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CIE 9709 2023 November Paper 53 Q3

A Level / CIE / S1

CIE 9709 2023 Nov Paper 53 Paper · Question 3

题目

Problem

Tim has two bags of marbles, AA and BB.

Bag AA contains 8 white, 4 red and 3 yellow marbles.

Bag BB contains 6 white, 7 red and 2 yellow marbles.

Tim also has an ordinary fair 6-sided dice. He rolls the dice. If he obtains a 1 or 2, he chooses two marbles at random from bag AA, without replacement. If he obtains a 3, 4, 5 or 6, he chooses two marbles at random from bag BB, without replacement.

(a) Find the probability that both marbles are white.

(3)

(b) Find the probability that the two marbles come from bag BB given that one is white and one is red.

(4)

题目中文翻译

Tim 有两袋弹珠,分别为 AA 和 BB。

袋 AA 中有 8 颗白色、4 颗红色和 3 颗黄色弹珠。

袋 BB 中有 6 颗白色、7 颗红色和 2 颗黄色弹珠。

Tim 还有一枚普通的均匀六面骰子。他掷这枚骰子。若掷出 1 或 2,他从袋 AA 中随机不放回地抽取两颗弹珠;若掷出 3、4、5 或 6,他从袋 BB 中随机不放回地抽取两颗弹珠。

(a) 求两颗弹珠均为白色的概率。

(3)

(b) 已知两颗弹珠一颗为白色、一颗为红色,求它们来自袋 BB 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

展开

先按骰子决定袋子,两袋概率分别1/3、2/3,袋内均不放回抽取。

答题过程

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The probabilities of choosing A and B are 26=13\frac26=\frac13 and 46=23\frac46=\frac23. Hence

P(WW)= 13815714+23615514= 56+60630=58315.\begin{align*} \mathrm{P}(WW) =&\,\frac13\frac8{15}\frac7{14} +\frac23\frac6{15}\frac5{14}\\[4mm] =&\,\frac{56+60}{630}=\boxed{\frac{58}{315}}. \end{align*}

(b)

解法一

思路

展开

白红可按WR或RW两种次序出现,分子只取B袋,分母包含A和B两袋的路径。

答题过程

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Let CC denote one white and one red marble. For each bag, WR and RW have equal probabilities, so

P(B∩C)=23(2615714)=415,\mathrm{P}(B\cap C)=\frac23\biggl(2\frac6{15}\frac7{14}\biggr)=\frac4{15},

and

P(A∩C)=13(2815414)=32315.\mathrm{P}(A\cap C)=\frac13\biggl(2\frac8{15}\frac4{14}\biggr)=\frac{32}{315}.

Therefore

P(C)=415+32315=116315,\mathrm{P}(C)=\frac4{15}+\frac{32}{315}=\frac{116}{315},

and

P(B∣C)=4/15116/315=2129.\mathrm{P}(B\mid C)=\frac{4/15}{116/315}=\boxed{\frac{21}{29}}.