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CIE 9231 2023 June Paper 43 Q5

A Level / CIE / FS

CIE 9231 2023 June Paper 43 Paper · Question 5

题目

Problem

The random variable XX has probability generating function GX(t)G_X(t) given by

GX(t)=k(1+3t+4t2),G_X(t) = k(1 + 3t + 4t^2),

where kk is a constant.

(a) Show that E(X)=118E(X) = \dfrac{11}{8}.

[3]

The random variable YY has probability generating function GY(t)G_Y(t) given by

GY(t)=13t2(1+2t).G_Y(t) = \frac{1}{3}t^2(1 + 2t).

The random variables XX and YY are independent and Z=X+YZ = X + Y.

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

[2]

(c) Use your answer to part (b) to find the value of Var(Z)\operatorname{Var}(Z).

[3]

(d) Write down the most probable value of ZZ.

[1]
题目中文翻译

随机变量 XX 的概率母函数 GX(t)G_X(t)

GX(t)=k(1+3t+4t2),G_X(t) = k(1 + 3t + 4t^2),

其中 kk 为常数。

(a) 证明 E(X)=118E(X) = \dfrac{11}{8}

随机变量 YY 的概率母函数 GY(t)G_Y(t)

GY(t)=13t2(1+2t).G_Y(t) = \frac{1}{3}t^2(1 + 2t).

随机变量 XXYY 相互独立,且 Z=X+YZ = X + Y

(b) 求 ZZ 的概率母函数,并将答案表示为关于 tt 的多项式。

(c) 利用你在 part (b) 中的答案,求 Var(Z)\operatorname{Var}(Z) 的值。

(d) 写出 ZZ 最可能的取值。

解答