Question

A bag contains tickets numbered 11, 12, 13……..30. A ticket is taken out from the bag at random. Find the probability that the number on the drawn ticket:
(a) is a multiple of 7.
(b) is greater than 15 and a multiple of 5.

Answer 1

Hint: The possibilities of selecting one ticket at random from the tickets numbered 11, 12, 13……..30 is ${}^{20}{{C}_{1}}$. Total outcomes are 20. Now, we are asked to find the probability of a drawn ticket having a number which is multiple of 7. We know that probability is the division of favorable outcomes to the total outcomes. In drawing a ticket numbered as a multiple of 7, the favorable outcomes are the number of tickets having numbers as multiples of 7. In option (b), the favorable outcomes are the number of tickets having numbers greater than 15 and also a multiple of 5 so using these numbers of favorable outcomes we can find the probability in option (b).

Complete step-by-step answer :
It is given that a bag contains the tickets numbered as:
11, 12, 13……..30
From the above number of tickets, the total number of tickets in the bag is equal to 20.
It is given that we have drawn one ticket at random from these 20 tickets so the possible number of ways of drawing one ticket at random from the 20 tickets is:
${}^{20}{{C}_{1}}$
We know that:
${}^{n}{{C}_{1}}=n$
Using the above relation, we can write ${}^{20}{{C}_{1}}$ as 20.
Hence, the total possible outcomes of drawing one ticket at random are equal to 20.
Now, in the option (a) we are asked to find the probability of drawing a ticket which is the multiple of 7.
We have to select a ticket from the tickets which are numbered as multiples of 7. The tickets which are numbered as the multiples of 7 are as follows:
14, 21 and 28
So, we have to select one ticket from these three tickets which would be:
$\begin{aligned} & {}^{3}{{C}_{1}} \\\ & =3 \\\ \end{aligned}$
We know that probability of a certain outcomes is equal to:
$\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$
Now, total outcomes we have calculated above as 20 and the favorable outcomes in this option (a) are 3.
Hence, the probability of drawing a ticket which is a multiple of 7 is equal to $\dfrac{3}{20}$.
In option (b), we have to find the probability that the drawn ticket has a number which is greater than 15 and the multiple of 5 so we have to draw a ticket from the tickets which have numbers greater than 15 and are multiple of 5.
20, 25 and 30
Hence, the possible ways to draw a ticket from the above three tickets are:
$\begin{aligned} & {}^{3}{{C}_{1}} \\\ & =3 \\\ \end{aligned}$
In finding the probability that the number on the drawn ticket is greater than 15 and the multiple of 5, the favorable outcomes are 3 and the total outcomes that we have calculated above is 20. Hence, the probability is equal to $\dfrac{3}{20}$.
Hence, the probability that the number on the drawn ticket is greater than 15 and the multiple of 5 is equal to $\dfrac{3}{20}$.

Note: The plausible mistake that could happen in this problem is that you would have wrongly calculated the number of tickets. As the tickets given in the above problem are:
11, 12, 13……..30
To calculate the number of tickets, you might have subtracted 11 from 30 which will give you the number 19 which is the wrong number. But when you manually count the tickets you will find the number is coming as 20. So, here is a trick to total the number of tickets without actually counting them by subtracting the first tem from the last term of the consecutive series and then add 1 to the result of this subtraction. The trick is applicable only when the series contains consecutive numbers or the consecutive numbers having a difference of 1.