2006 AMC 12B 第 17 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

17.

有一对特殊的骰子,每个骰子掷出 1,2,3,4,51, 2, 3, 4, 566 的概率之比为 1:2:3:4:5:61 : 2 : 3 : 4 : 5 : 6。掷这两个骰子,总和为 77 的概率是多少?

For a particular peculiar pair of dice, the probabilities of rolling 1,2,3,4,5,1, 2, 3, 4, 5, and 66 on each die are in the ratio 1:2:3:4:5:6.1 : 2 : 3 : 4 : 5 : 6. What is the probability of rolling a total of 77 on the two dice?

463\dfrac{4}{63}

18\dfrac{1}{8}

863\dfrac{8}{63}

16\dfrac{1}{6}

27\dfrac{2}{7}

答案:C
知识点:骰子(概率)基本概率
难度评级:1740
解答:

因为权重之和为 2121,掷出 kk 的概率为 k21\dfrac{k}{21}

总和为 77 来自 (1,6),(2,5),,(6,1)(1,6), (2,5), \ldots, (6,1),所以概率为 16+25+34+43+52+61212=56441=863. \begin{aligned} &\scriptsize \frac{1 \cdot 6 + 2 \cdot 5 + 3 \cdot 4 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1}{21^2} \\ &= \frac{56}{441} = \frac{8}{63}. \end{aligned}

因此,正确答案是 C

Since the weights sum to 21,21, the probability of rolling kk is k21.\dfrac{k}{21}.

A total of 77 comes from (1,6),(2,5),,(6,1),(1,6), (2,5), \ldots, (6,1), so the probability is 16+25+34+43+52+61212=56441=863. \begin{aligned} &\scriptsize \frac{1 \cdot 6 + 2 \cdot 5 + 3 \cdot 4 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1}{21^2} \\ &= \frac{56}{441} = \frac{8}{63}. \end{aligned}

Thus, the correct answer is C.

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