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CIE 9231 2022 June Paper 43 Q3

A Level / CIE / FS

CIE 9231 2022 June Paper 43 Paper · Question 3

题目

Problem

George throws two coins, A and B, at the same time. Coin A is biased so that the probability of obtaining a head is aa. Coin B is biased so that the probability of obtaining a head is bb, where b<ab < a. The probability generating function of XX, the number of heads obtained by George, is GX(t)G_X(t). The coefficients of tt and t2t^2 in GX(t)G_X(t) are 512\frac{5}{12} and 112\frac{1}{12} respectively.

(a) Find the value of aa.

[2]

The random variable YY is the sum of two independent observations of XX.

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

[3]

(c) Find Var⁡(Y)\operatorname{Var}(Y).

[3]
题目中文翻译

George 同时抛掷两枚硬币 A 和 B。硬币 A 是不均匀的,得到正面的概率为 aa。硬币 B 是不均匀的,得到正面的概率为 bb,其中 b<ab < a。随机变量 XX 表示 George 得到的正面数,其概率母函数为 GX(t)G_X(t)。在 GX(t)G_X(t) 中,tt 和 t2t^2 的系数分别为 512\frac{5}{12} 和 112\frac{1}{12}。

(a) 求 aa 的值。

(2)

随机变量 YY 是 XX 的两个独立观测值之和。

(b) 求 YY 的概率母函数,并将答案表示为 tt 的多项式。

(3)

(c) 求 Var⁡(Y)\operatorname{Var}(Y)。

(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

将两个正面指示变量的生成函数相乘。比较一次项与二次项系数,得到 a+ba+b 和 abab。

答题过程

展开

The generating function is ((1−a)+at)((1−b)+bt)((1-a)+at)((1-b)+bt). Thus ab=1/12ab=1/12 and a+b−2ab=5/12a+b-2ab=5/12, giving a+b=7/12a+b=7/12. Therefore aa and bb are roots of

 u2−712u+112= (u−13)(u−14)=0.\begin{align*} &\,u^2-\frac7{12}u+\frac1{12}\\ =&\,\Big(u-\frac13\Big)\Big(u-\frac14\Big)=0. \end{align*}

Since b<ab<a, we have a=1/3a=1/3 and b=1/4b=1/4.

(b)

解法一

思路

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由概率和为 11 得出常数项,再把一次观察的生成函数平方并展开。

答题过程

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The constant term of GXG_X is 1−5/12−1/12=1/21-5/12-1/12=1/2. Using independence,

 GY(t)=GX(t)2=(6+5t+t2)2144= 36+60t+37t2+10t3+t4144.\begin{align*} &\,G_Y(t)=G_X(t)^2=\frac{(6+5t+t^2)^2}{144}\\ =&\,\frac{36+60t+37t^2+10t^3+t^4}{144}. \end{align*}

(c)

解法一

思路

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用 (b) 的生成函数求均值和二阶阶乘矩,再代入方差公式。

答题过程

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Differentiating the generating function gives

GY′(t)= 60+74t+30t2+4t3144,GY′′(t)= 74+60t+12t2144.\begin{align*} G_Y'(t)=&\,\frac{60+74t+30t^2+4t^3}{144},\\ G_Y''(t)=&\,\frac{74+60t+12t^2}{144}. \end{align*}

The derivatives at 11 give

 GY′(1)=60+74+30+4144=76, GY′′(1)=74+60+12144=7372.\begin{align*} &\,G_Y'(1)=\frac{60+74+30+4}{144}=\frac76,\\ &\,G_Y''(1)=\frac{74+60+12}{144}=\frac{73}{72}. \end{align*}

Using Var⁡(Y)=GY′′(1)+GY′(1)−(GY′(1))2\operatorname{Var}(Y)=G_Y''(1)+G_Y'(1)-\big(G_Y'(1)\big)^2, we obtain Var⁡(Y)=73/72+7/6−(7/6)2=59/72\operatorname{Var}(Y)=73/72+7/6-(7/6)^2=59/72.

解法二

思路

展开

也可以把 YY 看成四次独立抛掷的正面总数。两枚硬币各投两次,因此分别求指标方差再相加。

答题过程

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For each of the two throws of coin A the variance is (1/3)(2/3)(1/3)(2/3), and for each throw of coin B it is (1/4)(3/4)(1/4)(3/4). Independence gives

Var⁡(Y)=2(29+316)=5972.\operatorname{Var}(Y)=2\Big(\frac29+\frac3{16}\Big)=\frac{59}{72}.