Question 861132
I need help with this question please. 
Three dice are tossed and they show 3 different numbers. Calculate the probability that you get at least one six.
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# of triples with three different numbers:: 6*5*4 = 120
# of triples with no 6 with three different numbers: 5*4*3 = 60
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P(at least one six | 3 different numbers) = 1 - P(no six | 3 diff #'s)
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= 1 - 60/120 = 1/2
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cheers,
Stan H.
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