Re: Two fair dice are thrown.
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10 Sep 2022, 22:18
here total number of outcomes = (1,1), (1,2), (1,3), ......., (6,4), (6,5), (6,6) = 36
desired outcomes that result in a sum to be 7 = (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) = 6
P (sum 7) = 6/36 = 1/6
outcomes with a 3 = (1,3), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,3), (5,3), (6,3) = 11
(3,4) and (4,3) are already included so we do not count them to calculate the probability of getting a 3.
therefore, desired outcomes = 11-2 = 9
P (a 3) = 9/36 = 1/4
now, P (sum 7 or a 3) = 1/6 + 1/4 = 10/24 = 5/12