题目

Problem

(a) Two ponds, $A$ and $B$ , each contain a large number of fish. It is known that $2.4\%$ of fish in pond $A$ are carp and $1.8\%$ of fish in pond $B$ are carp. Random samples of $50$ fish from pond $A$ and $60$ fish from pond $B$ are selected.

Use appropriate Poisson approximations to find the following probabilities.

(i) The samples contain at least $2$ carp from pond $A$ and at least $2$ carp from pond $B$ .

(3)

(ii) The samples contain at least $4$ carp altogether.

(3)

(b) The random variables $X$ and $Y$ have the distributions $\mathrm{Po}(\lambda)$ and $\mathrm{Po}(\mu)$ respectively. It is given that

$\mathrm{P}(X=0)=[\mathrm{P}(Y=0)]^2$ ,
$\mathrm{P}(X=2)=k[\mathrm{P}(Y=1)]^2$ , where $k$ is a non-zero constant.

Find the value of $k$ .

(4)

题目中文翻译

(a) 两个池塘 $A$ 和 $B$ 各有许多鱼。已知池塘 $A$ 有 $2.4\%$ 的鱼是鲤鱼，池塘 $B$ 有 $1.8\%$ 的鱼是鲤鱼。从池塘 $A$ 随机抽取 $50$ 条鱼，从池塘 $B$ 随机抽取 $60$ 条鱼。

用适当的泊松近似求下列事件的概率。

(i) 样本中来自池塘 $A$ 的鲤鱼至少有 $2$ 条，来自池塘 $B$ 的鲤鱼至少有 $2$ 条。

(3)

(ii) 两个样本中合计至少有 $4$ 条鲤鱼。

(3)

(b) 随机变量 $X$ 和 $Y$ 分别服从 $\mathrm{Po}(\lambda)$ 和 $\mathrm{Po}(\mu)$ 分布。已知

$\mathrm{P}(X=0)=[\mathrm{P}(Y=0)]^2$ ，
$\mathrm{P}(X=2)=k[\mathrm{P}(Y=1)]^2$ ，其中 $k$ 为非零常数。

求 $k$ 的值。

(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)(i)

解法一

思路

展开

两份样本中的鲤鱼数分别用Po1.2及Po1.08近似；各自至少两条的概率相乘。

答题过程

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Let $A$ and $B$ be the numbers of carp in the respective samples. The populations are large and the carp proportions are small, so use the independent Poisson approximations

\begin{align*} A\sim&\,\mathrm{Po}(50(0.024))=\mathrm{Po}(1.2),\\[4mm] B\sim&\,\mathrm{Po}(60(0.018))=\mathrm{Po}(1.08). \end{align*}

Hence

\begin{align*} &\,\mathrm{P}(A\geq2,\ B\geq2)\\[4mm] \approx&\,\bigl[1-e^{-1.2}(1+1.2)\bigr]\\[4mm] &\,\times\bigl[1-e^{-1.08}(1+1.08)\bigr]\\[4mm] =&\,\boxed{0.0991}\quad\text{(3 s.f.).} \end{align*}

(a)(ii)

解法一

思路

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独立Poisson变量之和仍为Poisson，参数相加；至少四条用前三条的补集。

答题过程

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The total $C=A+B$ is approximately $\mathrm{Po}(1.2+1.08)=\mathrm{Po}(2.28)$ . Therefore

\begin{align*} \mathrm{P}(C\geq4) \approx&\,1-e^{-2.28}\bigg(1+2.28+\frac{2.28^2}{2!}\bigg)\\[4mm] &\,-e^{-2.28}\frac{2.28^3}{3!}\\[4mm] =&\,\boxed{0.197}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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第一条概率关系给λ=2μ，代入第二条并消去共同的正因子。

答题过程

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The first given condition gives

\begin{align*} e^{-\lambda}=&\,(e^{-\mu})^2=e^{-2\mu},\\[4mm] \lambda=&\,2\mu. \end{align*}

The second condition gives

e^{-\lambda}\frac{\lambda^2}{2} =k(e^{-\mu}\mu)^2.

Substituting $\lambda=2\mu$ ,

2e^{-2\mu}\mu^2=ke^{-2\mu}\mu^2.

Cancelling the common positive factor, $\boxed{k=2}$ .

CIE 9709 2022 March Paper 62 Q7

题目

解答

(a)(i)

解法一

思路

答题过程

(a)(ii)

解法一

思路

答题过程

(b)

解法一

思路

答题过程