Question

A die has its six faces marked 0, 1, 1, 1, 6, 6. If two such dice are thrown together, then the probability of getting a total of 7 is _________.

Answer 1

The outcomes when two such dice are thrown together are

(0, 0), (0, 1), (0, 1), (0, 1), (0, 6), (0, 6)

(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)

(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)

(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)

(6, 0), (6, 1), (6, 1), (6, 1), (6, 6), (6, 6)

(6, 0), (6, 1), (6, 1), (6, 1), (6, 6), (6, 6)

Total number of outcomes = 36

Let E be the event of getting a total of 7. The outcomes in favour of event E are (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1) and (6, 1).

Favourable number of outcomes = 12

∴ Probability of getting a total of 7 = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{12}{36}=\frac{1}{3}$

A die has its six faces marked 0, 1, 1, 1, 6, 6. If two such dice are thrown together, then the probability of getting a total of 7 is $\underline{\frac{1}{3}}$.