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CIE 9709 2025 June Paper 51 Q2

A Level / CIE / S1

CIE 9709 2025 June Paper 51 Paper · Question 2

题目

Problem

(a) Find the number of different arrangements of the 8 letters in the word KANGAROO in which the two As are together and the two Os are not together.

(3)

A fair 8-sided dice has faces labelled K, A, N, G, A, R, O, O. The dice is rolled repeatedly.

(b) Find the probability that fewer than 6 rolls of this dice are required to obtain an A.

(2)

(c) Find the probability that the second A is obtained on the 6th roll of the dice.

(2)
题目中文翻译

(a) 将单词 KANGAROO 中的 8 个字母排列,要求两个 A 相邻,两个 O 不相邻。求不同排列的个数。

(3)

一枚均匀的八面骰子的各面分别标有 K、A、N、G、A、R、O、O。反复掷这枚骰子。

(b) 求掷出 A 所需的次数少于 6 次的概率。

(2)

(c) 求第 6 次掷骰子时才第二次掷出 A 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

将AA作为一个块,先算全部排列,再减去OO也相连的排列。

答题过程

展开

Treat AA as a single block. Together with K, N, G, R, O and O, there are seven objects with two identical Os, giving 7!2!\frac{7!}{2!} arrangements.

If OO is also a block, there are six distinct objects, giving 6!6! arrangements. Hence

7!2!−6!=2520−720=1800.\frac{7!}{2!}-6!=2520-720=\boxed{1800}.

解法二

思路

展开

也可以先排AA块和K、N、G、R五个对象,再从六个空隙中选两个,每处放一个O。

答题过程

展开

Arrange AA, K, N, G and R in 5!5! ways, creating six gaps including the end gaps. Place the identical Os in two distinct gaps so they are not together. Thus

5!(62)=120×15=1800.5!\binom62=120\times15=\boxed{1800}.

(b)

解法一

思路

展开

骰子有两个A面,成功概率为四分之一。少于6次即前5次至少出现一个A,用补集计算。

答题过程

展开

On each roll,

p=P(A)=28=14.p=\mathrm{P}(A)=\frac28=\frac14.

Therefore,

P(fewer than 6 rolls)=1−(34)5=7811024.\mathrm{P}(\text{fewer than 6 rolls}) =1-\biggl(\frac34\biggr)^5 =\boxed{\frac{781}{1024}}.

解法二

思路

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另一条路线按首次A出现于第1至第5次分类,分别乘之前没有A的概率再求和。

答题过程

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Let TT be the number of rolls to the first A. Then

P(T<6)= ∑k=15(34)k−114= 141−(3/4)51−3/4= 7811024.\begin{align*} \mathrm{P}(T<6)=&\,\sum_{k=1}^{5} \biggl(\frac34\biggr)^{k-1}\frac14\\[4mm] =&\,\frac14\frac{1-(3/4)^5}{1-3/4}\\[4mm] =&\,\boxed{\frac{781}{1024}}. \end{align*}

(c)

解法一

思路

展开

第6次是第二个A,前5次必须恰有一个A,第6次再成功。组合数安排前5次中的成功位置。

答题过程

展开 P(second A on roll 6)= (51)14(34)414= 4054096≈ 0.0989(3 s.f.).\begin{align*} \mathrm{P}(\text{second A on roll 6}) =&\,\binom51\frac14\biggl(\frac34\biggr)^4\frac14\\[4mm] =&\,\frac{405}{4096}\\[4mm] \approx&\,\boxed{0.0989}\quad\text{(3 s.f.)}. \end{align*}