Question

Tickets numbered 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?

(a)$\dfrac{1}{{15}}$
(b)$\dfrac{1}{2}$
(c). $\dfrac{2}{5}$
(d)$\dfrac{7}{{20}}$

Answer 1

Hint: Find the number of multiples of 3 and 7 between the numbers 1 to 20. Thus favourable outcome is the total number multiples of 3 and 7 between 1 to 20. Required probability equals favourable outcome by total number of tickets.

Complete step-by-step answer:
It is said that the tickets are numbered from 1 to 20. Thus we can say that the total number of tickets are 20.
Now these 20 tickets are mixed up and one ticket among these 20 tickets is drawn randomly. Now we need to find the probability that the ticket number drawn is a multiple of 3 or 7.
Let us write down the multiples of 3 and 7 from 1 to 20 are,
Multiples of 3 = $3,6,9,12,15,18$.
Multiples of 7 = $7,14$.
Thus there are 6 numbers that are multiple of three from 1 to 20. Similarly, there are 2 numbers that are multiple of 7. Therefore, the favourable outcome is equal to,
Favourable outcome = Number of multiples of 3 + Number of multiples of 7
= $6 + 2 = 8$
Hence favourable outcome = 8.
Thus the required probability is equal to favourable outcome by total number of tickets.
Required probability = total favourable outcome
Total number of tickets
Thus required probability = $\dfrac{8}{{20}} = \dfrac{2}{5}$.
Thus we got the required probability as, $\dfrac{2}{5}$.
Thus the probability that the ticket drawn has a number which is a multiple of 3 or 7 = $\dfrac{2}{5}$.
Therefore, option $(c)$ is the correct answer.

Note: You can also solve this with the help of set theory, let S be the total number of tickets then,
$S = \\{ 1,2,3,....,20\\}$
Let E be the event of getting a multiple of 3 or 7. Thus we get E as,
$E = \\{ 3,6,7,9,12,14,15,18\\}$.
Now required probability = $P(E)$
$P(E) = \dfrac{{n(E)}}{{n(S)}} = \dfrac{8}{{20}} = \dfrac{2}{5}$
Hence required probability = $P(E) = \dfrac{2}{5}$.