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CIE 9231 2023 June Paper 42 Q5

A Level / CIE / FS

CIE 9231 2023 June Paper 42 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。

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。

(c) 使用 ZZ 的概率母函数求 E(Z)\mathrm{E}(Z)

解答