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CIE 9709 2024 November Paper 52 Q3

A Level / CIE / S1

CIE 9709 2024 Nov Paper 52 Paper · Question 3

题目

Problem

A fair coin and an ordinary fair six-sided dice are thrown at the same time. The random variable XX is defined as follows.

  • If the coin shows a tail, XX is twice the score on the dice.
  • If the coin shows a head, XX is the score on the dice if the score is even and XX is 00 otherwise.

(a) Draw up the probability distribution table for XX.

(3)

(b) Find Var⁡(X)\operatorname{Var}(X).

(3)
题目中文翻译

同时抛掷一枚均匀硬币和一颗普通的均匀六面骰子。随机变量 XX 定义如下。

  • 如果硬币出现反面,则 XX 为骰子点数的两倍。
  • 如果硬币出现正面,且骰子点数为偶数,则 XX 为骰子点数;否则 XX 为 00。

(a) 列出 XX 的概率分布表。

(3)

(b) 求 Var⁡(X)\operatorname{Var}(X)。

(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

硬币与骰子共有12个等概率结果,先逐一对应X,再合并相同取值。

答题过程

展开

Each coin–dice outcome has probability 112\frac1{12}. The values of XX are

Dice score112233445566
Tail2244668810101212
Head002200440066

Combining equal values gives

xx002244668810101212
P(X=x)\mathrm{P}(X=x)312\frac3{12}212\frac2{12}212\frac2{12}212\frac2{12}112\frac1{12}112\frac1{12}112\frac1{12}

(b)

解法一

思路

展开

从分布表分别求一阶矩、二阶矩,再相减。

答题过程

展开 E(X)=2(2)+4(2)+6(2)+8+10+1212=5412=4.5.\mathrm{E}(X)=\frac{2(2)+4(2)+6(2)+8+10+12}{12} =\frac{54}{12}=4.5.

Also,

E(X2)= 22(2)+42(2)+62(2)12 +82+102+12212= 42012=35.\begin{align*} \mathrm{E}(X^2)=&\,\frac{2^2(2)+4^2(2)+6^2(2)}{12}\\[4mm] &\,\hspace{2pt}+\frac{8^2+10^2+12^2}{12}\\[4mm] =&\,\frac{420}{12}=35. \end{align*}

Hence

Var⁡(X)=35−(4.5)2=14.75.\operatorname{Var}(X)=35-(4.5)^2=\boxed{14.75}.