Question

Tickets numbered from 1 to 20 are mixed up together then drawn at random. What is the probability that the ticket has as a number which is a multiple of 3 or 7?

Answer 1

Hint: Probability of any event is defined as
$\text{P}=\dfrac{\text{Number of Favourable cases}}{\text{Total number of cases}}$

Complete step-by-step answer:
Get the favourable cases by calculating the total number of tickets which has numbers with multiple of 3 or 7. Total cases for the given problem is 20. Put the required values to the above equation to get the answer.
As we are given the tickets with numbers (1, 2, 3, 4, 5, 6……………..20) and we need to determine the probability of selecting a ticket which has a multiple of 3 or 7. Now we know the probability of any event can be defined mathematically as
$\text{P}=\dfrac{\text{Number of Favourable cases}}{\text{Total number of cases}}.....................\left( i \right)$
So, let us calculate the total number of tickets which has numbers with multiple of 3:
3, 6, 9, 12, 15, 18
Similarly, tickets which has numbers multiple of 7 are given as
7, 14
So, total numbers on the tickets which are multiple of 3 or 7 is given as
= 6 + 2
= 8
Hence, the favourable cases in the equation (i) is 8 and the total number of tickets are 20. So, probability to get a number of 3 or 7 is given as
$P=\dfrac{8}{20}=\dfrac{2}{5}$
So, $\dfrac{2}{5}$ is the answer to the given problem.

Note: Another approach to solve the problem would be given as
P = Probability to get the numbers which are multiple of 3 + probability to get numbers which are multiple of 7.

$\begin{aligned} & P=\dfrac{6}{20}+\dfrac{2}{20} \\\ & P=\dfrac{8}{20}=\dfrac{2}{5} \\\ \end{aligned}$
Hence, we can get the required answer by adding probabilities to get the numbers which are multiple of 3 and numbers which are multiple of 7. One may go wrong with the above approach, if he or she multiplies both the calculated probabilities.