Question

A box contains 20 balls bearing numbers 1, 2, 3,..., 20 respectively. A ball is drawn at random from the box. What is the probability that the number on the ball is
(i) an odd number
(ii) divisible by 2 or 3
(iii) a prime number
(iv) not divisible by 10

Answer 1

The possible outcomes are 1, 2, 3, 4, 5................20.
Number of all possible outcomes = 20
(i) Out of the given numbers, the odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19. Let E₁ be the event of getting an odd number.
Then, number of favourable outcomes = 10​
∴ P (getting an odd number) = $\frac{10}{20}=\frac{1}{2}$

(ii) Out of the given numbers, numbers divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20.

​ Out of the given numbers, numbers which are divisible by 3 are 3, 6, 9, 12, 15 and 18.

Let E₂ be the event of getting a number divisible by 2 or 3 .

Then, number of favourable outcomes = (10 + 6 − 3) = 13
∴ P (getting a number divisible by 2 or 3 ) = $\frac{13}{20}$

(iii) Out of the given numbers, the prime numbers are 2, 3, 5, 7, 11, 13, 17 and 19.

Let E₃ be the event of getting a prime number.
Then, number of favourable outcomes =​ 8
∴ P (getting a prime number) = $\frac{8}{20}=\frac{2}{5}$

(iv) Out of the given numbers, the numbers divisible by 10 are 10 and 20.
Let E₄ be the event of getting a number not divisible by 10.
Then, number of favourable outcomes = 20 − 2 = 18​
∴ P (getting a number not divisible by 10) = $\frac{18}{20}=\frac{9}{10}$