Integers 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? Choices
A) 3/8	B) 7/16	C) 1/2	D) 9/16	E) 5/8
First of all notice that half of the numbers from from 0 to 2007, inclusive are odd and half even.   Next, let's find the probability of ad - bc being odd and subtract it from 1.   The difference of two integers to be odd we should have either ODD-EVENOREVEN-ODD:   ODD-EVEN = ad - bc = odd*odd-even*evenorodd*odd-odd*evenorodd*odd-even*odd. Three cases. The probability of each is 1/2^4, so the probability of 3 cases is 3/2^4.   The same for EVEN-ODD: p=3/2^4.   P(even) = 1 - P(odd) = 1 - (3/2^4 + 3/2^4) = 10/16 = 5/8.   Answer: 5/8.
<result>E</result>
Integers 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?