Question

On the 6 faces of a dice, the numbers 0, 1, 1, 1, 6, 6 are marked. Such these both dice are thrown at a time and the sum of the digits on the top faces is noted.

(i) How many total number of possible outcomes.

(ii) What is the probability of getting a sum of 7 on the top faces.

Answer 1

Two dices contain the numbers 0, 1, 1, 1, 6, 6 on the faces each.

The all possible outcomes will be:

(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)

(i) The possible numbers which have different outcomes are (0, 0), (0, 1), (0, 6), (1, 0), (1, 1), (1, 6), (6, 6)

Hence, the possible outcomes are : (0, 1, 2, 6, 7, 12)

(ii) Total number of all possible outcomes = 36

The numbers that have a sum 7 are

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

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

The number of favourable outcomes = 12

∴ The probability of getting a sum of 7 = \(\frac { 12 }{ 36 }\) = \(\frac { 1 }{ 3 }\)