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CIE 9231 2025 June Paper 44 Q5

A Level / CIE / FS

CIE 9231 2025 June Paper 44 Paper · Question 5

题目

Problem

Eric has three identical coins, each of which is biased so that the probability of obtaining a head when it is thrown is 13\frac{1}{3}. The random variable XX is the number of heads obtained when Eric throws the three coins at the same time.

(a) Find the probability generating function GX(t)G_X(t) of XX.

[2]

Eric also has two fair 6-sided dice with faces numbered 1 to 6. The random variable YY is the number of sixes obtained when Eric throws the two dice at the same time. It is given that the probability generating function of YY is 2536+1036t+136t2\frac{25}{36}+\frac{10}{36}t+\frac{1}{36}t^2.

Eric throws the three coins and the two dice. The random variable ZZ is the sum of the number of heads obtained and the number of sixes obtained.

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

[3]

(c) Use GZ(t)G_Z(t) to find E(Z)E(Z) and Var⁡(Z)\operatorname{Var}(Z).

[5]
题目中文翻译

Eric 有三枚相同的硬币，每枚硬币都是不公平的，掷出正面的概率为 13\frac{1}{3}。随机变量 XX 表示 Eric 同时掷这三枚硬币时得到的正面个数。

(a) 求 XX 的概率生成函数 GX(t)G_X(t)。

(2)

Eric 还有两枚公平的 6 面骰子，各面编号为 1 到 6。随机变量 YY 表示 Eric 同时掷这两枚骰子时得到 6 的次数。已知 YY 的概率生成函数为 2536+1036t+136t2\frac{25}{36}+\frac{10}{36}t+\frac{1}{36}t^2。

Eric 掷这三枚硬币和两枚骰子。随机变量 ZZ 表示得到的正面个数与得到 6 的次数之和。

(b) 求 ZZ 的概率生成函数 GZ(t)G_Z(t)，并将答案表示为关于 tt 的多项式。

(3)

(c) 使用 GZ(t)G_Z(t) 求 E(Z)E(Z) 和 Var⁡(Z)\operatorname{Var}(Z)。

(5)
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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把每枚硬币的正面指示变量写成一次生成函数。三个独立结果相加对应三次幂。

答题过程

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The PGF is

 GX(t)=(23+13t)3= 8+12t+6t2+t327.\begin{align*} &\,G_X(t)=\Big(\frac23+\frac13t\Big)^3\\ =&\,\frac{8+12t+6t^2+t^3}{27}. \end{align*}

解法二

思路

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先分别求得到零、一、二、三个正面的概率，再用概率生成函数的定义组成多项式。

答题过程

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Since XX is binomial with parameters 33 and 1/31/3,

Pr⁡(X=0)= (23)3=827,Pr⁡(X=1)= 3(13)(23)2=1227,Pr⁡(X=2)= 3(13)2(23)=627,Pr⁡(X=3)= (13)3=127.\begin{align*} \Pr(X=0)=&\,\Big(\frac23\Big)^3=\frac8{27},\\ \Pr(X=1)=&\,3\Big(\frac13\Big)\Big(\frac23\Big)^2=\frac{12}{27},\\ \Pr(X=2)=&\,3\Big(\frac13\Big)^2\Big(\frac23\Big)=\frac6{27},\\ \Pr(X=3)=&\,\Big(\frac13\Big)^3=\frac1{27}. \end{align*}

Therefore

GX(t)=827+1227t+627t2+127t3.G_X(t)=\frac8{27}+\frac{12}{27}t+\frac6{27}t^2+\frac1{27}t^3.

(b)

解法一

思路

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将硬币和骰子的生成函数相乘。按 tt 的次数收集系数。

答题过程

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Independence gives

 GZ(t)= (8+12t+6t2+t3)(25+10t+t2)972= 200+380t+278t2+97t3+16t4+t5972.\begin{align*} &\,G_Z(t)\\ =&\,\frac{(8+12t+6t^2+t^3)(25+10t+t^2)}{972}\\ =&\,\frac{200+380t+278t^2+97t^3+16t^4+t^5}{972}. \end{align*}

(c)

解法一

思路

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按题意使用 GZG_Z。在 11 处求导，分别得到均值和二阶阶乘矩。

答题过程

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Differentiating the polynomial gives

GZ′(t)= 380+556t+291t2+64t3+5t4972,GZ′′(t)= 556+582t+192t2+20t3972.\begin{align*} G_Z'(t)=&\,\frac{380+556t+291t^2+64t^3+5t^4}{972},\\ G_Z''(t)=&\,\frac{556+582t+192t^2+20t^3}{972}. \end{align*}

Hence

GZ′(1)=380+2(278)+3(97)+4(16)+5972=43,G_Z'(1)=\frac{380+2(278)+3(97)+4(16)+5}{972}=\frac43,

and

GZ′′(1)=2(278)+6(97)+12(16)+20972=2518.G_Z''(1)=\frac{2(278)+6(97)+12(16)+20}{972}=\frac{25}{18}.

Hence E⁡[Z]=4/3\operatorname E[Z]=4/3 and

 Var⁡(Z)= GZ′′(1)+GZ′(1)−(GZ′(1))2= 2518+43−(43)2=1718.\begin{align*} &\,\operatorname{Var}(Z)\\ =&\,G_Z''(1)+G_Z'(1)-\big(G_Z'(1)\big)^2\\ =&\,\frac{25}{18}+\frac43-\Big(\frac43\Big)^2=\frac{17}{18}. \end{align*}