题目

Problem

Eric has three coins. One of the coins is fair. The other two coins are each biased so that the probability of obtaining a head on any throw is $\frac{1}{4}$ , independently of all other throws. Eric throws all three coins at the same time.

Events $A$ and $B$ are defined as follows.

$A$ : all three coins show the same result

$B$ : at least one of the biased coins shows a head

(a) Show that $\mathrm{P}(B)=\frac{7}{16}$ .

(2)

(b) Find $\mathrm{P}(A\mid B)$ .

(2)

The random variable $X$ is the number of heads obtained when Eric throws the three coins.

(3)

题目中文翻译

Eric 有三枚硬币。其中一枚是均匀硬币，另外两枚各自有偏，每次抛掷出现正面的概率均为 $\frac{1}{4}$ ，且与所有其他抛掷相互独立。Eric 同时抛掷这三枚硬币。

事件 $A$ 和 $B$ 定义如下。

$A$ ：三枚硬币均出现相同的结果

$B$ ：至少一枚有偏硬币出现正面

(a) 证明 $\mathrm{P}(B)=\frac{7}{16}$ 。

(2)

(b) 求 $\mathrm{P}(A\mid B)$ 。

(2)

随机变量 $X$ 表示 Eric 抛掷这三枚硬币时得到的正面数。

(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

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事件B只涉及两枚偏币，恰一正或两正共三条路径，公平币不必分类。

答题过程

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For the two biased coins, B occurs on HT, TH or HH. Therefore

\mathrm{P}(B)=\frac14\frac34+\frac34\frac14+\frac14\frac14 =\frac{3+3+1}{16}=\boxed{\frac7{16}}.

解法二

思路

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也可取两枚偏币都反面的补集。

答题过程

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The complement of B is that both biased coins show tails. Hence

\mathrm{P}(B)=1-\biggl(\frac34\biggr)^2=1-\frac9{16}=\boxed{\frac7{16}}.

(b)

解法一

思路

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A且B只可能三正面，三反面不符合B。

答题过程

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Since $A\cap B$ occurs only when all three coins show heads,

\mathrm{P}(A\cap B)=\frac12\frac14\frac14=\frac1{32}.

Using (a),

\mathrm{P}(A\mid B)=\frac{1/32}{7/16}=\boxed{\frac1{14}}.

(c)

解法一

思路

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按公平币为正或反分类，偏币正面数服从两次二项分布。

答题过程

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Let $Y$ count heads from the two biased coins. Then

\begin{align*} \mathrm{P}(Y=0)=&\,\frac9{16},\\[4mm] \mathrm{P}(Y=1)=&\,\frac6{16},\\[4mm] \mathrm{P}(Y=2)=&\,\frac1{16}. \end{align*}

Conditioning on the fair coin,

\mathrm{P}(X=0)=\frac12\frac9{16}=\frac9{32},

\mathrm{P}(X=1)=\frac12\frac9{16}+\frac12\frac6{16}=\frac{15}{32},

\mathrm{P}(X=2)=\frac12\frac6{16}+\frac12\frac1{16}=\frac7{32},

\mathrm{P}(X=3)=\frac12\frac1{16}=\frac1{32}.

Thus

$x$	$0$	$1$	$2$	$3$
$\mathrm{P}(X=x)$	$\frac9{32}$	$\frac{15}{32}$	$\frac7{32}$	$\frac1{32}$

CIE 9709 2022 November Paper 52 Q5

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

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(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程