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2007 AMC 12A Problems/Problem 12

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The following problem is from both the 2007 AMC 12A #12 and 2007 AMC 10A #16, so both problems redirect to this page.

Problem

Integers and , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that is even?

\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58

Solution

The only times when is even is when and are of the same parity. The chance of being odd is \frac 12 \cdot \frac 12 = \frac 14, so it has a probability of being even. Therefore, the probability that will be even is \left(\frac 14\right)^2+\left(\frac 34\right)^2=\frac 58\ \mathrm{(E)}.

See also

2007 AMC 12A (Problems)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
2007 AMC 10A (Problems)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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