Question

A die is thrown twice. What is the probability that (i) the difference between two numbers obtained is 3? (ii) the sum of the numbers obtained is 8?

Answer 1

Total possible outcomes when a die is thrown twice = 36

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

(i) Favourable outcomes with a difference of 3 = 6

{(1, 4), (2, 5), (3, 6), (4, 1), (5, 2), (6, 3)}

Probability = `"Favourable outcomes"/"Possible outcomes"`

= `6/36`

= `1/6`

(ii) Favourable outcomes with sum of the numbers obtained = 5

{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

Probability = `"Favourable outcomes"/"Possible outcomes"`

= `5/36`