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CIE 9231 2026 June Paper 44 Q2

A Level / CIE / FS

CIE 9231 2026 June Paper 44 Paper · Question 2

题目

Problem

The probability generating function GX(t)\mathrm{G}_X(t) of a random variable XX is given by

GX(t)=16(t+1)(t2+2).\mathrm{G}_X(t)=\frac{1}{6}(t+1)(t^2+2).

(a) Use GX(t)\mathrm{G}_X(t) to find E(X)\mathrm{E}(X) and Var⁡(X)\operatorname{Var}(X).

(4)

(b) The random variable YY is the sum of three independent observations of XX.

Find P(Y≤2)\mathrm{P}(Y\le2).

(3)
题目中文翻译

随机变量 XX 的概率生成函数 GX(t)\mathrm{G}_X(t) 为

GX(t)=16(t+1)(t2+2).\mathrm{G}_X(t)=\frac{1}{6}(t+1)(t^2+2).

(a) 利用 GX(t)\mathrm{G}_X(t) 求 E(X)\mathrm{E}(X) 和 Var⁡(X)\operatorname{Var}(X)。

(4)

(b) 随机变量 YY 是对 XX 进行三次相互独立的观测所得观测值的和。

求 P(Y≤2)\mathrm{P}(Y\le2)。

(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

展开生成函数,求两阶导数在 11 处的值,再代入方差公式。

答题过程

展开

The expanded PGF is GX(t)=(2+2t+t2+t3)/6G_X(t)=(2+2t+t^2+t^3)/6, so

GX′(t)=2+2t+3t26,GX′′(t)=2+6t6.G_X'(t)=\frac{2+2t+3t^2}{6},\qquad G_X''(t)=\frac{2+6t}{6}.

Hence

 GX′(1)=2+2+36=76, GX′′(1)=2+66=43.\begin{align*} &\,G_X'(1)=\frac{2+2+3}{6}=\frac76,\\ &\,G_X''(1)=\frac{2+6}{6}=\frac43. \end{align*}

Thus E⁡[X]=7/6\operatorname E[X]=7/6 and

Var⁡(X)= GX′′(1)+GX′(1)−(GX′(1))2= 43+76−(76)2=4136.\begin{align*} \operatorname{Var}(X)=&\,G_X''(1)+G_X'(1)-\big(G_X'(1)\big)^2\\ =&\,\frac43+\frac76-\Big(\frac76\Big)^2=\frac{41}{36}. \end{align*}

(b)

解法一

思路

展开

用独立性将生成函数取三次幂。只保留常数、一次项、二次项,不必展开更高次项。

答题过程

展开

We have GY(t)=GX(t)3=1216(1+t)3(2+t2)3G_Y(t)=G_X(t)^3=\frac1{216}(1+t)^3(2+t^2)^3. Up to degree two, the two factors are 1+3t+3t21+3t+3t^2 and 8+12t28+12t^2. Thus the coefficients of 1,t,t21,t,t^2 in their product are 8,24,368,24,36. Therefore

Pr⁡(Y≤2)=8+24+36216=1754.\Pr(Y\le2)=\frac{8+24+36}{216}=\frac{17}{54}.

解法二

思路

展开

也可以枚举三个观察值相加不超过 22 的组合,再计算每种排列的概率。

答题过程

展开

The probabilities for X=0,1,2X=0,1,2 are 1/3,1/3,1/61/3,1/3,1/6. The allowed triples, up to ordering, are (0,0,0)(0,0,0), (1,0,0)(1,0,0), (2,0,0)(2,0,0) and (1,1,0)(1,1,0). Independence gives

 Pr⁡(Y≤2)= (13)3+ 3(13)3+ 316(13)2+ 3(13)3= 1754.\begin{align*} &\,\Pr(Y\le2)\\ =&\,\Big(\frac13\Big)^3\\ +&\,3\Big(\frac13\Big)^3\\ +&\,3\frac16\Big(\frac13\Big)^2\\ +&\,3\Big(\frac13\Big)^3\\ =&\,\frac{17}{54}. \end{align*}