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CIE 9709 2025 June Paper 53 Q2

A Level / CIE / S1

CIE 9709 2025 June Paper 53 Paper · Question 2

题目

Problem

At a large college, all students who study Science also study exactly one of Art or Drama or Music. 20%20\% of these students study Art, 45%45\% study Drama and 35%35\% study Music.

(a) 3 students are selected at random from the students who study Science.

Find the probability that at least 1 of these students studies Drama.

(2)

(b) 10 students are selected at random from the students who study Science.

Find the probability that more than 7 study Art or Music.

(3)
题目中文翻译

在一所规模很大的学校中，所有学习科学的学生也都恰好选学美术、戏剧或音乐中的一门。这些学生中，20%20\% 学习美术，45%45\% 学习戏剧，35%35\% 学习音乐。

(a) 从学习科学的学生中随机选取 3 人。

求其中至少 1 人学习戏剧的概率。

(2)

(b) 从学习科学的学生中随机选取 10 人。

求其中超过 7 人学习美术或音乐的概率。

(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

至少一位学习戏剧的补集是三位都不学戏剧，概率为0.55的三次方。

答题过程

展开  P(at least one Drama student)= 1−(1−0.45)3= 1−(0.55)3= 0.833625.\begin{align*} &\,\mathrm{P}(\text{at least one Drama student})\\[4mm] =&\,1-(1-0.45)^3\\[4mm] =&\,1-(0.55)^3\\[4mm] =&\,\boxed{0.833625}. \end{align*}

解法二

思路

展开

也可以按恰有1、2、3位戏剧学生分类，使用三次二项分布的概率相加。

答题过程

展开

Let X∼B(3,0.45)X\sim\mathrm{B}(3,0.45) be the number studying Drama. Then

P(X≥1)= 3(0.45)(0.55)2+3(0.45)2(0.55)+(0.45)3= 0.833625.\begin{align*} \mathrm{P}(X\geq1) =&\,3(0.45)(0.55)^2 +3(0.45)^2(0.55)+(0.45)^3\\[4mm] =&\,\boxed{0.833625}. \end{align*}

(b)

解法一

思路

展开

美术和音乐互斥，合计概率0.55。超过7位即8、9、10位，直接相加三项。

答题过程

展开

Let XX be the number studying Art or Music. Then p=0.20+0.35=0.55p=0.20+0.35=0.55, so X∼B(10,0.55)X\sim\mathrm{B}(10,0.55).

P(X>7)= 45(0.55)8(0.45)2 +10(0.55)9(0.45)+(0.55)10= 0.0995596…≈ 0.0996.\begin{align*} \mathrm{P}(X>7) =&\,45(0.55)^8(0.45)^2\\[4mm] &\,\hspace{2pt}+10(0.55)^9(0.45)+(0.55)^{10}\\[4mm] =&\,0.0995596\ldots\\[4mm] \approx&\,\boxed{0.0996}. \end{align*}