题目

Problem

Alisha has four coins. One of these coins is biased so that the probability of obtaining a head is 0.6. The other three coins are fair. Alisha throws the four coins at the same time. The random variable $X$ denotes the number of heads obtained.

(a) Show that the probability of obtaining exactly one head is 0.225.

(3)

(b) Complete the following probability distribution table for $X$ .

(2)

$x$	$0$	$1$	$2$	$3$	$4$
$\mathrm{P}(X=x)$	$0.05$	$0.225$			$0.075$

(2)

题目中文翻译

Alisha 有四枚硬币。其中一枚有偏，出现正面的概率为 0.6，另外三枚是均匀硬币。Alisha 同时抛掷这四枚硬币。随机变量 $X$ 表示得到的正面数。

(a) 证明恰好得到一个正面的概率为 0.225。

(3)

(b) 补全下面 $X$ 的概率分布表。

(2)

$x$	$0$	$1$	$2$	$3$	$4$
$\mathrm{P}(X=x)$	$0.05$	$0.225$			$0.075$

(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

展开

按偏币是正面还是反面分类：偏币正则其余三枚全反，偏币反则其余三枚恰一正。

答题过程

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If the biased coin is a head, all three fair coins must be tails, with probability $0.6(0.5)^3$ .

If the biased coin is a tail, exactly one fair coin must be a head, with probability $0.4\binom31(0.5)^3$ .

Thus

\mathrm{P}(X=1)=0.6(0.5)^3+0.4(3)(0.5)^3 =0.075+0.150=\boxed{0.225}.

(b)

解法一

思路

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同样按偏币结果分类，分别计算总正面数为2与3的概率。

答题过程

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\begin{align*} \mathrm{P}(X=2) =&\,0.6\binom31(0.5)^3+0.4\binom32(0.5)^3\\[4mm] =&\,0.225+0.150=0.375, \end{align*}

\begin{align*} \mathrm{P}(X=3) =&\,0.6\binom32(0.5)^3+0.4(0.5)^3\\[4mm] =&\,0.225+0.050=0.275. \end{align*}

The completed distribution is

$x$	$0$	$1$	$2$	$3$	$4$
$\mathrm{P}(X=x)$	$0.05$	$0.225$	$0.375$	$0.275$	$0.075$

(c)

解法一

思路

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先算二阶矩，均值直接用题给2.1。

答题过程

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\begin{align*} \mathrm{E}(X^2)=&\,0.225+4(0.375)\\[4mm] &\,\hspace{2pt}+9(0.275)+16(0.075)\\[4mm] =&\,5.4. \end{align*}

Hence

\operatorname{Var}(X)=5.4-(2.1)^2=\boxed{0.99}.

CIE 9709 2023 March Paper 52 Q2

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