Question

A die is thrown once. Find the probability of getting
(i) an even number
(ii) a number less than 5
(iii) a number greater than 2
(iv) a number between 3 and 6
(v) a number other than 3
(vi) the number 5.

Answer 1

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.
Total number of possible outcomes = 6

(i)
Let E be the event of getting an even number.
Then, the favourable outcomes are 2, 4 and 6.
Number of favourable outcomes = 3
∴ Probability of getting an even number = P (E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{6}=\frac{1}{2}$
(ii)
Let E be the event of getting a number less than 5.
Then, the favourable outcomes are 1, 2, 3, 4
Number of favourable outcomes = 4
∴ Probability of getting a number less than 5 = P (E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{4}{6}=\frac{2}{3}$

(iii)
Let E be the event of getting a number greater than 2.
Then, the favourable outcomes are 3, 4, 5 and 6.
Number of favourable outcomes = 4
∴ Probability of getting a number greater than 2 = P (E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{4}{6}=\frac{2}{3}$
(iv)
Let E be the event of getting a number between 3 and 6.

Then, the favourable outcomes are 4, 5

Number of favourable outcomes = 2
∴ Probability of getting a number between 3 and 6 = P (E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{2}{6}=\frac{1}{3}$
(v)
Let E be the event of getting a number other than 3.

Then, the favourable outcomes are 1, 2, 4, 5 and 6.

Number of favourable outcomes = 5
∴ Probability of getting a number other than 3 = P (E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{5}{6}$
(vi)
Let E be the event of getting the number 5.

Then, the favourable outcome is 5.

Number of favourable outcomes = 1
∴ Probability of getting the number 5 = P (E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{1}{6}$