Question

20 cards are numbered from 1 to 20. One card is drawn at random. What is the probability that the number on the cards is:
(i) a multiple of 4?
(ii) not a multiple of 4?
(iii) odd?
(iv) greater than 12?
(v) divisible by 5?
(vi) not a multiple of 6?

Answer 1

Clearly, the sample space is given by S = {1, 2, 3, 4, 5........19, 20}.
i.e. n(S) = 20
(i)
Let E₁ = event of getting a multiple of 4
Then E₁ = {4, 8, 12, 16, 20}
i.e. n(E₁) = 5
Hence, required probability = P(E₁) = $\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{5}{20}=\frac{1}{4}$

(ii)
Let E₂ = event of getting a non-multiple of 4
Then P(non-multiple of 4) = P(E₂) = 1 $-$ P(multiple of 4)
= $1-\frac{1}{4}=\frac{3}{4}$

(iii)
Let E₃ = event of getting an odd number
Then E₃ = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
i.e. n(E₃) = 10
Hence, required probability = P(E₃) = $\frac{10}{20}=\frac{1}{2}$

(iv)
Let E₄ = event of getting a number greater than 12
Then E₄ = {13, 14, 15, 16, 17, 18, 19, 20}
i.e. n(E₄) = 8
Hence, required probability = P(E₄) = $\frac{8}{20}=\frac{2}{5}$

(v)
Let E₅ = event of getting a number divisible by 5
Then E₅ = {5, 10, 15, 20}
i.e. n(E₅) = 4
Hence, required probability = P(E₅) = $\frac{4}{20}=\frac{1}{5}$

(vi)
Let E₆ = event of getting a number which is not a multiple of 6
Then E₆' = event of getting a number which is a multiple of 6
E₆' = {6, 12, 18}
i.e. n(E₆' ) = 3
Now, P(E₆') = $\frac{3}{20}$
Hence, required probability P(E₆) = 1 − P(E₆')
= $1-\frac{3}{20}=\frac{20-3}{20}=\frac{17}{20}$