Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2026 June Paper 52 Q4

A Level / CIE / S1

CIE 9709 2026 June Paper 52 Paper · Question 4

题目

Problem

Every day Haz attempts to complete a particular type of puzzle in less than 10 minutes. On any attempt, the probability that she will succeed is 0.6.

(a) Find the probability that on 9 randomly chosen days, Haz completes the puzzle in less than 10 minutes on at least 7 of these 9 days.

(3)

120 days are chosen at random.

(b) Use an approximation to find the probability that Haz completes the puzzle in less than 10 minutes on fewer than 66 of these 120 days.

(5)
题目中文翻译

Haz 每天尝试在少于 10 分钟的时间内完成某一类型的谜题。每次尝试成功的概率为 0.6。

(a) 随机选取 9 天,求 Haz 在这 9 天中至少有 7 天用少于 10 分钟完成谜题的概率。

(3)

随机选取 120 天。

(b) 利用近似方法,求 Haz 在这 120 天中用少于 10 分钟完成谜题的天数少于 66 天的概率。

(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

把每天是否成功看作一次二项试验,至少7次即7、8、9次。直接相加只有三项。

答题过程

展开

Let XX be the number of successes on the 9 days. Then X∼B(9,0.6)X\sim\mathrm{B}(9,0.6), so

P(X≥7)= (97)(0.6)7(0.4)2 +(98)(0.6)8(0.4)+(0.6)9= 0.231787008≈ 0.232.\begin{align*} \mathrm{P}(X\geq7)=&\,\binom97(0.6)^7(0.4)^2\\[4mm] &\,\hspace{2pt}+\binom98(0.6)^8(0.4)+(0.6)^9\\[4mm] =&\,0.231787008\\[4mm] \approx&\,\boxed{0.232}. \end{align*}

(b)

解法一

思路

展开

先求二项分布的均值与方差,用正态分布近似。“少于66”包含65及以下,连续性修正后的上界为65.5。

答题过程

展开

Let XX be the number of successes in 120 days. Then X∼B(120,0.6)X\sim\mathrm{B}(120,0.6).

Since np=72np=72 and n(1−p)=48n(1-p)=48 are both greater than 5, use the normal approximation

Y∼N(72,28.8),Y\sim\mathrm{N}(72,28.8),

where 28.8=120(0.6)(0.4)28.8=120(0.6)(0.4) is the variance.

Applying a continuity correction,

P(X<66)= P(X≤65)≈ P(Y<65.5)= P(Z<65.5−7228.8)= Φ(−1.2112…)= 0.112908…≈ 0.113.\begin{align*} \mathrm{P}(X<66)=&\,\mathrm{P}(X\leq65)\\[4mm] \approx&\,\mathrm{P}(Y<65.5)\\[4mm] =&\,\mathrm{P}\biggl(Z<\frac{65.5-72}{\sqrt{28.8}}\biggr)\\[4mm] =&\,\Phi(-1.2112\ldots)\\[4mm] =&\,0.112908\ldots\\[4mm] \approx&\,\boxed{0.113}. \end{align*}