Question

There are 25 tickets numbered as 1, 2, 3, 4,..., 25 respectively. One ticket is drawn at random. What is the probability that the number on the ticket is a multiple of 3 or 5?
(a) $\frac{2}{5}$ (b) $\frac{11}{25}$ (c) $\frac{12}{25}$ (d) $\frac{13}{25}$

Answer 1

The total outcomes be 25

A be event of getting a ticket which has a number which is a multiple of 3 or 5

no. of outcomes fav to A = 12(3,5,6,9,10,12,15,18,20,21,24,25 )

probablity of A = 12/25

answer :- (c) 12/25