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CIE 9709 2023 November Paper 51 Q2

A Level / CIE / S1

CIE 9709 2023 Nov Paper 51 Paper · Question 2

题目

Problem

Hazeem repeatedly throws two ordinary fair 6-sided dice at the same time. On each occasion, the score is the sum of the two numbers that she obtains.

(a) Find the probability that it takes exactly 5 throws of the two dice for Hazeem to obtain a score of 8 or more.

(2)

(b) Find the probability that it takes no more than 4 throws of the two dice for Hazeem to obtain a score of 8 or more.

(2)

(c) For 8 randomly chosen throws of the two dice, find the probability that Hazeem obtains a score of 8 or more on fewer than 3 occasions.

(3)
题目中文翻译

Hazeem 反复同时抛掷两颗普通的均匀六面骰子。每次的得分是她得到的两个点数之和。

(a) 求 Hazeem 恰好需要抛掷这两颗骰子 5 次才能得到至少 8 分的概率。

(2)

(b) 求 Hazeem 抛掷这两颗骰子不超过 4 次便能得到至少 8 分的概率。

(2)

(c) 随机选取抛掷这两颗骰子的 8 次结果,求 Hazeem 得到至少 8 分的次数少于 3 次的概率。

(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

两骰和8至12的结果数分别为5、4、3、2、1,总成功概率15/36。首次成功第5次需要前4次失败。

答题过程

展开

There are 36 equally likely ordered dice outcomes. Scores 8, 9, 10, 11 and 12 have respectively 5, 4, 3, 2 and 1 outcomes. Thus

p=P(score≥8)=5+4+3+2+136=512.p=\mathrm{P}(\text{score}\geq8)=\frac{5+4+3+2+1}{36}=\frac5{12}.

The required probability is

(1−p)4p=(712)4512=0.0482454…≈0.0482.(1-p)^4p=\biggl(\frac7{12}\biggr)^4\frac5{12} =0.0482454\ldots\approx\boxed{0.0482}.

(b)

解法一

思路

展开

不超过4次即前四次至少一次成功,取四次全部失败的补集。

答题过程

展开 P=1−(712)4=0.884211…≈0.884.\mathrm{P}=1-\biggl(\frac7{12}\biggr)^4 =0.884211\ldots\approx\boxed{0.884}.

解法二

思路

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也可按首次成功在第1至第4次分类求和。

答题过程

展开

Writing p=512p=\frac5{12} and q=712q=\frac7{12},

P=p+qp+q2p+q3p=p1−q41−q≈0.884.\mathrm{P}=p+qp+q^2p+q^3p =p\frac{1-q^4}{1-q}\approx\boxed{0.884}.

(c)

解法一

思路

展开

八次中成功次数服从二项分布,少于3次直接求0、1、2三项。

答题过程

展开

Let S∼B(8,512)S\sim\mathrm{B}\bigl(8,\frac5{12}\bigr) and write p=512p=\frac5{12}, q=712q=\frac7{12}. Then

P(S<3)= q8+8pq7+(82)p2q6= 0.281548777…≈ 0.282.\begin{align*} \mathrm{P}(S<3)=&\,q^8+8pq^7+\binom82p^2q^6\\[4mm] =&\,0.281548777\ldots\\[4mm] \approx&\,\boxed{0.282}. \end{align*}