Question

A dice is thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3.

Answer 1

The sample space of the given experiment is given by
S = {1, 2, 3, 4, 5, 6}
∴ n (S) = 6
(i) Let A be the event of occurrence of a prime number.
Then A = {2, 3, 5}
i.e. n (A) = 3
$\therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}\mathrm{A}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{\mathrm{n}\left(\mathrm{A}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{3}{6}=\frac{1}{2}$

(ii) Let B be the event of occurrence of the number 2 or 4.
Then B = {2,4}
i.e. n (B) = 2
$\therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}\mathrm{B}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{\mathrm{n}\left(\mathrm{B}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{2}{6}=\frac{1}{3}$

(iii) Let C be the event of occurrence of a multiple of 2 or 3.
Then C = {2, 3, 4, 6}
i.e. n (C) = 4
$\therefore \mathrm{P}\left(\mathrm{C}\right)=\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}\mathrm{C}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{\mathrm{n}\left(\mathrm{C}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{4}{6}=\frac{2}{3}$