# Two different dic

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

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