2019 AMC 12B 第 13 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。

13.

一个红球和一个绿球随机且独立地被投入编号为正整数的箱子中。对每个球来说,投进第 kk 号箱子的概率为 2k2^{-k},其中 k=1,2,3,k=1,2,3,\ldots。红球被投入编号比绿球更大的箱子的概率是多少?

A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin kk is 2k2^{-k} for k=1,2,3,k=1,2,3,\ldots What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?

14\dfrac{1}{4}

27\dfrac{2}{7}

13\dfrac{1}{3}

38\dfrac{3}{8}

37\dfrac{3}{7}

答案:C
知识点:基本概率对称性对立事件概率
难度评级:1440
解答:

两球落入同一箱子的概率为 k=1(2k)2=k=14k=1/411/4=13. \begin{gathered} \sum_{k=1}^\infty \left(2^{-k}\right)^2=\sum_{k=1}^\infty 4^{-k} \\ =\dfrac{1/4}{1-1/4}=\dfrac13. \end{gathered}

由对称性,红球编号更大和绿球编号更大的概率相等,所以各自概率为 1132=13. \dfrac{1-\tfrac13}{2}=\dfrac13.

所以正确答案是 C

The probability the balls land in the same bin is k=1(2k)2=k=14k=1/411/4=13. \begin{gathered} \sum_{k=1}^\infty \left(2^{-k}\right)^2=\sum_{k=1}^\infty 4^{-k} \\ =\dfrac{1/4}{1-1/4}=\dfrac13. \end{gathered}

By symmetry, the red ball being higher and the green ball being higher are equally likely, so each has probability 1132=13. \dfrac{1-\tfrac13}{2}=\dfrac13.

Thus, C is the correct answer.

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