题目

Problem

The numbers of customers arriving at service desks $A$ and $B$ during a $10$ -minute period have the independent distributions $\mathrm{Po}(1.8)$ and $\mathrm{Po}(2.1)$ respectively.

(a) Find the probability that during a randomly chosen $15$ -minute period more than $2$ customers will arrive at desk $A$ .

(2)

(b) Find the probability that during a randomly chosen $5$ -minute period the total number of customers arriving at both desks is less than $4$ .

(3)

(c) An inspector waits at desk $B$ . She wants to wait long enough to be $90\%$ certain of seeing at least one customer arrive at the desk.

Find the minimum time for which she should wait, giving your answer correct to the nearest minute.

(4)

题目中文翻译

在一段 $10$ 分钟的时间内，到达服务台 $A$ 和 $B$ 的顾客人数相互独立，分别服从 $\mathrm{Po}(1.8)$ 和 $\mathrm{Po}(2.1)$ 分布。

(a) 随机选取一段 $15$ 分钟的时间，求到达服务台 $A$ 的顾客超过 $2$ 人的概率。

(2)

(b) 随机选取一段 $5$ 分钟的时间，求到达两个服务台的顾客总人数少于 $4$ 人的概率。

(3)

(c) 一名检查员在服务台 $B$ 等候。她希望等待足够长的时间，使她看到至少一名顾客到达该服务台的概率达到 $90\%$ 。

求她应等候的最短时间，答案精确到最接近的整分钟。

(4)

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

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将泊松参数按15分钟缩放，超过2人的补集为不超过2人。

答题过程

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Let $A_{15}$ be the number arriving at desk $A$ in 15 minutes. Then

A_{15}\sim\mathrm{Po}\bigg(1.8\frac{15}{10}\bigg)=\mathrm{Po}(2.7).

Therefore

\begin{align*} \mathrm{P}(A_{15}>2)=&\,1-e^{-2.7}\bigg(1+2.7+\frac{2.7^2}{2!}\bigg)\\[4mm] \approx&\,\boxed{0.506}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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先将两个服务台的参数都缩放至5分钟，再利用独立泊松变量的可加性求总数。

答题过程

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The total number $C$ arriving at both desks in 5 minutes has distribution

C\sim\mathrm{Po}\bigg((1.8+2.1)\frac{5}{10}\bigg)=\mathrm{Po}(1.95).

Hence

\begin{align*} \mathrm{P}(C<4)=&\,\mathrm{P}(C\leq3)\\[4mm] =&\,e^{-1.95}(1+1.95)\\[4mm] &\,+e^{-1.95}\bigg(\frac{1.95^2}{2!}+\frac{1.95^3}{3!}\bigg)\\[4mm] =&\,\boxed{0.866}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

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以零人到达为补事件，建立至少一人的概率不小于0.9的不等式，取对数求最短时间。

答题过程

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Let $t$ be the waiting time in minutes. The number arriving at desk $B$ during this time has distribution $\mathrm{Po}(0.21t)$ .

To have probability at least 0.90 of seeing at least one arrival,

\begin{align*} 1-e^{-0.21t}\geq&\,0.90,\\[4mm] e^{-0.21t}\leq&\,0.10,\\[4mm] -0.21t\leq&\,\ln0.10,\\[4mm] t\geq&\,\frac{-\ln0.10}{0.21}=10.96469\ldots. \end{align*}

The minimum waiting time, to the nearest minute, is $\boxed{11\text{ minutes}}$ .

CIE 9709 2024 November Paper 61 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程