 {3,6,9,12,15,18,7,14}

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{8}{20}\) = \(\frac{2}5\)