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CIE 9709 2024 June Paper 62 Q5

A Level / CIE / S2

CIE 9709 2024 June Paper 62 Paper · Question 5

题目

Problem

The number of goals scored by a sports team in the first half of any match has the distribution X∼Po(3.1)X\sim\mathrm{Po}(3.1). The number of goals scored by the same team in the second half of any match has the distribution Y∼Po(2.4)Y\sim\mathrm{Po}(2.4). You may assume that the distributions of XX and YY are independent.

(a) Find P(X<4)\mathrm{P}(X<4).

(2)

(b) Find the probability that, in a randomly chosen match, the team scores at least 55 goals.

(3)

(c) Given that the team scores a total of 55 goals in a randomly chosen match, find the probability that they score exactly 33 goals in the first half.

(4)
题目中文翻译

某支球队在任意一场比赛中上半场的进球数 XX 服从 Po(3.1)\mathrm{Po}(3.1) 分布。该球队在任意一场比赛中下半场的进球数 YY 服从 Po(2.4)\mathrm{Po}(2.4) 分布。可假设 XX 与 YY 相互独立。

(a) 求 P(X<4)\mathrm{P}(X<4)。

(2)

(b) 随机选取一场比赛，求该球队至少进 55 球的概率。

(3)

(c) 已知在随机选取的一场比赛中，该球队共进了 55 球，求上半场恰好进 33 球的概率。

(4)
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

上半场少于4球即0至3球，累加对应泊松概率。

答题过程

展开 P(X<4)= e−3.1(1+3.1) +e−3.1(3.122!+3.133!)= 0.624840…= 0.625(3 s.f.).\begin{align*} \mathrm{P}(X<4)=&\,e^{-3.1}(1+3.1)\\[4mm] &\,+e^{-3.1}\bigg(\frac{3.1^2}{2!}+\frac{3.1^3}{3!}\bigg)\\[4mm] =&\,0.624840\ldots\\[4mm] =&\,\boxed{0.625}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

展开

全场总进球是独立泊松变量之和，参数为5.5。至少5球取不超过4球的补集。

答题过程

展开

Since XX and YY are independent,

S=X+Y∼Po(3.1+2.4)=Po(5.5).S=X+Y\sim\mathrm{Po}(3.1+2.4)=\mathrm{Po}(5.5).

Thus

P(S≥5)= 1−e−5.5(1+5.5) −e−5.5(5.522!+5.533!) −e−5.55.544!= 0.642482…= 0.642(3 s.f.).\begin{align*} \mathrm{P}(S\geq5)=&\,1-e^{-5.5}(1+5.5)\\[4mm] &\,-e^{-5.5}\bigg(\frac{5.5^2}{2!}+\frac{5.5^3}{3!}\bigg)\\[4mm] &\,\hspace{2pt}-e^{-5.5}\frac{5.5^4}{4!}\\[4mm] =&\,0.642482\ldots\\[4mm] =&\,\boxed{0.642}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

展开

条件总数为5且上半场3球，等价于上半场3球且下半场2球。用独立性求联合概率，再除以总数为5的概率。

答题过程

展开  P(X=3∣X+Y=5)= P(X=3,Y=2)P(X+Y=5)= e−5.53.13(2.4)23! 2!e−5.55.555!= 3.13(2.4)2 5!3! 2! 5.55= 0.340953…= 0.341(3 s.f.).\begin{align*} &\,\mathrm{P}(X=3\mid X+Y=5)\\[4mm] =&\,\frac{\mathrm{P}(X=3,Y=2)}{\mathrm{P}(X+Y=5)}\\[4mm] =&\,\frac{e^{-5.5}\dfrac{3.1^3(2.4)^2}{3!\,2!}} {e^{-5.5}\dfrac{5.5^5}{5!}}\\[4mm] =&\,\frac{3.1^3(2.4)^2\,5!}{3!\,2!\,5.5^5}\\[4mm] =&\,0.340953\ldots\\[4mm] =&\,\boxed{0.341}\quad\text{(3 s.f.).} \end{align*}