题目

Problem

There are a large number of students at Greenfield college. Each student travels to college by car, by bus or on foot, independently of any other student. The probability that any student travels by car is 0.4. The probability that any student travels by bus is 0.35.

(a) 3 students from Greenfield college are selected at random.

Find the probability that none of these 3 students travel to college by car.

(1)

(b) 11 students from Greenfield college are selected at random.

Find the probability that fewer than 9 of these 11 students travel to college by car or by bus.

(3)

题目中文翻译

Greenfield 学院有很多学生。每名学生乘小汽车、乘公共汽车或步行前往学院，其出行方式与其他任何学生相互独立。任一学生乘小汽车的概率为 0.4，乘公共汽车的概率为 0.35。

(a) 从 Greenfield 学院的学生中随机选取 3 人。

求这 3 人中没有人乘小汽车前往学院的概率。

(1)

(b) 从 Greenfield 学院的学生中随机选取 11 人。

求这 11 人中乘小汽车或公共汽车前往学院的人数少于 9 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

不乘小汽车的概率是1减0.4，三名学生相互独立，概率连乘。

答题过程

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\mathrm{P}(\text{none by car})=(1-0.4)^3 =\boxed{0.216}.

(b)

解法一

思路

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乘汽车或巴士互斥，合计概率0.75。11人中满足条件的人数服从二项分布；少于9人的补集只有9、10、11人三项。

答题过程

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Let $X$ be the number who travel by car or bus. Since $p=0.4+0.35=0.75$ , we have $X\sim\mathrm{B}(11,0.75)$ .

\begin{align*} \mathrm{P}(X<9)=&\,1-\mathrm{P}(X\geq9)\\[4mm] =&\,1-\bigl[\binom{11}{9}(0.75)^9(0.25)^2\\[4mm] &\,\hspace{2pt}+11(0.75)^{10}(0.25)+(0.75)^{11}\bigr]\\[4mm] =&\,0.544799\ldots\\[4mm] \approx&\,\boxed{0.545}. \end{align*}

CIE 9709 2025 November Paper 53 Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程