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In a single throw of two dice, determine the probability of not getting the same number on the two dice.

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Best answer

In a single throw of two dice.

Possible outcomes are as follow:

(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 number of outcomes = 36

Favorable outcomes (i.e. not getting the same number) = All outcomes except (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

We know,

Probability of occurrence of an event = (Total number of favorable outcomes) / (Total number of outcomes)

Therefore, the probability of at least one boy in the family = 30/36 = 5/6

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