题目

Problem

Rachel has three coins. The first coin is biased so that the probability of obtaining a head when it is thrown is $\frac{1}{3}$ . The second coin is biased so that the probability of obtaining a head when it is thrown is $\frac{1}{4}$ . The third coin is fair.

Rachel throws the three coins at the same time. The random variable $X$ is the number of tails that she obtains.

Draw up the probability distribution table for $X$ .

(3)

题目中文翻译

Rachel 有三枚硬币。第一枚硬币不均匀，掷出正面的概率为 $\frac{1}{3}$ 。第二枚硬币不均匀，掷出正面的概率为 $\frac{1}{4}$ 。第三枚硬币均匀。

Rachel 同时掷这三枚硬币。随机变量 $X$ 表示她掷出的反面个数。

列出 $X$ 的概率分布表。

(3)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

解法一

思路

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三枚硬币的正面概率不同，不能使用同一成功概率的二项分布。按反面个数分类，逐条路径相乘再相加。

答题过程

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The head probabilities for the three coins are $\frac13,\frac14,\frac12$ , and the tail probabilities are $\frac23,\frac34,\frac12$ . Writing outcomes in coin order,

\begin{align*} \mathrm{P}(X=0)=&\,\frac13\frac14\frac12=\frac1{24},\\[4mm] \mathrm{P}(X=1)=&\,\frac23\frac14\frac12 +\frac13\frac34\frac12+\frac13\frac14\frac12\\[4mm] =&\,\frac{2+3+1}{24}=\frac6{24},\\[4mm] \mathrm{P}(X=2)=&\,\frac23\frac34\frac12 +\frac23\frac14\frac12+\frac13\frac34\frac12\\[4mm] =&\,\frac{6+2+3}{24}=\frac{11}{24},\\[4mm] \mathrm{P}(X=3)=&\,\frac23\frac34\frac12=\frac6{24}. \end{align*}

Thus

$x$	$0$	$1$	$2$	$3$
$\mathrm{P}(X=x)$	$\frac1{24}$	$\frac6{24}$	$\frac{11}{24}$	$\frac6{24}$

CIE 9709 2025 June Paper 52 Q1

题目

解答

解法一

思路

答题过程