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1) intergers a,b,c, and d, not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that ad-bc is even?
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the answer is 5/8
ad- bc = EVEN means we must have ad, bc both ODD or ad, bc both EVEN in the first case we have a,b,c,d all odd in the second case we just need one in each pair to be EVEN
there are 1004 odd numbers from 0 to 2007 inclusive there are 1004 even
the probability of picking 4 odd numbers \[(1/2)^4\] the probability of picking at least 1 even in each pair \[= (1- (1/2))^2\] \[=(3/4)^2\] so we add them and get \[\frac{1}{16} + \frac{9}{16} = \frac{10}{16} = \frac{5}{8}\]
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