Question:
There are 25 tickets numbered 1, 2, 3, 4,..., 25 respectively. One ticket is draw 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) $\frac{13}{25}$
Solution:
(c) $\frac{12}{25}$
Explanation:
Total number of tickets = 25
Let E be the event of getting a multiple of 3 or 5.
Then,
Multiples of 3 are 3, 6, 9, 12, 15, 18, 21 and 24.
Multiples of 5 are 5, 10, 15, 20 and 25.
Number of favourable outcomes = ( 8 + 5 − 1) = 12
$\therefore P($ getting a multiple of 3 or 5$)=P(\mathrm{E})=\frac{12}{25}$