Question

A box contains 36 tickets numbered from 1 to 36. One ticket is random. Find the probability that the number on the ticket is either by 3 or is a perfect square.
A. $\dfrac{1}{9}$
B. $\dfrac{1}{3}$
C. $\dfrac{2}{9}$
D. $\dfrac{4}{9}$

Answer 1

Probability of an event is given by the ratio of the favourable cases for that event and total cases (sample space) for the given problem. Total cases will be all the numbers from 1 to 36. Next, find the numbers, which are multiple of 3 or a perfect square. Perfect square is a number, which has rational square roots.

Complete step-by-step answer:
Here, we have a box which contains 36 tickets numbered from 1 to 36 and we need to determine the probability of drawing one ticket such that the number on that ticket is either divisible by 3 or is a perfect square number.
So, we know that probability of an event is the ratio of favourable number of cases for that event and total number of cases for the sample. Hence, it is given as
$P=\dfrac{\text{Favourable cases}}{\text{Total cases}}................(i)$
As we know, the total number of cases is also termed as sample space. So, sample space for the given problem would be the numbers from 1 to 36. Hence sample space (s) can be given as
s ={1,2,3,4,5,6…………………36}
Hence, the number of elements in sample space is 36. It means the total number of cases for the given problem is 36. Now, we need to find the numbers which are divisible by 3 or a perfect square number. So, let us calculate one by one. Numbers which are divisible by ‘3’ of the numbers from 1 to 36 are given as;
{3,6,9,12,15,18,21,24,27,30,33,36}
And the numbers which are perfect square numbers (numbers whose square root is a rational number) of the numbers is 1 to 36 are given as,
{1,4,9,16,25,36}
Hence, we can observe that (9, 36) are the number numbers belonging to the both groups i.e. divisibility by 3 and perfect square. So, we can’t include them twice. Hence, the numbers which are divisible by 3 and a perfect square can be given as
{1,3,4,6,9,12,15,16,18,21,24,25,27,30,33,36}
Hence, the total number of numbers which are divisible by 3 or a perfect square is 16. Hence, the probability of getting a ticket with a number which is divisible by 3 or perfect square can be given from the equation (i) is.
$P=\dfrac{\text{Favourable cases for given event}}{\text{Total cases}\left( \text{sample space} \right)}$
Hence, we get
$P=\dfrac{16}{36}=\dfrac{4}{9}$
So, option (d) is the correct answer.

Note: One may calculate favourable cases for the given event by adding the number of numbers which are divisible by 3 and the numbers which are perfect square, which is wrong as 9 and 36 are the numbers which are common to both so, we can’t include them twice. So, be careful with this part of the problem. Favourable cases and the total cases are the numbers which are divisible by 3 or a perfect square and numbers from 1 to 36 respectively. We don’t need to add the numbers from 1 to 36 for total cases or the number for favourable cases. We need to count these numbers and use the counts to get probability. So, be clear with the fundamental definition of the probability. Don’t miss any number for the divisibility by 3 and perfect square numbers, other probability will not be accurate.