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CIE 9231 2023 June Paper 41 Q5

A Level / CIE / FS

CIE 9231 2023 June Paper 41 Paper · Question 5

题目

Problem

Harry has three coins.

  • One coin is biased so that, when it is thrown, the probability of obtaining a head is 13\frac{1}{3}.
  • The second coin is biased so that, when it is thrown, the probability of obtaining a head is 14\frac{1}{4}.
  • The third coin is biased so that, when it is thrown, the probability of obtaining a head is 15\frac{1}{5}.

The random variable XX is the number of heads that Harry obtains when he throws all three coins together.

(a) Find the probability generating function of XX.

[3]

Isaac has two fair coins. The random variable YY is the number of heads that Isaac obtains when he throws both of his coins together. The random variable ZZ is the total number of heads obtained when Harry throws his three coins and Isaac throws his two coins.

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

[4]

(c) Use the probability generating function of ZZ to find E(Z)\mathrm{E}(Z).

[2]
题目中文翻译

Harry 有三枚硬币。

  • 第一枚硬币是不均匀的，抛掷时得到正面的概率为 13\frac{1}{3}。
  • 第二枚硬币是不均匀的，抛掷时得到正面的概率为 14\frac{1}{4}。
  • 第三枚硬币是不均匀的，抛掷时得到正面的概率为 15\frac{1}{5}。

随机变量 XX 表示 Harry 同时抛掷这三枚硬币时得到的正面数。

(a) 求 XX 的概率母函数。

(3)

Isaac 有两枚公平硬币。随机变量 YY 表示 Isaac 同时抛掷这两枚硬币时得到的正面数。随机变量 ZZ 表示 Harry 抛掷三枚硬币且 Isaac 抛掷两枚硬币时得到的正面总数。

(b) 求 ZZ 的概率母函数，并将答案表示为 tt 的多项式。

(4)

(c) 使用 ZZ 的概率母函数求 E(Z)\mathrm{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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每枚硬币的正面指示变量有一个一次生成函数。用独立性将三个生成函数相乘。

答题过程

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The three factors are (2+t)/3(2+t)/3, (3+t)/4(3+t)/4 and (4+t)/5(4+t)/5. Thus

 GX(t)=(2+t)(3+t)(4+t)60= 24+26t+9t2+t360.\begin{align*} &\,G_X(t)=\frac{(2+t)(3+t)(4+t)}{60}\\ =&\,\frac{24+26t+9t^2+t^3}{60}. \end{align*}

解法二

思路

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分别列出得到零、一、二、三个正面的概率，再把这些概率作为母函数的系数。

答题过程

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Writing H for a head and T for a tail, independence gives

P(X=0)= 23⋅34⋅45=2460,P(X=1)= P(HTT)+P(THT)+P(TTH)= 12+8+660=2660,P(X=2)= P(HHT)+P(HTH)+P(THH)= 4+3+260=960,P(X=3)= 13⋅14⋅15=160.\begin{align*} P(X=0)=&\,\frac23\cdot\frac34\cdot\frac45=\frac{24}{60},\\ P(X=1)=&\,P(\mathrm{HTT})+P(\mathrm{THT})+P(\mathrm{TTH})\\ =&\,\frac{12+8+6}{60}=\frac{26}{60},\\ P(X=2)=&\,P(\mathrm{HHT})+P(\mathrm{HTH})+P(\mathrm{THH})\\ =&\,\frac{4+3+2}{60}=\frac9{60},\\ P(X=3)=&\,\frac13\cdot\frac14\cdot\frac15=\frac1{60}. \end{align*}

Therefore

GX(t)=24+26t+9t2+t360.G_X(t)=\frac{24+26t+9t^2+t^3}{60}.

(b)

解法一

思路

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两枚公平硬币给出 ((1+t)/2)2((1+t)/2)^2。将它与 (a) 相乘，并展开成多项式。

答题过程

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For Isaac’s two fair coins, GY(t)=(1+t)2/4=(1+2t+t2)/4G_Y(t)=(1+t)^2/4=(1+2t+t^2)/4. By independence,

 GZ(t)= (24+26t+9t2+t3)(1+2t+t2)240= 24+74t+85t2+45t3+11t4+t5240.\begin{align*} &\,G_Z(t)\\ =&\,\frac{(24+26t+9t^2+t^3)(1+2t+t^2)}{240}\\ =&\,\frac{24+74t+85t^2+45t^3+11t^4+t^5}{240}. \end{align*}

(c)

解法一

思路

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按题意使用 (b) 的多项式。其一阶导数在 t=1t=1 处给出均值。

答题过程

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Differentiating the probability generating function from (b),

GZ′(t)=74+170t+135t2+44t3+5t4240.G_Z'(t)=\frac{74+170t+135t^2+44t^3+5t^4}{240}.

The mean is

 GZ′(1)=74+2(85)+3(45)+4(11)+5240= 428240=10760.\begin{align*} &\,G_Z'(1)=\frac{74+2(85)+3(45)+4(11)+5}{240}\\ =&\,\frac{428}{240}=\frac{107}{60}. \end{align*}