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CIE 9709 2025 November Paper 52 Q6

A Level / CIE / S1

CIE 9709 2025 Nov Paper 52 Paper · Question 6

题目

Problem

For a randomly chosen person, their next birthday is equally likely to occur on any day of the week, independently of any other person’s birthday.

(a) Find the probability that, out of 10 randomly chosen people, none of them will have their next birthday on a Saturday or Sunday.

(1)

(b) Find the probability that, out of 10 randomly chosen people, fewer than 3 will have their next birthday on a Wednesday.

(3)

(c) Use a suitable approximation to find the probability that, out of 392 randomly chosen people, more than 65 will have their next birthday on a Friday.

(5)
题目中文翻译

对于随机选取的一个人，其下一次生日落在一周中任意一天的可能性相同，并且与其他任何人的生日相互独立。

(a) 随机选取 10 个人，求他们中没有人的下一次生日落在星期六或星期日的概率。

(1)

(b) 随机选取 10 个人，求其中下一次生日落在星期三的人数少于 3 的概率。

(3)

(c) 利用适当的近似方法，求随机选取的 392 人中，下一次生日落在星期五的人数多于 65 的概率。

(5)
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

每人下一次生日不在周末的概率为七分之五，10人之间独立，概率直接连乘。

答题过程

展开 P(none on Saturday or Sunday)= (57)10≈ 0.0346(3 s.f.).\begin{align*} \mathrm{P}(\text{none on Saturday or Sunday}) =&\,\biggl(\frac57\biggr)^{10}\\[4mm] \approx&\,\boxed{0.0346}\quad\text{(3 s.f.)}. \end{align*}

(b)

解法一

思路

展开

10人中生日为周三的人数服从成功概率七分之一的二项分布。少于3人即0、1、2人。

答题过程

展开

Let X∼B(10,17)X\sim\mathrm{B}(10,\frac17) be the number whose next birthday is on a Wednesday. Then

P(X<3)= (67)10+1017(67)9 +45(17)2(67)8= 0.838395…≈ 0.838.\begin{align*} \mathrm{P}(X<3) =&\,\biggl(\frac67\biggr)^{10} +10\frac17\biggl(\frac67\biggr)^9\\[4mm] &\,\hspace{2pt}+45\biggl(\frac17\biggr)^2 \biggl(\frac67\biggr)^8\\[4mm] =&\,0.838395\ldots\\[4mm] \approx&\,\boxed{0.838}. \end{align*}

(c)

解法一

思路

展开

用392次二项分布的正态近似。超过65即至少66，连续性修正后的下界为65.5；最后取右尾概率。

答题过程

展开

Let X∼B(392,17)X\sim\mathrm{B}(392,\frac17) be the number whose next birthday is on a Friday.

Here np=56np=56 and n(1−p)=336n(1-p)=336 are both greater than 5, so use the normal approximation

Y∼N(56,48),48=3921767.Y\sim\mathrm{N}(56,48), \qquad 48=392\frac17\frac67.

Applying a continuity correction,

P(X>65)= P(X≥66)≈ P(Y>65.5)= 1−Φ(65.5−5648)= 1−Φ(1.371206…)= 0.085155…≈ 0.0852(3 s.f.).\begin{align*} \mathrm{P}(X>65)=&\,\mathrm{P}(X\geq66)\\[4mm] \approx&\,\mathrm{P}(Y>65.5)\\[4mm] =&\,1-\Phi\biggl(\frac{65.5-56}{\sqrt{48}}\biggr)\\[4mm] =&\,1-\Phi(1.371206\ldots)\\[4mm] =&\,0.085155\ldots\\[4mm] \approx&\,\boxed{0.0852}\quad\text{(3 s.f.)}. \end{align*}