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CIE 9709 2024 November Paper 52 Q7

A Level / CIE / S1

CIE 9709 2024 Nov Paper 52 Paper · Question 7

题目

Problem

In a game, players attempt to score a goal by kicking a ball into a net. The probability that Leno scores a goal is 0.40.4 on any attempt, independently of all other attempts. The random variable XX denotes the number of attempts that it takes Leno to score a goal.

(a) Find P(X=5)\mathrm{P}(X=5).

(1)

(b) Find P(3≤X≤7)\mathrm{P}(3\leq X\leq 7).

(2)

(c) Find the probability that Leno scores his second goal on or before his 5th attempt.

(3)

Leno has 75 attempts to score a goal.

(d) Use a suitable approximation to find the probability that Leno scores more than 28 goals but fewer than 35 goals.

(5)
题目中文翻译

在一项游戏中，玩家尝试把球踢入球网得分。Leno 每次尝试进球的概率为 0.40.4，且各次尝试相互独立。随机变量 XX 表示 Leno 进一个球所需的尝试次数。

(a) 求 P(X=5)\mathrm{P}(X=5)。

(1)

(b) 求 P(3≤X≤7)\mathrm{P}(3\leq X\leq 7)。

(2)

(c) 求 Leno 在第 5 次尝试时或之前进第二个球的概率。

(3)

Leno 有 75 次尝试进球的机会。

(d) 使用适当的近似方法，求 Leno 进球数大于 28 但小于 35 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

前四次不进球，第五次进球。

答题过程

展开 P(X=5)=(0.6)4(0.4)=0.05184.\mathrm{P}(X=5)=(0.6)^4(0.4)=\boxed{0.05184}.

(b)

解法一

思路

展开

首次进球在第3至第7次，等价于前两次均未进，但前七次并非全未进。

答题过程

展开 P(3≤X≤7)= P(X>2)−P(X>7)= (0.6)2−(0.6)7= 0.3320064≈ 0.332.\begin{align*} \mathrm{P}(3\leq X\leq7) =&\,\mathrm{P}(X>2)-\mathrm{P}(X>7)\\[4mm] =&\,(0.6)^2-(0.6)^7\\[4mm] =&\,0.3320064\\[4mm] \approx&\,\boxed{0.332}. \end{align*}

解法二

思路

展开

也可以按首次进球在第3、4、5、6、7次分类，将五项等比数列求和。

答题过程

展开 P(3≤X≤7)= 0.4[(0.6)2+(0.6)3+(0.6)4 +(0.6)5+(0.6)6]= 0.144(1−0.65)1−0.6= 0.3320064≈ 0.332.\begin{align*} \mathrm{P}(3\leq X\leq7) =&\,0.4\bigl[(0.6)^2+(0.6)^3+(0.6)^4\\[4mm] &\,\hspace{2pt}+(0.6)^5+(0.6)^6\bigr]\\[4mm] =&\,\frac{0.144(1-0.6^5)}{1-0.6}\\[4mm] =&\,0.3320064\\[4mm] \approx&\,\boxed{0.332}. \end{align*}

(c)

解法一

思路

展开

第五次或之前进第二球，等价于前五次至少进两球，用二项分布的0、1球补集。

答题过程

展开

Let S∼B(5,0.4)S\sim\mathrm{B}(5,0.4) count goals in the first five attempts. Then

P(S≥2)= 1−[(0.6)5+5(0.4)(0.6)4]= 0.66304≈ 0.663.\begin{align*} \mathrm{P}(S\geq2) =&\,1-\bigl[(0.6)^5+5(0.4)(0.6)^4\bigr]\\[4mm] =&\,0.66304\\[4mm] \approx&\,\boxed{0.663}. \end{align*}

解法二

思路

展开

另一条路线按第二球在第2至第5次分类。若第二球在第r次，前r−1次恰有一次进球，第r次进球。

答题过程

展开

The second goal may occur on attempt 2, 3, 4 or 5. These disjoint events have total probability

P= (0.4)2+2(0.4)2(0.6) +3(0.4)2(0.6)2+4(0.4)2(0.6)3= 0.16+0.192+0.1728+0.13824= 0.66304≈ 0.663.\begin{align*} \mathrm{P}=&\,(0.4)^2+2(0.4)^2(0.6)\\[4mm] &\,\hspace{2pt}+3(0.4)^2(0.6)^2+4(0.4)^2(0.6)^3\\[4mm] =&\,0.16+0.192+0.1728+0.13824\\[4mm] =&\,0.66304\\[4mm] \approx&\,\boxed{0.663}. \end{align*}

(d)

解法一

思路

展开

进球数服从二项分布，用正态近似。严格界限28与35对应整数29至34，修正后为28.5至34.5。

答题过程

展开

Let G∼B(75,0.4)G\sim\mathrm{B}(75,0.4). Since np=30np=30 and n(1−p)=45n(1-p)=45 are both greater than 5, use Y∼N(30,18)Y\sim\mathrm{N}(30,18).

With continuity corrections,

P(28<G<35)= P(29≤G≤34)≈ P(28.5<Y<34.5)= Φ(34.5−3018)−Φ(28.5−3018)= 0.493741…≈ 0.494.\begin{align*} \mathrm{P}(28<G<35) =&\,\mathrm{P}(29\leq G\leq34)\\[4mm] \approx&\,\mathrm{P}(28.5<Y<34.5)\\[4mm] =&\,\Phi\biggl(\frac{34.5-30}{\sqrt{18}}\biggr) -\Phi\biggl(\frac{28.5-30}{\sqrt{18}}\biggr)\\[4mm] =&\,0.493741\ldots\\[4mm] \approx&\,\boxed{0.494}. \end{align*}