AMC12 2007 A

AMC12 2007 A · Q12

AMC12 2007 A · Q12. It mainly tests Probability (basic), Parity (odd/even).

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?

从 0 到 2007（含端点）中独立且随机选取整数 $a, b, c,$ 和 $d$（不一定互不相同）。$ad-bc$ 为偶数的概率是多少？

(A) \frac{3}{8} \frac{3}{8}

(B) \frac{7}{16} \frac{7}{16}

(D) \frac{9}{16} \frac{9}{16}

(E) \frac{5}{8} \frac{5}{8}

Answer

Correct choice: (E)

正确答案：（E）

Solution

The only time when $ad-bc$ is even is when $ad$ and $bc$ are of the same parity. The chance of $ad$ being odd is $\frac 12 \cdot \frac 12 = \frac 14$, since the only way to have $ad$ be odd is to have both $a$ and $d$ be odd. As a result, $ad$ has a $\frac 34$ probability of being even. $bc$ also has a $\frac 14$ chance of being odd and a $\frac34$ chance of being even. Therefore, the probability that $ad-bc$ will be even is $\left(\frac 14\right)^2+\left(\frac 34\right)^2=\boxed {\mathrm{(E )} \frac 58\ }$.

只有当 $ad$ 与 $bc$ 同奇偶时，$ad-bc$ 才为偶数。$ad$ 为奇数的概率是 $\frac 12 \cdot \frac 12 = \frac 14$，因为 $ad$ 为奇数的唯一方式是 $a$ 与 $d$ 都为奇数。因此，$ad$ 为偶数的概率为 $\frac 34$。同理，$bc$ 为奇数的概率为 $\frac 14$，为偶数的概率为 $\frac34$。因此，$ad-bc$ 为偶数的概率为 $\left(\frac 14\right)^2+\left(\frac 34\right)^2=\boxed {\mathrm{(E )} \frac 58\ }$。

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