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CIE 9709 2024 November Paper 51 Q4

A Level / CIE / S1

CIE 9709 2024 Nov Paper 51 Paper · Question 4

题目

Problem

Rahul has two bags, XX and YY. Bag XX contains 4 red marbles and 2 blue marbles. Bag YY contains 3 red marbles and 4 blue marbles. Rahul also has a coin which is biased so that the probability of obtaining a head when it is thrown is 14\frac{1}{4}.

Rahul throws the coin.

  • If he obtains a head, he chooses at random a marble from bag XX. He notes the colour and replaces the marble in bag XX. He then chooses at random a second marble from bag XX.
  • If he obtains a tail, he chooses at random a marble from bag YY. He notes the colour and discards the marble. He then chooses at random a second marble from bag YY.

(a) Find the probability that the two marbles that Rahul chooses are the same colour.

(3)

(b) Find the probability that the two marbles that Rahul chooses are both from bag YY given that both marbles are blue.

(3)
题目中文翻译

Rahul 有两个袋子 XX 和 YY。袋 XX 中有 4 颗红色弹珠和 2 颗蓝色弹珠。袋 YY 中有 3 颗红色弹珠和 4 颗蓝色弹珠。Rahul 还有一枚不均匀的硬币,掷出正面的概率为 14\frac{1}{4}。

Rahul 掷这枚硬币。

  • 如果掷出正面,他从袋 XX 中随机抽取一颗弹珠,记下颜色,并将弹珠放回袋 XX 中。然后他再从袋 XX 中随机抽取第二颗弹珠。
  • 如果掷出反面,他从袋 YY 中随机抽取一颗弹珠,记下颜色,并将弹珠丢弃。然后他再从袋 YY 中随机抽取第二颗弹珠。

(a) 求 Rahul 抽到的两颗弹珠颜色相同的概率。

(3)

(b) 已知两颗弹珠都是蓝色的,求 Rahul 抽到的两颗弹珠都来自袋 YY 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

展开

先按硬币结果分类。X袋有放回,第二次概率不变;Y袋不放回,第二次分母减1,同色时相应颜色数也减1。

答题过程

展开

Let HH denote a head and SS the event that the marbles have the same colour. Then

P(S∣H)=(46)2+(26)2=59,\mathrm{P}(S\mid H)=\biggl(\frac46\biggr)^2+\biggl(\frac26\biggr)^2 =\frac59, P(S∣H′)=3726+4736=37.\mathrm{P}(S\mid H')=\frac37\frac26+\frac47\frac36=\frac37.

Therefore

P(S)=1459+3437=536+928=2963.\mathrm{P}(S)=\frac14\frac59+\frac34\frac37 =\frac5{36}+\frac9{28}=\boxed{\frac{29}{63}}.

(b)

解法一

思路

展开

已知两蓝,分母包括从X和从Y抽到两蓝的两条路径;分子只取Y路径。

答题过程

展开

Let BB denote the event that both marbles are blue. Then

P(Y∩B)=344736=314,\mathrm{P}(Y\cap B)=\frac34\frac47\frac36=\frac3{14},

while

P(B)=142626+314=136+314=61252.\mathrm{P}(B)=\frac14\frac26\frac26+\frac3{14} =\frac1{36}+\frac3{14}=\frac{61}{252}.

Hence

P(Y∣B)=3/1461/252=5461.\mathrm{P}(Y\mid B)=\frac{3/14}{61/252}=\boxed{\frac{54}{61}}.