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Old 07-25-2007   #4
polol
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First questiona. No. of ways = 12C4 = 485 waysb. No. of ways = 12C1 x 11C3 = 1980 waysc. No. of ways = 12C1 x 11C1 x 10C1 x 9C1 = 11880 waysSecond questionEvent: A 4 appears on at least one of the dice.P(A 4 appears on at least one of the dice.) = P(4,1) + P(4,2) + P(4,3) + P( 4,4) + P (4,5) + (4,6) + P(1,4) + P(2,4) + ....= (1/6 x 1/6)x11 = 11/36.