题目

Problem

Jacob throws three coins at the same time.

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 biased so that the probability of obtaining a head when it is thrown is $\frac{1}{5}$ .

The random variable $X$ is the number of heads obtained.

(a) Show that $\mathrm{P}(X=2)=\frac{3}{20}$ .

(1)

(b) Draw up the probability distribution table for $X$ .

(3)

(2)

题目中文翻译

Jacob 同时掷三枚硬币。

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 $\frac{1}{4}$ 。

第三枚硬币不均匀，掷出正面的概率为 $\frac{1}{5}$ 。

随机变量 $X$ 表示掷出的正面个数。

(a) 证明 $\mathrm{P}(X=2)=\frac{3}{20}$ 。

(1)

(b) 列出 $X$ 的概率分布表。

(3)

(2)

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

(a)

解法一

思路

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恰有两正面有三种互斥次序，三枚硬币正面概率不同，必须保持硬币的顺序分别计算。

答题过程

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The possible outcomes, in the order of the three coins, are HHT, HTH and THH. Thus

\begin{align*} \mathrm{P}(X=2) =&\,\frac13\frac14\frac45+\frac13\frac34\frac15 \\[4mm] &\,\hspace{2pt}+\frac23\frac14\frac15\\[4mm] =&\,\frac{4+3+2}{60}\\[4mm] =&\,\boxed{\frac3{20}}. \end{align*}

(b)

解法一

思路

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零正面和三正面分别只有一种次序；一正面有三种，二正面使用(a)的结果。

答题过程

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For no heads and three heads,

\begin{align*} \mathrm{P}(X=0)=&\,\frac23\frac34\frac45=\frac{24}{60},\\[4mm] \mathrm{P}(X=3)=&\,\frac13\frac14\frac15=\frac1{60}. \end{align*}

For exactly one head,

\begin{align*} \mathrm{P}(X=1) =&\,\frac13\frac34\frac45+\frac23\frac14\frac45 \\[4mm] &\,\hspace{2pt}+\frac23\frac34\frac15\\[4mm] =&\,\frac{12+8+6}{60}=\frac{26}{60}. \end{align*}

Using (a), the distribution is

$x$	$0$	$1$	$2$	$3$
$\mathrm{P}(X=x)$	$\frac{24}{60}$	$\frac{26}{60}$	$\frac9{60}$	$\frac1{60}$

(c)

解法一

思路

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先算二阶矩，减去题目给出的均值平方。

答题过程

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\mathrm{E}(X^2)=\frac{0+26+4(9)+9(1)}{60} =\frac{71}{60}.

Hence

\begin{align*} \operatorname{Var}(X) =&\,\frac{71}{60}-\biggl(\frac{47}{60}\biggr)^2\\[4mm] =&\,\frac{4260-2209}{3600}\\[4mm] =&\,\boxed{\frac{2051}{3600}}. \end{align*}

CIE 9709 2025 March Paper 52 Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程