Question

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

Answer 1

SOLUTION :  

Given : Tickets marked with numbers from 1 to 20

Total number of outcomes = 20

Let E = Event of getting a multiple of 3

Tickets marked with multiple of 3 are : 3, 6, 9, 12, 15, 18

Number of outcome favourable to E = 6

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

P(E) = 6/20 = 3/10

Hence, the required probability of getting a multiple of 3,  P(E) = 3/10 .

HOPE THIS ANSWER WILL HELP YOU ...

Answer 2

p(e)=no of favourable outcomes/total no of possible outcomes
p(getting multiple of 3)=6/20
or 3/10
or o.3 answer...hope it helps..(TT)