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If the die is tossed twice, find the probability that the sum of the numbers appearing is at least 6
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total outcome=6*6=36 i.e <1,1><1,2><1,3>.......<6,4><6,5><6.6> Now no of outcome with sum at least 6 = 36- no of outcomes having sum less than 6 Outcomes having sum less than 6 = {1,1}, {1,2}, {1,3}, {1,4}, {2,1}, {2,2},{2,3}, {3,1}, {3,2}, {4,1} no of outcome having sum less than 6=10 no of outcome having sum atleast 6 i.e favourable outcome= 36-10=26 probability= no. of favourable outcome/ total no of outcome = 26/36 =13/18
Thank you (Mike, Clifton Park, N.Y.)
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