<PRE><B>Question 1070161</B>
Three distinct integers are chosen at random from the first 20 positive integers compute the probability that
a- their sum is even::: 
all three are even:: 10C3 = 120
2 are odd and one is even:: 10C2*10 = 450
P(sum is even) = (120+450)/20C3 = 370/1140 = 37/114
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b- their product is even 
all three are odd:: 10C3 = 120
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P(product is even) = (20C3-120)/20C3 = 1020/1140 =  51/57 = 17/19
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Cheers,
Stan H.
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