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CIE 9709 2025 June Paper 52 Q4

A Level / CIE / S1

CIE 9709 2025 June Paper 52 Paper · Question 4

题目

Problem

Vehicles approaching a certain road junction from Bromley must go either left, right or straight on. Over time, it is known that 30%30\% turn left, 25%25\% turn right and 45%45\% go straight on. The driver of each vehicle chooses a direction independently of all other drivers.

(a) Find the probability that the next three vehicles approaching this junction from Bromley all go in different directions.

(2)

(b) Find the probability that, from the vehicles approaching this junction from Bromley today, the 1st vehicle to go left is before the 9th vehicle.

(2)

(c) Find the probability that, from the vehicles approaching this junction from Bromley today, the 2nd vehicle to go left is the 7th vehicle.

(2)
题目中文翻译

从 Bromley 驶向某个道路交叉口的车辆必须选择左转、右转或直行。长期以来，已知 30%30\% 的车辆左转，25%25\% 右转，45%45\% 直行。每辆车的驾驶员选择方向时都与其他所有驾驶员相互独立。

(a) 求接下来从 Bromley 驶向该交叉口的三辆车各自选择不同方向的概率。

(2)

(b) 在今天从 Bromley 驶向该交叉口的车辆中，求第一辆左转的车出现在第 9 辆车之前的概率。

(2)

(c) 在今天从 Bromley 驶向该交叉口的车辆中，求第二辆左转的车恰好是第 7 辆车的概率。

(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

三个方向各出现一次，指定次序的概率为三概率相乘，次序共有3的阶乘种。

答题过程

展开

One specified order has probability 0.30(0.25)(0.45)0.30(0.25)(0.45). There are 3!=63!=6 different orders, so

P=6(0.30)(0.25)(0.45)=0.2025.\mathrm{P}=6(0.30)(0.25)(0.45)=\boxed{0.2025}.

(b)

解法一

思路

展开

第9辆之前首次左转，就是前8辆至少一次左转，取补集。

答题过程

展开  P(first left before vehicle 9)= 1−(0.7)8= 0.94235199≈ 0.942(3 s.f.).\begin{align*} &\,\mathrm{P}(\text{first left before vehicle 9})\\[4mm] =&\,1-(0.7)^8\\[4mm] =&\,0.94235199\\[4mm] \approx&\,\boxed{0.942}\quad\text{(3 s.f.)}. \end{align*}

解法二

思路

展开

也可以按首次左转的序号分类，从第1至第8辆逐项相加。

答题过程

展开

Let TT be the position of the first vehicle to turn left. Then

P(T<9)= ∑k=18(0.7)k−1(0.3)= 0.31−(0.7)81−0.7≈ 0.942(3 s.f.).\begin{align*} \mathrm{P}(T<9)=&\,\sum_{k=1}^{8}(0.7)^{k-1}(0.3)\\[4mm] =&\,0.3\frac{1-(0.7)^8}{1-0.7}\\[4mm] \approx&\,\boxed{0.942}\quad\text{(3 s.f.)}. \end{align*}

(c)

解法一

思路

展开

前6辆恰好一次左转，第7辆再左转。组合数只安排前6辆中的那次成功。

答题过程

展开 P= (61)(0.3)(0.7)5(0.3)= 0.0907578≈ 0.0908.\begin{align*} \mathrm{P}=&\,\binom61(0.3)(0.7)^5(0.3)\\[4mm] =&\,0.0907578\\[4mm] \approx&\,\boxed{0.0908}. \end{align*}