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 5 ? \[\]
A.$\dfrac{1}{2}$\[\]
B.$\dfrac{7}{20}$\[\]
C.$\dfrac{9}{20}$\[\]
D. None of these \[\]

Answer 1

Hint: We find the number of tickets with number either multiple or 3 or 5 by counting the multiples of 3 or 5 under or equal to 20 as the number of favourable outcomes$n\left( A \right)$. We find the number of possible outcomes $n\left( S \right)$as the total number of tickets. The required probability is $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$.\[\]

Complete step by step answer:
We know from definition of probability that if there is $n\left( A \right)$ number of ways of event $A$ occurring (or number of favourable outcomes) and $n\left( S \right)$ is the size of the sample space (number of all possible outcomes) then the probability of the event $A$ occurring is given by
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}\]
We are given the question that tickets numbered from 1 to 20 are mixed up and a ticket is drawn at random. We can select any one ticket out of 20 numbered tickets. So the number of all possible outcomes is
\[n\left( S \right)=20\]
We are asked to find out the probability that the ticket drawn has a number which is a multiple of 3 or 5. Let us denote the event of getting a ticket with a number which is a multiple of 3 or 5 as $A$. We see that the multiples of 3 under or equal to 20 are 3, 6, 9,12,15,18 and multiples 5 under or equal to 20 are 5, 10, 15 and 20. So the event $A$ can happen if we have a ticket with number 3, 5, 6, 9, 10,12,15,18, 20. So the number of favourable outcome is
\[n\left( A \right)=9\]
So the required probability is
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{9}{20}\]

So, the correct answer is “Option C”.

Note: We note that we should use the common multiples only once like 15 in this problem. We can alternatively solve by finding the ticket has number 3 as $P\left( A \right)$, the ticket has number 5 as $P\left( B \right)$ , then the required probability $P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$ where the $P\left( A\bigcap B \right)$ is the probability of getting the ticket with a number that is a multiple of both 3 and 5.