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From 20 tickets marked with the first 20 numerals, one is drawn in random: find the chance that it is a multiple of 3 or of 7.

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Hint: Determine the multiples of 3 & 7.
Let us suppose, we are having 20 tickets which are marked with the first 20 numbers such that the first ticket is numbered as 1, second ticket as 2 and so on up to 20 tickets.
So, the numbers marked on the tickets are 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19 and 20.
Clearly the numbers which are multiple of 3 are 3,6,9,12,15,18 and the numbers which are multiple of 7 are 7,14.
It is given that one ticket is drawn from 20 tickets, we have to find the probability that the drawn ticket has a number which is a multiple of either 3 or 7.
As we know that the general formula for probability is given as
${\text{Probability of occurrence of an event}} = \dfrac{{{\text{Number of favourable outcomes }}}}{{{\text{Total number of outcomes}}}}$
Here, a favourable case is getting a number which is multiple of 3 or 7.
Tickets on which the number marked is a multiple of 3 or 7 are 3,6,7,9,12,14,15,18.
$ \Rightarrow {\text{Number of favourable outcomes}} = 8$
Also, Total number of outcomes = Total number of tickets = 20
Therefore, ${\text{Probability of drawing a ticket with its number as a multiple of 3 or 7}} = \dfrac{8}{{20}} = \dfrac{2}{5}.$

Note- In these types of problems, we use the general formula of probability where the favourable event is the event whose probability is asked and the total number of outcomes involves all the possible cases which can exist when a random draw occurs.

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