CalcGospel 國際數學圖譜
返回

CIE 9709 2025 March Paper 52 Q2

A Level / CIE / S1

CIE 9709 2025 March Paper 52 Paper · Question 2

题目

Problem

Last year, an online store sold a large number of computers. 55%55\% of the computers were made by company FF, 30%30\% were made by company GG and 15%15\% were made by company HH.

A random sample of 3 customers who each bought a computer from this store is chosen.

(a) Find the probability that the 3 customers bought computers all made by different companies.

(1)

A random sample of 12 customers who each bought a computer from this store is chosen.

(b) Find the probability that fewer than 10 of these customers bought a computer made by company FF.

(3)

A random sample of 140 customers who each bought a computer from this store is chosen.

(c) Use a suitable approximation to find the probability that more than 24 of these customers bought a computer made by company HH.

(5)
题目中文翻译

去年，一家网店售出了大量电脑。其中 55%55\% 由公司 FF 生产，30%30\% 由公司 GG 生产，15%15\% 由公司 HH 生产。

从曾在这家网店各购买一台电脑的顾客中随机抽取 3 人作为样本。

(a) 求这 3 名顾客所购电脑全部由不同公司生产的概率。

(1)

从曾在这家网店各购买一台电脑的顾客中随机抽取 12 人作为样本。

(b) 求其中购买公司 FF 生产的电脑的顾客少于 10 人的概率。

(3)

从曾在这家网店各购买一台电脑的顾客中随机抽取 140 人作为样本。

(c) 使用适当的近似方法，求其中购买公司 HH 生产的电脑的顾客超过 24 人的概率。

(5)
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

三位顾客各买不同公司的电脑，一种指定公司次序的概率为三概率相乘，共六种次序。

答题过程

展开 P=3!(0.55)(0.30)(0.15)=0.1485.\mathrm{P}=3!(0.55)(0.30)(0.15)=\boxed{0.1485}.

(b)

解法一

思路

展开

12人中买F电脑的人数服从二项分布。少于10的补集为10、11、12，只有三项。

答题过程

展开

Let X∼B(12,0.55)X\sim\mathrm{B}(12,0.55). Then

P(X<10)= 1−[(1210)(0.55)10(0.45)2 +12(0.55)11(0.45)+(0.55)12]= 0.957858…≈ 0.958.\begin{align*} \mathrm{P}(X<10) =&\,1-\bigl[\binom{12}{10}(0.55)^{10}(0.45)^2\\[4mm] &\,\hspace{2pt}+12(0.55)^{11}(0.45)+(0.55)^{12}\bigr]\\[4mm] =&\,0.957858\ldots\\[4mm] \approx&\,\boxed{0.958}. \end{align*}

(c)

解法一

思路

展开

H电脑的购买概率为0.15。用正态近似140次二项分布，超过24即至少25，修正边界为24.5。

答题过程

展开

Let X∼B(140,0.15)X\sim\mathrm{B}(140,0.15). Since np=21np=21 and n(1−p)=119n(1-p)=119 are both greater than 5, use Y∼N(21,17.85)Y\sim\mathrm{N}(21,17.85), where 17.85=140(0.15)(0.85)17.85=140(0.15)(0.85).

With a continuity correction,

P(X>24)= P(X≥25)≈ P(Y>24.5)= 1−Φ(24.5−2117.85)= 1−Φ(0.828416…)= 0.203717…≈ 0.204.\begin{align*} \mathrm{P}(X>24)=&\,\mathrm{P}(X\geq25)\\[4mm] \approx&\,\mathrm{P}(Y>24.5)\\[4mm] =&\,1-\Phi\biggl(\frac{24.5-21}{\sqrt{17.85}}\biggr)\\[4mm] =&\,1-\Phi(0.828416\ldots)\\[4mm] =&\,0.203717\ldots\\[4mm] \approx&\,\boxed{0.204}. \end{align*}