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CIE 9709 2023 November Paper 53 Q5

A Level / CIE / S1

CIE 9709 2023 Nov Paper 53 Paper · Question 5

题目

Problem

The probability that a driver passes an advanced driving test is 0.3 on any given attempt.

(a) Dipak keeps taking the test until he passes. The random variable XX denotes the number of attempts required for Dipak to pass the test.

(i) Find P(2≤X≤6)\mathrm{P}(2\leq X\leq 6).

(2)

(ii) Find E(X)\mathrm{E}(X).

(1)

Five friends will each take their advanced driving test tomorrow.

(b) Find the probability that at least three of them will pass tomorrow.

(3)

75 people will take their advanced driving test next week.

(c) Use an approximation to find the probability that more than 20 of them will pass next week.

(5)

题目中文翻译

一名驾驶员每次参加高级驾驶考试时，通过的概率为 0.3。

(a) Dipak 不断参加考试，直到通过为止。随机变量 XX 表示 Dipak 通过考试所需的尝试次数。

(i) 求 P(2≤X≤6)\mathrm{P}(2\leq X\leq 6)。

(2)

(ii) 求 E(X)\mathrm{E}(X)。

(1)

五位朋友明天将各自参加高级驾驶考试。

(b) 求他们中至少三人明天通过考试的概率。

(3)

下周将有 75 人参加高级驾驶考试。

(c) 用近似方法求他们中超过 20 人下周通过考试的概率。

(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)(i)

解法一

思路

展开

首次成功在第2至第6次，可用前1次失败的概率减去前6次全失败的概率。

答题过程

展开 P(2≤X≤6)= P(X>1)−P(X>6)= 0.7−(0.7)6=0.582351≈0.582.\begin{align*} \mathrm{P}(2\leq X\leq6) =&\,\mathrm{P}(X>1)-\mathrm{P}(X>6)\\[4mm] =&\,0.7-(0.7)^6=0.582351\approx\boxed{0.582}. \end{align*}

解法二

思路

展开

也可按首次成功发生于第2、3、4、5、6次分别求和。

答题过程

展开 P= 0.7(0.3)+(0.7)2(0.3)+(0.7)3(0.3) +(0.7)4(0.3)+(0.7)5(0.3)= 0.582351≈0.582.\begin{align*} \mathrm{P}=&\,0.7(0.3)+(0.7)^2(0.3)+(0.7)^3(0.3)\\[4mm] &\,\hspace{2pt}+(0.7)^4(0.3)+(0.7)^5(0.3)\\[4mm] =&\,0.582351\approx\boxed{0.582}. \end{align*}

(a)(ii)

解法一

思路

展开

几何分布的首次成功次数均值为成功概率的倒数。

答题过程

展开

Since X∼Geo⁡(0.3)X\sim\operatorname{Geo}(0.3),

E(X)=10.3=103.\mathrm{E}(X)=\frac1{0.3}=\boxed{\frac{10}{3}}.

(b)

解法一

思路

展开

五人至少三人通过，直接算3、4、5三项二项和。

答题过程

展开

Let S∼B(5,0.3)S\sim\mathrm{B}(5,0.3). Then

P(S≥3)= (53)(0.3)3(0.7)2 +5(0.3)4(0.7)+(0.3)5= 0.1323+0.02835+0.00243= 0.16308≈0.163.\begin{align*} \mathrm{P}(S\geq3) =&\,\binom53(0.3)^3(0.7)^2\\[4mm] &\,\hspace{2pt}+5(0.3)^4(0.7)+(0.3)^5\\[4mm] =&\,0.1323+0.02835+0.00243\\[4mm] =&\,0.16308\approx\boxed{0.163}. \end{align*}

(c)

解法一

思路

展开

通过人数服从二项分布，正态近似均值22.5、方差15.75，超过20修正为20.5。

答题过程

展开

Let S∼B(75,0.3)S\sim\mathrm{B}(75,0.3). Since np=22.5np=22.5 and n(1−p)=52.5n(1-p)=52.5 are both greater than 5, use Y∼N(22.5,15.75)Y\sim\mathrm{N}(22.5,15.75).

With a continuity correction,

P(S>20)= P(S≥21)≈ P(Y>20.5)= 1−Φ(20.5−22.515.75)= 0.692852667…≈0.693.\begin{align*} \mathrm{P}(S>20) =&\,\mathrm{P}(S\geq21)\\[4mm] \approx&\,\mathrm{P}(Y>20.5)\\[4mm] =&\,1-\Phi\biggl(\frac{20.5-22.5}{\sqrt{15.75}}\biggr)\\[4mm] =&\,0.692852667\ldots\approx\boxed{0.693}. \end{align*}