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CIE 9709 2023 March Paper 52 Q3

A Level / CIE / S1

CIE 9709 2023 March Paper 52 Paper · Question 3

题目

Problem

80% of the residents of Kinwawa are in favour of a leisure centre being built in the town.

20 residents of Kinwawa are chosen at random and asked, in turn, whether they are in favour of the leisure centre.

(a) Find the probability that more than 17 of these residents are in favour of the leisure centre.

(3)

(b) Find the probability that the 5th person asked is the first person who is not in favour of the leisure centre.

(1)

(c) Find the probability that the 7th person asked is the second person who is not in favour of the leisure centre.

(2)

题目中文翻译

Kinwawa 的居民中有 80% 赞成在镇上建造一座休闲中心。

随机选择 Kinwawa 的 20 名居民，依次询问他们是否赞成建造这座休闲中心。

(a) 求这些居民中超过 17 人赞成建造休闲中心的概率。

(3)

(b) 求第 5 名被询问者是第一位不赞成建造休闲中心的人的概率。

(1)

(c) 求第 7 名被询问者是第二位不赞成建造休闲中心的人的概率。

(2)

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

赞成人数服从二项分布，超过17即18、19、20三项。

答题过程

展开

Let S∼B(20,0.8)S\sim\mathrm{B}(20,0.8). Then

P(S>17)= (2018)(0.8)18(0.2)2 +20(0.8)19(0.2)+(0.8)20= 0.206084718…≈0.206.\begin{align*} \mathrm{P}(S>17) =&\,\binom{20}{18}(0.8)^{18}(0.2)^2\\[4mm] &\,\hspace{2pt}+20(0.8)^{19}(0.2)+(0.8)^{20}\\[4mm] =&\,0.206084718\ldots\approx\boxed{0.206}. \end{align*}

(b)

解法一

思路

展开

首次反对在第5人，前4人全赞成，第5人反对。

答题过程

展开 P=(0.8)4(0.2)=0.08192.\mathrm{P}=(0.8)^4(0.2)=\boxed{0.08192}.

(c)

解法一

思路

展开

前6人恰1人反对，第7人反对，首次反对有6个位置。

答题过程

展开 P=(61)(0.2)(0.8)5(0.2)=0.0786432≈0.0786.\mathrm{P}=\binom61(0.2)(0.8)^5(0.2) =0.0786432\approx\boxed{0.0786}.