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CIE 9709 2024 June Paper 53 Q5

A Level / CIE / S1

CIE 9709 2024 June Paper 53 Paper · Question 5

题目

Problem

Salah decides to attempt the crossword puzzle in his newspaper each day. The probability that he will complete the puzzle on any given day is 0.650.65, independent of other days.

(a) Find the probability that Salah completes the puzzle for the first time on the 5th day.

(1)

(b) Find the probability that Salah completes the puzzle for the second time on the 5th day.

(2)

(c) Find the probability that Salah completes the puzzle fewer than 5 times in a week (7 days).

(3)

(d) Use a suitable approximation to find the probability that Salah completes the puzzle more than 50 times in a period of 84 days.

(5)
题目中文翻译

Salah 决定每天尝试做报纸上的纵横填字游戏。他在任意一天完成填字游戏的概率为 0.650.65，且各天相互独立。

(a) 求 Salah 在第 5 天首次完成填字游戏的概率。

(1)

(b) 求 Salah 在第 5 天第二次完成填字游戏的概率。

(2)

(c) 求 Salah 在一周（7 天）内完成填字游戏少于 5 次的概率。

(3)

(d) 使用适当的近似方法，求 Salah 在 84 天内完成填字游戏超过 50 次的概率。

(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

前四天失败，第五天成功。

答题过程

展开 P=(0.35)4(0.65)=0.0097540625≈0.00975.\mathrm{P}=(0.35)^4(0.65)=0.0097540625\approx\boxed{0.00975}.

(b)

解法一

思路

展开

前四天恰一次成功，第五天成功，共4个首次成功位置。

答题过程

展开 P=(41)(0.65)(0.35)3(0.65)=0.07245875≈0.0725.\mathrm{P}=\binom41(0.65)(0.35)^3(0.65) =0.07245875\approx\boxed{0.0725}.

(c)

解法一

思路

展开

一周成功次数服从二项分布，少于5的补集为5、6、7次。

答题过程

展开

Let S∼B(7,0.65)S\sim\mathrm{B}(7,0.65). Then

P(S<5)= 1−[(75)(0.65)5(0.35)2 +7(0.65)6(0.35)+(0.65)7]= 0.467716675…≈ 0.468.\begin{align*} \mathrm{P}(S<5) =&\,1-\bigl[\binom75(0.65)^5(0.35)^2\\[4mm] &\,\hspace{2pt}+7(0.65)^6(0.35)+(0.65)^7\bigr]\\[4mm] =&\,0.467716675\ldots\\[4mm] \approx&\,\boxed{0.468}. \end{align*}

(d)

解法一

思路

展开

正态近似二项分布，超过50即至少51，连续性修正50.5。

答题过程

展开

Let X∼B(84,0.65)X\sim\mathrm{B}(84,0.65). Since np=54.6np=54.6 and n(1−p)=29.4n(1-p)=29.4 are both greater than 5, use Y∼N(54.6,19.11)Y\sim\mathrm{N}(54.6,19.11).

With a continuity correction,

P(X>50)= P(X≥51)≈ P(Y>50.5)= 1−Φ(50.5−54.619.11)= 0.825850420…≈ 0.826.\begin{align*} \mathrm{P}(X>50) =&\,\mathrm{P}(X\geq51)\\[4mm] \approx&\,\mathrm{P}(Y>50.5)\\[4mm] =&\,1-\Phi\biggl(\frac{50.5-54.6}{\sqrt{19.11}}\biggr)\\[4mm] =&\,0.825850420\ldots\\[4mm] \approx&\,\boxed{0.826}. \end{align*}