AMC10 2004 B

AMC10 2004 B · Q11

AMC10 2004 B · Q11. It mainly tests Manipulating equations, Probability (basic).

Two eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?

两个八面骰子，每面分别编号1到8。掷骰子时，每面出现在上面的概率相等。两个上面数字的乘积大于它们的和的概率是多少？

(A) \frac{1}{2} \frac{1}{2}

(B) \frac{47}{64} \frac{47}{64}

(D) \frac{55}{64} \frac{55}{64}

(E) \frac{7}{8} \frac{7}{8}

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) There are $8\cdot 8=64$ ordered pairs that can represent the top numbers on the two dice. Let $m$ and $n$ represent the top numbers on the dice. Then $mn>m+n$ implies that $mn-m-n>0$, that is, \[ 1<mn-m-n+1=(m-1)(n-1). \] This inequality is satisfied except when $m=1$, $n=1$, or when $m=n=2$. There are $16$ ordered pairs $(m,n)$ excluded by these conditions, so the probability that the product is greater than the sum is \[ \frac{64-16}{64}=\frac{48}{64}=\frac{3}{4}. \]

（C）两个骰子朝上的点数共有 $8\cdot 8=64$ 个有序对。设 $m$ 和 $n$ 表示两个骰子朝上的点数。则 $mn>m+n$ 推出 $mn-m-n>0$，即 \[ 1<mn-m-n+1=(m-1)(n-1). \] 除非 $m=1$、$n=1$，或者 $m=n=2$，否则该不等式成立。由这些条件排除的有序对 $(m,n)$ 有 $16$ 个，因此“积大于和”的概率为 \[ \frac{64-16}{64}=\frac{48}{64}=\frac{3}{4}. \]

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