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CIE 9709 2024 March Paper 52 Q2

A Level / CIE / S1

CIE 9709 2024 March Paper 52 Paper · Question 2

题目

Problem

Sam is a member of a soccer club. She is practising scoring goals. The probability that Sam will score a goal on any attempt is 0.70.7, independently of all other attempts.

(a) Sam makes 10 attempts at scoring goals.

Find the probability that Sam will score goals on fewer than 8 of these attempts.

(3)

(b) Find the probability that Sam’s first successful attempt will be before her 5th attempt.

(2)

(c) Wei is a member of the same soccer club. He is also practising scoring goals. The probability that Wei will score a goal on any attempt is 0.60.6, independently of all other attempts.

Wei is going to keep making attempts until he scores 3 goals.

Find the probability that he scores his third goal on his 7th attempt.

(3)
题目中文翻译

Sam 是一家足球俱乐部的会员。她正在练习射门。Sam 每次射门进球的概率为 0.70.7，且各次射门相互独立。

(a) Sam 进行 10 次射门。

求 Sam 在这些射门中进球少于 8 次的概率。

(3)

(b) 求 Sam 在第 5 次射门之前首次进球的概率。

(2)

(c) Wei 是同一家足球俱乐部的会员。他也正在练习射门。Wei 每次射门进球的概率为 0.60.6，且各次射门相互独立。

Wei 将持续射门，直到进满 3 球为止。

求他在第 7 次射门时进第三个球的概率。

(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

成功次数服从二项分布，少于8的补集为8、9、10。

答题过程

展开

Let X∼B(10,0.7)X\sim\mathrm{B}(10,0.7). Then

P(X<8)= 1−[(108)(0.7)8(0.3)2 +10(0.7)9(0.3)+(0.7)10]= 0.6172172136≈ 0.617.\begin{align*} \mathrm{P}(X<8) =&\,1-\bigl[\binom{10}{8}(0.7)^8(0.3)^2\\[4mm] &\,\hspace{2pt}+10(0.7)^9(0.3)+(0.7)^{10}\bigr]\\[4mm] =&\,0.6172172136\\[4mm] \approx&\,\boxed{0.617}. \end{align*}

(b)

解法一

思路

展开

第五次之前即前四次内，取四次均失败的补集。

答题过程

展开 P=1−(0.3)4=0.9919.\mathrm{P}=1-(0.3)^4=\boxed{0.9919}.

解法二

思路

展开

也可以按首次成功在第1、2、3、4次分类。

答题过程

展开 P= 0.7+0.3(0.7)+(0.3)2(0.7) +(0.3)3(0.7)= 0.9919.\begin{align*} \mathrm{P}=&\,0.7+0.3(0.7)+(0.3)^2(0.7)\\[4mm] &\,\hspace{2pt}+(0.3)^3(0.7)\\[4mm] =&\,\boxed{0.9919}. \end{align*}

(c)

解法一

思路

展开

前6次恰两球，第7次第三球。前6次的两次成功位置有6选2种。

答题过程

展开 P=(62)(0.6)2(0.4)4(0.6)=0.082944≈0.0829.\mathrm{P}=\binom62(0.6)^2(0.4)^4(0.6) =0.082944\approx\boxed{0.0829}.