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CIE 9709 2022 June Paper 52 Q7

A Level / CIE / S1

CIE 9709 2022 June Paper 52 Paper · Question 7

题目

Problem

Hanna buys 12 hollow chocolate eggs that each contain a sweet. The eggs look identical but Hanna knows that 3 contain a red sweet, 4 contain an orange sweet and 5 contain a yellow sweet. Each of Hanna’s three children in turn randomly chooses and eats one of the eggs, keeping the sweet it contained.

(a) Find the probability that all 3 eggs chosen contain the same colour sweet.

(4)

(b) Find the probability that all 3 eggs chosen contain a yellow sweet, given that all three children have the same colour sweet.

(2)

(c) Find the probability that at least one of Hanna’s three children chooses an egg that contains an orange sweet.

(3)

题目中文翻译

Hanna 买了 12 个空心巧克力蛋，每个蛋内都装有一颗糖果。这些蛋看起来完全相同，但 Hanna 知道其中 3 个装有红色糖果，4 个装有橙色糖果，5 个装有黄色糖果。Hanna 的三个孩子依次各随机选取并吃掉一个蛋，保留蛋内的糖果。

(a) 求选出的 3 个蛋内的糖果颜色相同的概率。

(4)

(b) 已知三个孩子获得的糖果颜色相同，求选出的 3 个蛋内都是黄色糖果的概率。

(2)

(c) 求 Hanna 的三个孩子中至少有一个选到装有橙色糖果的蛋的概率。

(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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三种同色事件互斥，每次蛋已吃掉，分母依次12、11、10。

答题过程

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The three possibilities are three red, three orange or three yellow sweets. As the eggs are not replaced,

P(same colour)= 312211110+412311210 +512411310= 6+24+601320=344.\begin{align*} \mathrm{P}(\text{same colour})=&\,\frac3{12}\frac2{11}\frac1{10} +\frac4{12}\frac3{11}\frac2{10}\\[4mm] &\,\hspace{2pt}+\frac5{12}\frac4{11}\frac3{10}\\[4mm] =&\,\frac{6+24+60}{1320}=\boxed{\frac3{44}}. \end{align*}

解法二

思路

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也可把三枚实物蛋视为不计顺序的等可能子集。

答题过程

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There are (123)\binom{12}{3} equally likely sets of three eggs. The favourable sets contain three red, three orange or three yellow sweets. Hence

P(same colour)= (33)+(43)+(53)(123)= 1+4+10220=344.\begin{align*} \mathrm{P}(\text{same colour}) =&\,\frac{\binom33+\binom43+\binom53}{\binom{12}{3}}\\[4mm] =&\,\frac{1+4+10}{220}=\boxed{\frac3{44}}. \end{align*}

(b)

解法一

思路

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全黄包含于同色，条件概率用全黄概率除以(a)。

答题过程

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The probability of three yellow sweets is

P(YYY)=512411310=122.\mathrm{P}(YYY)=\frac5{12}\frac4{11}\frac3{10}=\frac1{22}.

Since three yellow sweets imply the same colour, the required probability is

P(YYY)P(same colour)=1/223/44=23.\frac{\mathrm{P}(YYY)}{\mathrm{P}(\text{same colour})} =\frac{1/22}{3/44}=\boxed{\frac23}.

(c)

解法一

思路

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取完全没有橙色的补集，非橙色蛋共8枚。

答题过程

展开 P(at least one orange)= 1−812711610= 1−1455=4155.\begin{align*} \mathrm{P}(\text{at least one orange}) =&\,1-\frac8{12}\frac7{11}\frac6{10}\\[4mm] =&\,1-\frac{14}{55}=\boxed{\frac{41}{55}}. \end{align*}

解法二

思路

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按恰有1、2、3枚橙色蛋分类，用组合数处理不计顺序的三枚蛋。

答题过程

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There are four eggs with orange sweets and eight without. Therefore

P(at least one orange)= (41)(82)+(42)(81)+(43)(123)= 112+48+4220=4155.\begin{align*} \mathrm{P}(\text{at least one orange}) =&\,\frac{\binom41\binom82+\binom42\binom81+\binom43}{\binom{12}{3}}\\[4mm] =&\,\frac{112+48+4}{220}=\boxed{\frac{41}{55}}. \end{align*}

解法三

思路

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按第一枚橙色蛋由第几个孩子取到分类，后面的选择不必限制。

答题过程

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The first orange sweet may be chosen by the first, second or third child. These cases are disjoint, so

P(at least one orange)= 412+812411+812711410= 13+833+28165=4155.\begin{align*} \mathrm{P}(\text{at least one orange}) =&\,\frac4{12}+\frac8{12}\frac4{11} +\frac8{12}\frac7{11}\frac4{10}\\[4mm] =&\,\frac13+\frac8{33}+\frac{28}{165} =\boxed{\frac{41}{55}}. \end{align*}