Two different dice are tossed together. Find the probability

(i) That the number on each die is even.

(ii) That the sum of numbers appearing on the two dice is 5.

Given two different dice are tossed together. The number of possible outcomes are 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) Let E be the event ‘number on each die is even’.

The outcomes favourable to E are {2, 2}, {2, 4}, {2, 6}, {4, 2}, {4, 4}, {4, 6}, {6, 2}, {6, 4} and {6, 6}.

Thus, the number of outcomes favourable to E is 9.

We know that probability of an event E, P (E)

⇒ P (E) = 9/36 = 1/4

∴ P (getting a number on each die even) = 1/4

(ii) Let F be the event ‘sum of numbers is 5’.

The outcomes favourable to F are {1, 4}, {2, 3}, {3, 2}, {4, 1}.

Thus, the number of outcomes favourable to F is 4.

⇒ P (F) = 4/36 = 1/9

∴ P (sum of numbers is 5) = 1/9