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CIE 9231 2025 June Paper 43 Q6

A Level / CIE / FS

CIE 9231 2025 June Paper 43 Paper · Question 6

题目

Problem

A bag contains 7 red balls and 3 blue balls. Kieran selects 2 balls at random, without replacement. The number of red balls selected by Kieran is denoted by XX, and the number of different colours present in Kieran’s selection is denoted by YY.

(a) Find the probability generating functions, GX(t)G_X(t) of XX and GY(t)G_Y(t) of YY.

[4]

The random variable ZZ is the sum of the number of red balls and the number of different colours present in Kieran’s selection. Kieran claims that the probability generating function of ZZ is equal to GX(t)×GY(t)G_X(t) \times G_Y(t).

(b) Explain why Kieran is incorrect.

[1]

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

[4]

(d) Use the probability generating function of ZZ to find E(Z)E(Z).

[2]
题目中文翻译

一个袋子中有 7 个红球和 3 个蓝球。Kieran 随机取出 2 个球,不放回。Kieran 取出的红球个数记为 XX,Kieran 所取球中出现的不同颜色的种数记为 YY

(a) 求 XX 的概率母函数 GX(t)G_X(t)YY 的概率母函数 GY(t)G_Y(t)

随机变量 ZZ 是红球个数与 Kieran 所取球中出现的不同颜色种数之和。Kieran 声称 ZZ 的概率母函数等于 GX(t)×GY(t)G_X(t) \times G_Y(t)

(b) 解释为什么 Kieran 是错误的。

(c) 求 ZZ 的概率母函数,并将答案表示为 tt 的多项式。

(d) 使用 ZZ 的概率母函数求 E(Z)E(Z)

解答