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CIE 9709 2022 March Paper 52 Q6

A Level / CIE / S1

CIE 9709 2022 March Paper 52 Paper · Question 6

题目

Problem

A factory produces chocolates in three flavours: lemon, orange and strawberry in the ratio 3:5:73:5:7 respectively. Nell checks the chocolates on the production line by choosing chocolates randomly one at a time.

(a) Find the probability that the first chocolate with lemon flavour that Nell chooses is the 7th chocolate that she checks.

(1)

(b) Find the probability that the first chocolate with lemon flavour that Nell chooses is after she has checked at least 6 chocolates.

(2)

‘Surprise’ boxes of chocolates each contain 15 chocolates: 3 are lemon, 5 are orange and 7 are strawberry.

Petra has a box of Surprise chocolates. She chooses 3 chocolates at random from the box. She eats each chocolate before choosing the next one.

(c) Find the probability that none of Petra’s 3 chocolates has orange flavour.

(2)

(d) Find the probability that each of Petra’s 3 chocolates has a different flavour.

(3)

(e) Find the probability that at least 2 of Petra’s 3 chocolates have strawberry flavour given that none of them has orange flavour.

(4)

题目中文翻译

一家工厂生产柠檬、橙子和草莓三种口味的巧克力，数量之比依次为 3:5:73:5:7。Nell 在生产线上检查巧克力，每次随机选取一颗。

(a) 求 Nell 选到的第一颗柠檬味巧克力是她检查的第 7 颗巧克力的概率。

(1)

(b) 求 Nell 在已经检查了至少 6 颗巧克力之后，才首次选到柠檬味巧克力的概率。

(2)

每盒“Surprise”巧克力含有 15 颗巧克力：3 颗柠檬味、5 颗橙子味和 7 颗草莓味。

Petra 有一盒 Surprise 巧克力。她从盒中随机选取 3 颗巧克力，每选一颗就将其吃掉，然后才选下一颗。

(c) 求 Petra 选取的 3 颗巧克力中没有橙子味的概率。

(2)

(d) 求 Petra 选取的 3 颗巧克力的口味各不相同的概率。

(3)

(e) 已知 Petra 选取的 3 颗巧克力中没有橙子味，求其中至少有 2 颗草莓味的概率。

(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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生产线独立抽取，柠檬概率1/5，前六次非柠檬后第七次柠檬。

答题过程

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The lemon probability is 3/(3+5+7)=1/53/(3+5+7)=1/5. Thus

P(first lemon on check 7)= (45)615= 409678125.\begin{align*} \mathrm{P}(\text{first lemon on check 7}) =&\,\biggl(\frac45\biggr)^6\frac15\\[4mm] =&\,\boxed{\frac{4096}{78125}}. \end{align*}

(b)

解法一

思路

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首次柠檬在检查六颗之后，意味着前六颗全非柠檬，之后不必限定具体次数。

答题过程

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Let TT be the check on which the first lemon occurs. The event is T>6T>6, so the first six chocolates must all be non-lemon:

P(T>6)=(45)6=409615625.\mathrm{P}(T>6)=\biggl(\frac45\biggr)^6 =\boxed{\frac{4096}{15625}}.

解法二

思路

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也可用前六次内首次柠檬的六项几何概率之和的补集。

答题过程

展开 P(T>6)= 1−∑r=16(45)r−115= 1−151−(4/5)61−4/5= 409615625.\begin{align*} \mathrm{P}(T>6)=&\,1-\sum_{r=1}^{6}\biggl(\frac45\biggr)^{r-1}\frac15\\[4mm] =&\,1-\frac15\frac{1-(4/5)^6}{1-4/5}\\[4mm] =&\,\boxed{\frac{4096}{15625}}. \end{align*}

(c)

解法一

思路

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盒内不放回，非橙色共10颗，每次分子分母都减一。

答题过程

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There are 3+7=103+7=10 non-orange chocolates. Hence

P(no orange)=1015914813=2491.\mathrm{P}(\text{no orange})=\frac{10}{15}\frac9{14}\frac8{13} =\boxed{\frac{24}{91}}.

解法二

思路

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也可按等可能的三颗实物组合取非橙色子集。

答题过程

展开 P(no orange)=(103)(153)=120455=2491.\mathrm{P}(\text{no orange})=\frac{\binom{10}{3}}{\binom{15}{3}} =\frac{120}{455}=\boxed{\frac{24}{91}}.

(d)

解法一

思路

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三口味各一颗，固定一种顺序的概率乘3!。

答题过程

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For a fixed order, such as lemon, orange, strawberry, the probability is

315514713.\frac3{15}\frac5{14}\frac7{13}.

The six possible orders are disjoint and have the same probability. Thus

P(all different)=3!315514713=313.\mathrm{P}(\text{all different})=3!\frac3{15}\frac5{14}\frac7{13} =\boxed{\frac3{13}}.

解法二

思路

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也可直接每口味选一颗除以所有三颗的组合数。

答题过程

展开 P(all different)= (31)(51)(71)(153)= 105455=313.\begin{align*} \mathrm{P}(\text{all different}) =&\,\frac{\binom31\binom51\binom71}{\binom{15}{3}}\\[4mm] =&\,\frac{105}{455}=\boxed{\frac3{13}}. \end{align*}

(e)

解法一

思路

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在原15颗中求SSS及SSL三个顺序的交集概率，再除以(c)无橙色概率。

答题过程

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Use SS for strawberry and LL for lemon. The favourable paths are SSSSSS, SSLSSL, SLSSLS and LSSLSS. Their total probability is

p= 715614513+3315714613= 210+3782730=1465.\begin{align*} p=&\,\frac7{15}\frac6{14}\frac5{13} +3\frac3{15}\frac7{14}\frac6{13}\\[4mm] =&\,\frac{210+378}{2730}=\frac{14}{65}. \end{align*}

All these paths have no orange. Using (c), the required conditional probability is

pP(no orange)=14/6524/91=4960.\frac{p}{\mathrm{P}(\text{no orange})} =\frac{14/65}{24/91}=\boxed{\frac{49}{60}}.

解法二

思路

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已知三颗无橙色，条件样本空间就是十颗非橙色中的三颗组合。

答题过程

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Given no orange, the three chosen chocolates form an equally likely subset of the 10 lemon or strawberry chocolates. At least two strawberries means either two strawberries and one lemon, or three strawberries. Hence

P= (72)(31)+(73)(103)= 63+35120=4960.\begin{align*} \mathrm{P}=&\,\frac{\binom72\binom31+\binom73}{\binom{10}{3}}\\[4mm] =&\,\frac{63+35}{120}=\boxed{\frac{49}{60}}. \end{align*}