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CIE 9709 2026 June Paper 51 Q4

A Level / CIE / S1

CIE 9709 2026 June Paper 51 Paper · Question 4

题目

Problem

Jong has a bag containing 5 red marbles and 7 blue marbles. He chooses marbles from the bag one at a time at random. Each time he replaces the marble in the bag.

(a) Find the probability that the first red marble that Jong obtains is before his 6th choice.

(2)

Kai has a bag containing 5 green marbles and 7 yellow marbles. He chooses marbles from the bag one at a time at random. Each time he replaces the marble in the bag.

(b) Find the probability that the second green marble that Kai obtains is on his 9th choice.

(3)

Lam has a bag containing 5 black marbles and 7 white marbles. He chooses 10 marbles from the bag one at a time, at random and with replacement.

(c) Find the probability that Lam obtains fewer than 3 black marbles.

(3)
题目中文翻译

Jong 有一个袋子，里面装有 5 颗红色弹珠和 7 颗蓝色弹珠。他每次从袋中随机取出一颗弹珠，每次取出后都将弹珠放回袋中。

(a) 求 Jong 在第 6 次取珠之前首次取到红色弹珠的概率。

(2)

Kai 有一个袋子，里面装有 5 颗绿色弹珠和 7 颗黄色弹珠。他每次从袋中随机取出一颗弹珠，每次取出后都将弹珠放回袋中。

(b) 求 Kai 在第 9 次取珠时取到第 2 颗绿色弹珠的概率。

(3)

Lam 有一个袋子，里面装有 5 颗黑色弹珠和 7 颗白色弹珠。他从袋中逐颗随机取出弹珠，每次取出后放回，共取 10 次。

(c) 求 Lam 取到的黑色弹珠少于 3 颗的概率。

(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

第6次之前首次取到红珠，等价于前5次至少有一次红珠。用其对立事件“前5次全部为蓝珠”计算最短。

答题过程

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Each draw has probability 712\frac7{12} of being blue, independently of the other draws. Therefore,

P(first red before draw 6)= 1−P(first 5 all blue)= 1−(712)5= 0.932456…≈ 0.932.\begin{align*} \mathrm{P}(\text{first red before draw 6}) =&\,1-\mathrm{P}(\text{first 5 all blue})\\[4mm] =&\,1-\biggl(\frac7{12}\biggr)^5\\[4mm] =&\,0.932456\ldots\\[4mm] \approx&\,\boxed{0.932}. \end{align*}

解法二

思路

展开

也可以按首次红珠出现于第1至第5次分类。每一项先连续取蓝珠，再取红珠。

答题过程

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Let TT be the draw on which the first red marble is obtained. Then

P(T=k)=(712)k−1512.\mathrm{P}(T=k)=\biggl(\frac7{12}\biggr)^{k-1}\frac5{12}.

Consequently,

P(T<6)= 512[1+712+(712)2 +(712)3+(712)4]≈ 0.932.\begin{align*} \mathrm{P}(T<6)=&\,\frac5{12}\biggl[1+\frac7{12} +\biggl(\frac7{12}\biggr)^2\\[4mm] &\,\hspace{2pt}+\biggl(\frac7{12}\biggr)^3 +\biggl(\frac7{12}\biggr)^4\biggr]\\[4mm] \approx&\,\boxed{0.932}. \end{align*}

(b)

解法一

思路

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第9次是第2颗绿珠，要求前8次恰有1颗绿珠，且第9次为绿珠。前8次那颗绿珠可以出现在8个位置中的任意一个。

答题过程

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The first 8 draws must contain one green and seven yellow marbles, and draw 9 must be green. Thus

P= (81)512(712)7512= 8(512)2(712)7= 0.031921630…≈ 0.0319.\begin{align*} \mathrm{P}=&\,\binom81\frac5{12} \biggl(\frac7{12}\biggr)^7\frac5{12}\\[4mm] =&\,8\biggl(\frac5{12}\biggr)^2 \biggl(\frac7{12}\biggr)^7\\[4mm] =&\,0.031921630\ldots\\[4mm] \approx&\,\boxed{0.0319}. \end{align*}

(c)

解法一

思路

展开

有放回抽取使10次试验独立，每次取到黑珠的概率固定。用二项分布把0、1、2颗的概率相加。

答题过程

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Let XX be the number of black marbles obtained. Then

X∼B(10,512).X\sim\mathrm{B}\biggl(10,\frac5{12}\biggr).

Therefore,

P(X<3)= ∑x=02(10x)(512)x(712)10−x= (712)10+10512(712)9 +45(512)2(712)8= 0.141891…≈ 0.142.\begin{align*} \mathrm{P}(X<3)=&\,\sum_{x=0}^{2}\binom{10}{x} \biggl(\frac5{12}\biggr)^x \biggl(\frac7{12}\biggr)^{10-x}\\[4mm] =&\,\biggl(\frac7{12}\biggr)^{10} +10\frac5{12}\biggl(\frac7{12}\biggr)^9\\[4mm] &\,\hspace{2pt}+45\biggl(\frac5{12}\biggr)^2 \biggl(\frac7{12}\biggr)^8\\[4mm] =&\,0.141891\ldots\\[4mm] \approx&\,\boxed{0.142}. \end{align*}