Question

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two digit number (ii) a perfect square number (iii) a number divisible by 5.

Answer 1

Hint: To solve this question, we should use the basic principles of probability. Let us consider a sample space which has n elements and an event E is which has x favourable cases. Then the probability P(E) is given by the formula P(E) = $ \dfrac{x}{n} $ . Using this formula, the sample space in the question has 90 elements which is the number of discs available. We should take case by case and write favourable cases and divide them by 90 to get the required probability.

Complete step-by-step answer:
Let us consider a sample space which has n elements and an event E is which has x favourable cases. Then the probability P(E) is given by the formula
P(E) = $ \dfrac{x}{n}\to \left( 1 \right) $
In our case, we have a sample space of 90 numbers and the number of elements in the sample space is 90.
Let us consider case-1, the event of a 2 digit number coming when we randomly pick a disc.
The favourable cases are 10, 11, 12, ……..90. The number of favourable cases can be calculated by removing the cases of 1, 2, 3…9. We can see that the cases to be removed are 9 in number. So, the number favourable cases for the event 2 digit number is = 90 – 9 = 81.
Using equation-1 we can infer that x = 81 and n = 90
P(2 digit number) = $ \dfrac{81}{90}=\dfrac{9}{10} $
Let us consider case-2, the event of a perfect square number coming when we randomly pick a disc.
The favourable cases are 1, 4, 9, 16, 25, 36,49, 64, 81. We can see that the cases are 9 in number. So, the number of favourable cases for perfect square number is = 9.
Using equation-1 we can infer that x = 9 and n = 90
P(perfect square number) = $ \dfrac{9}{90}=\dfrac{1}{10} $
Let us consider case-3, the event of a number divisible by 5 coming when we randomly pick a disc.
The favourable cases are 5, 10, 15,…….90. We know that 90 is the eighteenth multiple of 5.So, there are 18 numbers in 1 to 90 which are divisible by 5. So, the number favourable cases for a number divisible by 5 is = 18.
Using equation-1 we can infer that x = 18 and n = 90
P(number divisible by 5) = $ \dfrac{18}{90}=\dfrac{2}{10} $
 $ \therefore $ P(2 digit number) = $ \dfrac{81}{90}=\dfrac{9}{10} $ , P(perfect square number) = $ \dfrac{9}{90}=\dfrac{1}{10} $ ,
P(number divisible by 5) = $ \dfrac{18}{90}=\dfrac{2}{10} $ .


Note: Students can make a mistake when calculating the number of 2 digit numbers. Two digit numbers range from 10 to 90. There is a chance of mistake by taking
$ 90-10=80 $, 2 digit numbers which leads to a wrong answer. In this calculation, we neglected the 2 digit number 10. A similar mistake can be done in calculating the multiples of 5.