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CIE 9709 2024 June Paper 52 Q2

A Level / CIE / S1

CIE 9709 2024 June Paper 52 Paper · Question 2

题目

Problem

Seva has a coin which is biased so that when it is thrown the probability of obtaining a head is 13\frac{1}{3}. He also has a bag containing 4 red marbles and 5 blue marbles.

Seva throws the coin. If he obtains a head, he selects one marble from the bag at random. If he obtains a tail, he selects two marbles from the bag at random and without replacement.

(a) Find the probability that Seva selects at least one red marble.

(3)

(b) Find the probability that Seva obtains a head given that he selects no red marbles.

(2)
题目中文翻译

Seva 有一枚不均匀硬币，抛掷时出现正面的概率为 13\frac{1}{3}。他还有一个袋子，内有 4 颗红色弹珠和 5 颗蓝色弹珠。

Seva 抛掷这枚硬币。如果出现正面，他从袋中随机取出一颗弹珠。如果出现反面，他从袋中随机不放回地取出两颗弹珠。

(a) 求 Seva 取到至少一颗红色弹珠的概率。

(3)

(b) 已知 Seva 没有取到红色弹珠，求硬币出现正面的概率。

(2)
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

有红的路径分为正面后红、反面后第一枚红、反面后先蓝再红，互不重复。

答题过程

展开

Write H,TH,T for the coin results and R,BR,B for marble colours. At least one red occurs on HR, TR (regardless of the second marble), or TBR. Thus

P= 1349+2349+235948= 427+827+527=1727.\begin{align*} \mathrm{P}=&\,\frac13\frac49+\frac23\frac49+\frac23\frac59\frac48\\[4mm] =&\,\frac4{27}+\frac8{27}+\frac5{27}=\boxed{\frac{17}{27}}. \end{align*}

解法二

思路

展开

也可以减去没有红的两条路径：正面后一蓝，或反面后两蓝。

答题过程

展开 P(at least one red)= 1−(1359+235948)= 1−527−527=1727.\begin{align*} \mathrm{P}(\text{at least one red}) =&\,1-\biggl(\frac13\frac59+\frac23\frac59\frac48\biggr)\\[4mm] =&\,1-\frac5{27}-\frac5{27}=\boxed{\frac{17}{27}}. \end{align*}

(b)

解法一

思路

展开

没有红时只可能HB或TBB，用交集概率除以它们的总概率。

答题过程

展开

Using (a),

P(no red)=1−1727=1027.\mathrm{P}(\text{no red})=1-\frac{17}{27}=\frac{10}{27}.

Also P(H∩no red)=1359=527\mathrm{P}(H\cap\text{no red})=\frac13\frac59=\frac5{27}. Hence

P(H∣no red)=5/2710/27=12.\mathrm{P}(H\mid\text{no red})=\frac{5/27}{10/27}=\boxed{\frac12}.