题目

Problem

A factory produces chocolates. $30\%$ of the chocolates are wrapped in gold foil, $25\%$ are wrapped in red foil and the remainder are unwrapped.

Indigo chooses 8 chocolates at random from the production line.

(a) Find the probability that she obtains no more than 2 chocolates that are wrapped in red foil.

(3)

Jake chooses chocolates one at a time at random from the production line.

(b) Find the probability that the first time he obtains a chocolate that is wrapped in red foil is before the 7th choice.

(2)

Keifa chooses chocolates one at a time at random from the production line.

(c) Find the probability that the second chocolate chosen is the first one wrapped in gold foil given that the fifth chocolate chosen is the first unwrapped chocolate.

(5)

题目中文翻译

一家工厂生产巧克力。其中 $30\%$ 用金色箔纸包装， $25\%$ 用红色箔纸包装，其余没有包装。

Indigo 从生产线上随机选取 8 块巧克力。

(a) 求她选到的用红色箔纸包装的巧克力不超过 2 块的概率。

(3)

Jake 从生产线上随机选取巧克力，每次选取一块。

(b) 求他在第 7 次选取之前首次选到用红色箔纸包装的巧克力的概率。

(2)

Keifa 从生产线上随机选取巧克力，每次选取一块。

(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)

解法一

思路

展开

红包装的块数服从二项分布，不超过2即0、1、2三项。

答题过程

展开

Let $R\sim\mathrm{B}(8,0.25)$ . Then

\begin{align*} \mathrm{P}(R\leq2)=&\,(0.75)^8+8(0.25)(0.75)^7\\[4mm] &\,\hspace{2pt}+\binom82(0.25)^2(0.75)^6\\[4mm] =&\,0.678543\ldots\\[4mm] \approx&\,\boxed{0.679}. \end{align*}

(b)

解法一

思路

展开

第7次之前即前6次内首次红，取前6次均不红的补集。

答题过程

展开

\mathrm{P}=1-(0.75)^6=0.822021484\ldots\approx\boxed{0.822}.

解法二

思路

展开

也可以按首次红在第1至第6次分类，求六项等比和。

答题过程

展开

\begin{align*} \mathrm{P}=&\,0.25\bigl[1+0.75+0.75^2\\[4mm] &\,\hspace{2pt}+0.75^3+0.75^4+0.75^5\bigr]\\[4mm] =&\,0.25\frac{1-0.75^6}{1-0.75}\\[4mm] =&\,0.822021484\ldots\approx\boxed{0.822}. \end{align*}

(c)

解法一

思路

展开

先构造两个事件的交集：第1块必须红，第2块金，第3、4块任意有包装，第5块无包装，再除以已知事件概率。

答题过程

展开

Let $A$ be the event that the second chocolate is the first gold-wrapped chocolate, and $B$ the event that the fifth chocolate is the first unwrapped chocolate.

The probability of a wrapped chocolate is $0.30+0.25=0.55$ . Thus

\mathrm{P}(B)=(0.55)^4(0.45).

For $A\cap B$ , the colour pattern is R, G, (R or G), (R or G), U. Hence

\mathrm{P}(A\cap B)=(0.25)(0.30)(0.55)^2(0.45).

Therefore

\begin{align*} \mathrm{P}(A\mid B) =&\,\frac{(0.25)(0.30)(0.55)^2(0.45)}{(0.55)^4(0.45)}\\[4mm] =&\,\frac{0.075}{0.3025}=\frac{30}{121} \approx\boxed{0.248}. \end{align*}

解法二

思路

展开

已知前四块都包装，可以在红与金之间重新归一化概率；独立抽取下，前两块必须先红后金。

答题过程

展开

Given that the fifth chocolate is the first unwrapped chocolate, the first four chocolates are wrapped. For each of these independent choices,

\mathrm{P}(\text{red}\mid\text{wrapped})=\frac{0.25}{0.55}, \qquad \mathrm{P}(\text{gold}\mid\text{wrapped})=\frac{0.30}{0.55}.

The second is the first gold chocolate precisely when the first is red and the second is gold. The third and fourth may be either colour. Hence

\mathrm{P}=\frac{0.25}{0.55}\frac{0.30}{0.55} =\frac{30}{121}\approx\boxed{0.248}.

CIE 9709 2024 November Paper 53 Q5

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

解法二

思路

答题过程