AMC12 2006 B

AMC12 2006 B · Q17

AMC12 2006 B · Q17. It mainly tests Basic counting (rules of product/sum), Probability (basic).

For a particular peculiar pair of dice, the probabilities of rolling $1$, $2$, $3$, $4$, $5$ and $6$ on each die are in the ratio $1:2:3:4:5:6$. What is the probability of rolling a total of $7$ on the two dice?

对于一对特殊的骰子，每个骰子上掷出 $1$、$2$、$3$、$4$、$5$ 和 $6$ 的概率之比为 $1:2:3:4:5:6$。两个骰子点数之和为 $7$ 的概率是多少？

(A) $\frac{4}{63}$ $\frac{4}{63}$

(B) $\frac{1}{8}$ $\frac{1}{8}$

(D) $\frac{1}{6}$ $\frac{1}{6}$

(E) $\frac{2}{7}$ $\frac{2}{7}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

The probability of getting an $x$ on one of these dice is $\frac{x}{21}$. The probability of getting $1$ on the first and $6$ on the second die is $\frac 1{21}\cdot\frac 6{21}$. Similarly we can express the probabilities for the other five ways how we can get a total $7$. (Note that we only need the first three, the other three are symmetric.) Summing these, the probability of getting a total $7$ is: \[2\cdot\left( \frac 1{21}\cdot\frac 6{21} + \frac 2{21}\cdot\frac 5{21} + \frac 3{21}\cdot\frac 4{21} \right) = \frac{56}{441} = \boxed{\frac{8}{63}}\] See also 2016 AIME I Problems/Problem 2

在这种骰子上掷出 $x$ 的概率为 $\frac{x}{21}$。第一个骰子掷出 $1$ 且第二个骰子掷出 $6$ 的概率是 $\frac 1{21}\cdot\frac 6{21}$。类似地，我们可以写出另外五种得到总和为 $7$ 的方式的概率。（注意我们只需要前三种，另外三种是对称的。）将这些相加，总和为 $7$ 的概率是： \[2\cdot\left( \frac 1{21}\cdot\frac 6{21} + \frac 2{21}\cdot\frac 5{21} + \frac 3{21}\cdot\frac 4{21} \right) = \frac{56}{441} = \boxed{\frac{8}{63}}\] 另见 2016 AIME I Problems/Problem 2

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