题目

Problem

Janice is playing a computer game. She has to complete level 1 and level 2 to finish the game. She is allowed at most two attempts at any level.

For level 1, the probability that Janice completes it at the first attempt is 0.6. If she fails at her first attempt, the probability that she completes it at the second attempt is 0.3.
If Janice completes level 1, she immediately moves on to level 2.
For level 2, the probability that Janice completes it at the first attempt is 0.4. If she fails at her first attempt, the probability that she completes it at the second attempt is 0.2.

(a) Show that the probability that Janice moves on to level 2 is 0.72.

(1)

(b) Find the probability that Janice finishes the game.

(3)

(4)

题目中文翻译

Janice 正在玩一个电脑游戏。她必须完成第 1 关和第 2 关才能通关。每一关最多允许她尝试两次。

对于第 1 关，Janice 在第一次尝试时完成该关的概率为 0.6。如果她第一次尝试失败，则她在第二次尝试时完成该关的概率为 0.3。
如果 Janice 完成第 1 关，她立即进入第 2 关。
对于第 2 关，Janice 在第一次尝试时完成该关的概率为 0.4。如果她第一次尝试失败，则她在第二次尝试时完成该关的概率为 0.2。

(a) 证明 Janice 进入第 2 关的概率为 0.72。

(1)

(b) 求 Janice 通关的概率。

(3)

(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

展开

第一关可首次完成或先失败后完成，两条路径相加。

答题过程

展开

\mathrm{P}(\text{reach level 2})=0.6+0.4(0.3)=\boxed{0.72}.

(b)

解法一

思路

展开

承接(a)，进入第二关后可首次通过或先失败再通过，合并条件概率相乘。

答题过程

展开

Given that Janice reaches level 2, its completion probability is $0.4+0.6(0.2)=0.52$ . Therefore

\mathrm{P}(\text{finish game})=0.72\bigl[0.4+0.6(0.2)\bigr] =\boxed{0.3744}.

解法二

思路

展开

也可列出两关各在第一次或第二次通过的四条互斥完整路径。

答题过程

展开

The four ways to finish are

Level 1	Level 2	Probability
Pass first time	Pass first time	$0.6(0.4)=0.24$
Fail, then pass	Pass first time	$0.4(0.3)(0.4)=0.048$
Pass first time	Fail, then pass	$0.6(0.6)(0.2)=0.072$
Fail, then pass	Fail, then pass	$0.4(0.3)(0.6)(0.2)=0.0144$

Hence

\begin{align*} \mathrm{P}(\text{finish})=&\,0.24+0.048+0.072+0.0144\\[4mm] =&\,\boxed{0.3744}. \end{align*}

(c)

解法一

思路

展开

通关且恰一次失败，只能第一关失败一次或第二关失败一次，取两条路径后除以(b)。

答题过程

展开

Finishing with exactly one failed attempt occurs on these disjoint paths:

Fail then pass level 1, and pass level 2 first time: $0.4(0.3)(0.4)=0.048$ .
Pass level 1 first time, then fail and pass level 2: $0.6(0.6)(0.2)=0.072$ .

Using (b), the conditional probability is

\frac{0.048+0.072}{0.3744}=\frac{0.12}{0.3744}=\boxed{\frac{25}{78}}.

CIE 9709 2022 June Paper 51 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

解法二

思路

答题过程

(c)

解法一

思路

答题过程