Question

A die is thrown once. What is the probability of getting a number more than 4?
(a) $\frac{1}{3}$
(b) $\frac{1}{6}$
(c) $\frac{1}{2}$
(d) $\frac{3}{4}$

Answer 1

(a) $\frac{1}{3}$

Explanation:
In a single throw of a die, all possible outcomes are:
1, 2, 3, 4, 5, 6
Total number of possible outcomes = 6
Let E be the event of getting a number more than 4.
Then, the favourable outcomes are 5 and 6.
Number of favourable outcomes = 2
∴ Probability of getting a number more than 4:
P (E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{2}{6}=\frac{1}{3}$