Question

A die is rolled twice. Find the probability that

(i) 5 will not come up either time,
(ii) 5 will come up exactly one time,
(iii) 5 will come up both the times.

Answer 1

Total number of outcomes = 36.

(i) Cases where 5 comes up on at least one time are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) and (6, 5).

Number of such cases = 11.

Number of cases where 5 will not come up either time = 36 − 11 = 25.

∴ P(5 will not come up either time) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{25}{36}$

Thus, the probability that 5 will not come up either time is $\frac{25}{36}$.

(ii) Cases where 5 comes up on exactly one time are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) and (6, 5).

Number of such cases = 10.

∴ P(5 will come up exactly one time) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{10}{36}=\frac{5}{18}$

Thus, the probability that 5 will come up exactly one time is $\frac{5}{18}$.

(iii) Cases where 5 comes up on exactly two times is (5, 5).

Number of such cases = 1.

∴ P(5 will come up both the times) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{1}{36}$

Thus, the probability that 5 will come up both the times is $\frac{1}{36}$.