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CIE 9709 2024 March Paper 62 Q3

A Level / CIE / S2

CIE 9709 2024 March Paper 62 Paper · Question 3

题目

Problem

In a certain lottery, on average 11 in every 10 00010\,000 tickets is a prize-winning ticket. An agent sells 60006000 tickets.

(a) Use a suitable approximating distribution to find the probability that at least 33 of the tickets sold by the agent are prize-winning tickets.

(3)

(b) Justify the use of your approximating distribution in this context.

(1)
题目中文翻译

在某种彩票中,平均每 10 00010\,000 张彩票有 11 张中奖。一名代理商售出 60006000 张彩票。

(a) 用适当的近似分布求该代理商售出的彩票中至少有 33 张中奖的概率。

(3)

(b) 结合题意说明采用该近似分布的理由。

(1)
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

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大量彩票、低中奖率,采用参数6000乘1/10000的泊松近似,至少3张取0至2张的补集。

答题过程

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Let WW be the number of prize-winning tickets sold. Using the Poisson approximation,

W≈Po(6000110000)=Po(0.6).W\approx\mathrm{Po}\bigg(6000\frac1{10000}\bigg)=\mathrm{Po}(0.6).

Therefore

P(W≥3)≈ 1−e−0.6(1+0.6+0.622!)= 0.0231153…= 0.0231(3 s.f.).\begin{align*} \mathrm{P}(W\geq3)\approx&\,1-e^{-0.6}\bigg(1+0.6+\frac{0.6^2}{2!}\bigg)\\[4mm] =&\,0.0231153\ldots\\[4mm] =&\,\boxed{0.0231}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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结合彩票情境给出具体的大n、小p参数。

答题过程

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The agent sells many tickets, n=6000>50n=6000>50, and the probability that any ticket wins is small, p=110000<0.1p=\frac1{10000}<0.1, so a Poisson approximation is suitable.