题目

Problem

Harry has three coins:

One 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 $\frac{1}{5}$ .

Harry throws the three coins. The random variable $X$ is the number of heads that he obtains.

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

(4)

Harry has two other coins, each of which is biased so that the probability of obtaining a head when it is thrown is $p$ . He throws all five coins at the same time. The random variable $Y$ is the number of heads that he obtains.

(b) Given that $\mathrm{P}(Y=0)=6\mathrm{P}(Y=5)$ , find the value of $p$ .

(3)

题目中文翻译

Harry 有三枚硬币：

第一枚是不均匀硬币，抛掷时出现正面的概率为 $\frac{1}{3}$ 。
第二枚是不均匀硬币，抛掷时出现正面的概率为 $\frac{1}{4}$ 。
第三枚是不均匀硬币，抛掷时出现正面的概率为 $\frac{1}{5}$ 。

Harry 抛掷这三枚硬币。随机变量 $X$ 为他得到的正面个数。

(a) 列出 $X$ 的概率分布表。

(4)

Harry 还有另外两枚硬币，每枚都是不均匀硬币，抛掷时出现正面的概率均为 $p$ 。他同时抛掷全部五枚硬币。随机变量 $Y$ 为他得到的正面个数。

(b) 已知 $\mathrm{P}(Y=0)=6\mathrm{P}(Y=5)$ ，求 $p$ 的值。

(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

(a)

解法一

思路

展开

三枚硬币正面概率不同，不可直接用二项分布。按各个有序结果相乘再相加。

答题过程

展开

\mathrm{P}(X=0)=\frac23\frac34\frac45=\frac{24}{60}, \qquad \mathrm{P}(X=3)=\frac13\frac14\frac15=\frac1{60}.

For exactly one head,

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

For exactly two heads,

\begin{align*} \mathrm{P}(X=2)=&\,\frac13\frac14\frac45\\[4mm] &\,\hspace{2pt}+\frac13\frac34\frac15\\[4mm] &\,\hspace{4pt}+\frac23\frac14\frac15\\[4mm] =&\,\frac{4+3+2}{60}=\frac9{60}. \end{align*}

Thus

$x$	$0$	$1$	$2$	$3$
$\mathrm{P}(X=x)$	$\frac25$	$\frac{13}{30}$	$\frac3{20}$	$\frac1{60}$

(b)

解法一

思路

展开

五枚全反与全正分别用前3枚的对应概率乘后2枚的概率，列方程并舍弃超出概率范围的根。

答题过程

展开

Using (a),

\mathrm{P}(Y=0)=\frac{24}{60}(1-p)^2, \qquad \mathrm{P}(Y=5)=\frac1{60}p^2.

The given condition implies

\begin{align*} 24(1-p)^2=&\,6p^2,\\[4mm] 4-8p+4p^2=&\,p^2,\\[4mm] 3p^2-8p+4=&\,0,\\[4mm] (3p-2)(p-2)=&\,0. \end{align*}

Thus $p=\frac23$ or $p=2$ . Since $0\leq p\leq1$ , reject 2, giving

p=\boxed{\frac23}.

CIE 9709 2024 June Paper 51 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程