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Question: Tickets numbered from 1 to 20 are mixed up and a ticket is drawn at random What is the probability t...

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

Explanation

Solution

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 outcomesn(A)n\left( A \right). We find the number of possible outcomes n(S)n\left( S \right)as the total number of tickets. The required probability is P(A)=n(A)n(S)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(A)n\left( A \right) number of ways of event AA occurring (or number of favourable outcomes) and n(S)n\left( S \right) is the size of the sample space (number of all possible outcomes) then the probability of the event AA occurring is given by
P(A)=n(A)n(S)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(S)=20n\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 AA. 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 AA 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(A)=9n\left( A \right)=9
So the required probability is
P(A)=n(A)n(S)=920P\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(A)P\left( A \right), the ticket has number 5 as P(B)P\left( B \right) , then the required probability P(A⋃B)=P(A)+P(B)−P(A⋂B)P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right) where the P(A⋂B)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.