Question

Tickets numbers 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

SOLUTION :  

GIVEN: Tickets are marked from 1 to 20 are mixed up.

Total number of outcomes = 20

Let E = Event  of getting multiple of 3 or 7.

Cards marked with multiple of 3 or 7 are 3, 6, 7, 9, 12, 14, 15 and 18

Number of outcome favourable to E = 8

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 8/20 = ⅖  

Hence, the required probability of getting a multiple of 3 or 7, P(E) = ⅖ .

HOPE THIS ANSWER WILL HELP  YOU….

Answer 2

multiple of 3=3,6,9,12,15,18
multiple of 7=7,14
so there are 6 and 2 multiples of 3 and 7 so probability =3+7=10
probability= 10/21
plz mark as brainlest and thanks me