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CIE 9709 2023 June Paper 51 Q7

A Level / CIE / S1

CIE 9709 2023 June Paper 51 Paper · Question 7

题目

Problem

A children’s wildlife magazine is published every Monday. For the next 12 weeks it will include a model animal as a free gift. There are five different models: tiger, leopard, rhinoceros, elephant and buffalo, each with the same probability of being included in the magazine.

Sahim buys one copy of the magazine every Monday.

(a) Find the probability that the first time that the free gift is an elephant is before the 6th Monday.

(2)

(b) Find the probability that Sahim will get more than two leopards in the 12 magazines.

(3)

(c) Find the probability that after 5 weeks Sahim has exactly one of each animal.

(3)

题目中文翻译

一本儿童野生动物杂志每周一出版。在接下来的 12 周里，每期杂志将免费附赠一个动物模型。共有五种不同的模型：老虎、豹、犀牛、大象和水牛，每种模型随杂志附赠的概率相同。

Sahim 每周一购买一本该杂志。

(a) 求免费赠品首次为大象模型是在第 6 个周一之前的概率。

(2)

(b) 求 Sahim 从这 12 本杂志中得到超过两个豹模型的概率。

(3)

(c) 求 5 周后 Sahim 恰好每种动物模型各有一个的概率。

(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

第6个周一之前即前5次出现大象，取五次都非大象的补集。

答题过程

展开 P=1−(0.8)5=0.67232≈0.672.\mathrm{P}=1-(0.8)^5=0.67232\approx\boxed{0.672}.

解法二

思路

展开

也可以按第一次大象在第1至第5周分类求等比和。

答题过程

展开 P= 0.2[1+0.8+0.82 +0.83+0.84]= 0.21−0.851−0.8= 0.67232≈0.672.\begin{align*} \mathrm{P}=&\,0.2\bigl[1+0.8+0.8^2\\[4mm] &\,\hspace{2pt}+0.8^3+0.8^4\bigr]\\[4mm] =&\,0.2\frac{1-0.8^5}{1-0.8}\\[4mm] =&\,0.67232\approx\boxed{0.672}. \end{align*}

(b)

解法一

思路

展开

豹模型数量服从二项分布，超过2的补集只有0、1、2三项。

答题过程

展开

Let L∼B(12,0.2)L\sim\mathrm{B}(12,0.2). Then

P(L>2)= 1−[(0.8)12+12(0.2)(0.8)11 +(122)(0.2)2(0.8)10]= 0.441654251…≈0.442.\begin{align*} \mathrm{P}(L>2) =&\,1-\bigl[(0.8)^{12}+12(0.2)(0.8)^{11}\\[4mm] &\,\hspace{2pt}+\binom{12}{2}(0.2)^2(0.8)^{10}\bigr]\\[4mm] =&\,0.441654251\ldots\approx\boxed{0.442}. \end{align*}

(c)

解法一

思路

展开

指定一种五种动物次序的概率为(1/5)^5，共有5!种次序。

答题过程

展开 P=5!(15)5=1203125=0.0384.\mathrm{P}=5!\biggl(\frac15\biggr)^5 =\frac{120}{3125}=\boxed{0.0384}.