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CIE 9231 2022 Nov Paper 43 Q4

A Level / CIE / FS

CIE 9231 2022 Nov Paper 43 Paper · Question 4

题目

Problem

Jason has three biased coins. For each coin the probability of obtaining a head when it is thrown is 23\frac{2}{3}. Jason throws all three coins. The number of heads obtained is denoted by XX.

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

[3]

Jason also has two unbiased coins. He throws all five coins. The number of heads obtained from the two unbiased coins is denoted by YY. It is given that GY(t)=14+12t+14t2G_Y(t)=\frac{1}{4}+\frac{1}{2}t+\frac{1}{4}t^2. The random variable ZZ is the total number of heads obtained when Jason throws all five coins.

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

[3]

(c) Find E⁡(Z)\operatorname{E}(Z).

[2]
题目中文翻译

Jason 有三枚有偏硬币。每枚硬币抛出正面的概率为 23\frac{2}{3}。Jason 抛掷全部三枚硬币。所得正面个数记为 XX。

(a) 求 XX 的概率生成函数 GX(t)G_X(t)。

(3)

Jason 还有两枚公平硬币。他抛掷全部五枚硬币。两枚公平硬币所得正面个数记为 YY。已知 GY(t)=14+12t+14t2G_Y(t)=\frac{1}{4}+\frac{1}{2}t+\frac{1}{4}t^2。随机变量 ZZ 表示 Jason 抛掷全部五枚硬币时得到的正面总数。

(b) 求 ZZ 的概率生成函数，并将答案表示为多项式。

(3)

(c) 求 E⁡(Z)\operatorname{E}(Z)。

(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

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将每枚偏硬币的正面指示变量写成一次生成函数。三次独立抛掷对应三次幂。

答题过程

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The PGF is

 GX(t)=(13+23t)3= 1+6t+12t2+8t327.\begin{align*} &\,G_X(t)=\Big(\frac13+\frac23t\Big)^3\\ =&\,\frac{1+6t+12t^2+8t^3}{27}. \end{align*}

解法二

思路

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先按正面个数计算二项概率，再把这些概率作为生成函数各项的系数。

答题过程

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The three independent coins give

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

Hence

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

(b)

解法一

思路

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把 (a) 和题给的生成函数相乘，再按 tt 的次数合并。

答题过程

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By independence,

 GZ(t)= (1+6t+12t2+8t3)(1+2t+t2)108= 1+8t+25t2+38t3+28t4+8t5108.\begin{align*} &\,G_Z(t)\\ =&\,\frac{(1+6t+12t^2+8t^3)(1+2t+t^2)}{108}\\ =&\,\frac{1+8t+25t^2+38t^3+28t^4+8t^5}{108}. \end{align*}

(c)

解法一

思路

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用生成函数的一阶导数在 t=1t=1 处求期望。

答题过程

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Differentiating the polynomial gives

E⁡[Z]=GZ′(1)=8+2(25)+3(38)+4(28)+5(8)108=3.\operatorname E[Z]=G_Z'(1)=\frac{8+2(25)+3(38)+4(28)+5(8)}{108}=3.

解法二

思路

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也可以把五枚硬币各自的正面概率相加。

答题过程

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The expected number of heads is the sum of the five indicator means, so E⁡[Z]=3(2/3)+2(1/2)=3\operatorname E[Z]=3(2/3)+2(1/2)=3.