题目

Problem

Drivers who wish to obtain an advanced driving qualification must pass two tests: a theory test and a skills test.

Drivers must pass the theory test first and they are allowed up to a maximum of two attempts to pass the theory test. The probability that any driver passes the theory test on their first attempt is 0.7 and the probability that any driver passes on their second attempt is 0.6.

Drivers who are successful in the theory test are then allowed up to a maximum of three attempts to pass the skills test. The probability that any driver passes the skills test on any attempt is 0.2.

(a) Complete the tree diagram to illustrate this information, showing all the probabilities.

(2)

(b) Show that the probability that a randomly chosen driver passes the theory test at the first attempt and the skills test at the second attempt is 0.112.

(1)

(c) Find the probability that a randomly chosen driver obtains the advanced driving qualification.

(3)

(d) Find the probability that a randomly chosen driver passes the theory test at the first attempt given that the driver obtained the advanced driving qualification.

(3)

题目中文翻译

希望取得高级驾驶资格的驾驶员必须通过两项考试：理论考试和技能考试。

驾驶员必须先通过理论考试，最多允许参加两次理论考试。任一驾驶员第一次参加理论考试就通过的概率为 0.7，第二次参加时通过的概率为 0.6。

通过理论考试的驾驶员随后最多允许参加三次技能考试。任一驾驶员每次参加技能考试时通过的概率均为 0.2。

(a) 补全树状图表示这些信息，并标明所有概率。

(2)

(b) 证明：随机选取的一名驾驶员第一次参加理论考试就通过、且第二次参加技能考试时通过的概率为 0.112。

(1)

(c) 求随机选取的一名驾驶员取得高级驾驶资格的概率。

(3)

(d) 已知随机选取的一名驾驶员取得了高级驾驶资格，求该驾驶员第一次参加理论考试就通过的概率。

(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

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理论测试最多两次，技能测试最多三次；每次通过后就结束该阶段。只从理论通过的两条分支延伸技能树，不能在已失去资格后继续画分支。

答题过程

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For the theory test, the first attempt has probabilities 0.7 for pass and 0.3 for fail. After a first failure, the second attempt has probabilities 0.6 for pass and 0.4 for fail.

From each theory-pass branch, extend the skills-test tree. Every skills attempt has probabilities 0.2 for pass and 0.8 for fail. After a failure, allow another attempt until three skills attempts have been made. Stop a branch as soon as the driver passes or exhausts the permitted attempts.

(b)

解法一

思路

展开

指定路径依次为理论首次通过、技能首次失败、技能第二次通过，将这三条分支概率相乘。

答题过程

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\begin{align*} \mathrm{P}(\text{theory pass first, skills pass second}) =&\,0.7\times0.8\times0.2\\[4mm] =&\,\boxed{0.112}. \end{align*}

(c)

解法一

思路

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先算在两次机会内通过理论的概率，再乘以通过理论后在三次技能机会内至少成功一次的条件概率。

答题过程

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The probability of passing the theory test within two attempts is

0.7+0.3(0.6)=0.88.

Once theory has been passed, the probability of passing skills within three attempts is

1-(0.8)^3=0.488.

Therefore the probability of obtaining the qualification is

0.88\times0.488=0.42944\approx\boxed{0.429}\quad\text{(3 s.f.)}.

解法二

思路

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也可以直接沿树图列出取得资格的六条互斥路径，按理论通过的次数分类，再分别考虑三种技能通过次数。

答题过程

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The successful paths and their probabilities are:

Theory passed on attempt	Skills passed on attempt	Probability
$1$	$1$	$0.7(0.2)=0.14$
$1$	$2$	$0.7(0.8)(0.2)=0.112$
$1$	$3$	$0.7(0.8)^2(0.2)=0.0896$
$2$	$1$	$0.3(0.6)(0.2)=0.036$
$2$	$2$	$0.3(0.6)(0.8)(0.2)=0.0288$
$2$	$3$	$0.3(0.6)(0.8)^2(0.2)=0.02304$

Adding these mutually exclusive probabilities gives

\begin{align*} &\,0.14+0.112+0.0896\\[4mm] &\,\hspace{2pt}+0.036+0.0288+0.02304\\[4mm] =&\,0.42944. \end{align*}

Thus the required probability is $\boxed{0.429}$ to 3 significant figures.

解法三

思路

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第三条路线用补集，失败只有两大类：理论两次失败，或理论最终通过但技能三次失败。

答题过程

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Failure to qualify occurs in either of two mutually exclusive cases:

failure on both theory attempts, with probability $0.3(0.4)=0.12$ ;
passing theory but failing all three skills attempts, with probability $0.88(0.8)^3=0.45056$ .

Hence

\mathrm{P}(\text{qualify}) =1-0.12-0.45056 =0.42944 \approx\boxed{0.429}\quad\text{(3 s.f.)}.

(d)

解法一

思路

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分子是理论首次通过并最终取得资格，分母用(c)未舍入的资格概率。分子需包含技能在第1、2、3次通过的全部情况。

答题过程

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Let $T_1$ denote passing theory at the first attempt and $Q$ denote obtaining the qualification. Then

\begin{align*} \mathrm{P}(T_1\cap Q) =&\,0.7\bigl[0.2+0.8(0.2)+(0.8)^2(0.2)\bigr]\\[4mm] =&\,0.3416. \end{align*}

Using the unrounded result from (c),

\mathrm{P}(T_1\mid Q)=\frac{0.3416}{0.42944} =\boxed{\frac{35}{44}}.

CIE 9709 2026 March Paper 52 Q5

题目

解答

(a)

解法一

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(b)

解法一

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(c)

解法一

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解法二

思路

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解法三

思路

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(d)

解法一

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