Q) In a two-dice game, a player throws two dice simultaneously. A player scores the sum of the two dice thrown and gradually reaches a higher score as they continue to roll. Answer the following questions:
i. Find the probability that the difference between the numbers on the two dice is 3.
ii. Find the probability that the product of the numbers on the two dice is more than 18.

Ans:

Total possible outcomes with two dice are thrown together:

(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),

∴ Total outcomes = 36

(i) For the difference of 3, the possible outcomes are:

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

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

∴ Favorable outcomes = 6

∴ The probability for the difference of 3 between numbers on two dice:

= \frac{favourable~outcomes}{Total~outcomes}

= \frac{6}{36} = \frac{1}{6}

Therefore, the probability for difference between the numbers on the two dice being 3 is \frac{1}{6}.

(ii) For product of numbers on two dice > 18, the possible outcomes are:
(4, 5), (4, 6),
(5,4), (5,5), (5, 6)
(6,4),(6,5),(6,6)

∴ Number of favorable outcomes = 8

∴ The probability for product of numbers on two dice > 18:

= \frac{favourable~outcomes}{Total~outcomes}

= \frac{8}{36} = \frac{2}{9}

Therefore, the probability for product of the numbers on the two dice being more than 18 is \frac{2}{9}.

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