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CIE 9231 2023 Nov Paper 42 Q3

A Level / CIE / FS

CIE 9231 2023 Nov Paper 42 Paper · Question 3

题目

Problem

Toby has a bag which contains 6 red marbles and 3 green marbles. He randomly chooses 3 marbles from the bag, without replacement. The random variable XX is the number of red marbles that Toby obtains.

(a) Find the probability generating function of XX.

[3]

Ling also has a bag which contains 6 red marbles and 3 green marbles. He randomly chooses 2 marbles from his bag, without replacement. The random variable YY is the number of red marbles that Ling obtains. It is given that the probability generating function of YY is 112(1+6t+5t2)\frac{1}{12}(1+6t+5t^2).

The random variable ZZ is the total number of red marbles that Toby and Ling obtain.

(b) Find the probability generating function of ZZ, expressing your answer as a polynomial in tt.

[3]

(c) Use the probability generating function of ZZ to find Var⁡(Z)\operatorname{Var}(Z).

[4]
题目中文翻译

Toby 有一个袋子,里面有 6 个红色弹珠和 3 个绿色弹珠。他从袋中随机取出 3 个弹珠,且不放回。随机变量 XX 表示 Toby 取到的红色弹珠个数。

(a) 求 XX 的概率生成函数。

(3)

Ling 也有一个袋子,里面有 6 个红色弹珠和 3 个绿色弹珠。他从袋中随机取出 2 个弹珠,且不放回。随机变量 YY 表示 Ling 取到的红色弹珠个数。已知 YY 的概率生成函数为 112(1+6t+5t2)\frac{1}{12}(1+6t+5t^2)。

随机变量 ZZ 表示 Toby 和 Ling 取到的红色弹珠总数。

(b) 求 ZZ 的概率生成函数,并将答案表示为关于 tt 的多项式。

(3)

(c) 使用 ZZ 的概率生成函数求 Var⁡(Z)\operatorname{Var}(Z)。

(4)
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

不放回抽样时,用组合数分别计算选到 jj 个红球和 3−j3-j 个绿球的方法数。

答题过程

展开

There are (93)=84\binom93=84 equally likely subsets. For j=0,1,2,3j=0,1,2,3,

Pr⁡(X=j)=(6j)(33−j)84.\Pr(X=j)=\frac{\binom6j\binom3{3-j}}{84}.

The numerators are 1,18,45,201,18,45,20, so

GX(t)=1+18t+45t2+20t384.G_X(t)=\frac{1+18t+45t^2+20t^3}{84}.

(b)

解法一

思路

展开

两个袋子的抽取独立,所以总数的生成函数是两个生成函数的乘积。展开并合并各次幂。

答题过程

展开

The generating function is

 GZ(t)= (1+18t+45t2+20t3)(1+6t+5t2)1008= 1+24t+158t2+380t3+345t4+100t51008.\begin{align*} &\,G_Z(t)\\ =&\,\frac{(1+18t+45t^2+20t^3)(1+6t+5t^2)}{1008}\\ =&\,\frac{1+24t+158t^2+380t^3+345t^4+100t^5}{1008}. \end{align*}

(c)

解法一

思路

展开

用多项式在 11 处的一阶和二阶导数求均值及二阶阶乘矩,再算方差。

答题过程

展开

From (b),

 GZ′(1)=24+316+1140+1380+5001008=103, GZ′′(1)=316+2280+4140+20001008=263.\begin{align*} &\,G_Z'(1)=\frac{24+316+1140+1380+500}{1008}=\frac{10}{3},\\ &\,G_Z''(1)=\frac{316+2280+4140+2000}{1008}=\frac{26}{3}. \end{align*}

Therefore

 Var⁡(Z)= GZ′′(1)+GZ′(1)−GZ′(1)2= 263+103−(103)2= 89.\begin{align*} &\,\operatorname{Var}(Z)\\ =&\,G_Z''(1)+G_Z'(1)-G_Z'(1)^2\\ =&\,\frac{26}{3}+\frac{10}{3}-\Big(\frac{10}{3}\Big)^2\\ =&\,\frac89. \end{align*}