题目

Problem

A bag contains 50 counters, each with one of the numbers 4, 7 or 10 written on it in the ratio 2 : 3 : 5 respectively.

A random sample of 2 counters is taken from the bag. The numbers on the 2 counters are recorded as D₁ and D₂

The random variable M represents the mean of D₁ and D₂

(a) Show that P(M = 4) = 9/245

(1)

(b) Find the sampling distribution of M

(6)

A random sample of n sets of 2 counters is taken. The random variable T represents the number of these n sets of 2 counters that have a mean of 4

Given that each set of 2 counters is replaced after it is drawn,

(3)

(Total for Question 4 is 10 marks)

题目中文翻译

一个袋子里有 50 个筹码，每个筹码上写有数字 4、7 或 10，比例分别为 2 : 3 : 5。

从袋中随机抽取 2 个筹码。2 个筹码上的数字分别记录为 D₁ 和 D₂。

随机变量 M 表示 D₁ 和 D₂ 的均值。

(a) 证明 P(M = 4) = 9/245

(b) 求 M 的抽样分布。

随机抽取 n 组 2 个筹码。随机变量 T 表示这 n 组中均值为 4 的组数。

已知每组 2 个筹码在抽取后被放回，

（第 4 题共 10 分）

解答

(a)

解法一

思路

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两个数的均值为 4，当且仅当抽到的两个筹码都写着 4。袋中共有 $50\times\frac{2}{10}=10$ 个数字 4 的筹码；同一组内是不放回抽样，因此第二次只剩 9 个。

答题过程

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There are

50\times\frac{2}{2+3+5}=10

counters numbered 4. Since the two counters are drawn without replacement,

P(M=4)=\frac{10}{50}\times\frac9{49} =\boxed{\frac9{245}}.

(b)

解法一

思路

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先由比例求出数字 4、7、10 的筹码分别有 10、15、25 个。列举两个数字的所有无序组合，可得 $M$ 的五个可能值；异数搭配有两种抽取顺序，同数搭配则第二次的可选数要减 1。把产生同一均值的组合概率相加。

答题过程

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The numbers of counters labelled 4, 7 and 10 are respectively

10,\qquad15,\qquad25.

The possible values of $M$ are

4,\qquad5.5,\qquad7,\qquad8.5,\qquad10.

Using sampling without replacement,

P(M=4)=\frac9{245},

\begin{align*} P(M=5.5)=&\,2\left(\frac{10}{50}\right) \left(\frac{15}{49}\right)\\ =&\,\frac6{49}, \end{align*}

\begin{align*} P(M=7)=&\,2\left(\frac{10}{50}\right) \left(\frac{25}{49}\right)\\ &\,\hspace{2pt}+\left(\frac{15}{50}\right) \left(\frac{14}{49}\right)\\ =&\,\frac{71}{245}, \end{align*}

\begin{align*} P(M=8.5)=&\,2\left(\frac{15}{50}\right) \left(\frac{25}{49}\right)\\ =&\,\frac{15}{49}, \end{align*}

and

\begin{align*} P(M=10)=&\,\left(\frac{25}{50}\right) \left(\frac{24}{49}\right)\\ =&\,\frac{12}{49}. \end{align*}

Hence the sampling distribution of $M$ is

$m$	4	5.5	7	8.5	10
$P(M=m)$	$\dfrac9{245}$	$\dfrac6{49}$	$\dfrac{71}{245}$	$\dfrac{15}{49}$	$\dfrac{12}{49}$

(c)

解法一

思路

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每组抽完后都放回，所以各组相互独立，且每组均值为 4 的概率恒为 $9/245$ 。因此 $T$ 服从二项分布；将 $P(T=0)$ 写成“每组都失败”的概率，解指数不等式，并取满足严格不等式的最小整数。

答题过程

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Since each set is replaced after it is drawn, the sets are independent and

T\sim\operatorname{B}\left(n,\frac9{245}\right).

Therefore,

\begin{align*} P(T=0)=&\,\left(1-\frac9{245}\right)^n\\ =&\,\left(\frac{236}{245}\right)^n. \end{align*}

The required inequality is

\left(\frac{236}{245}\right)^n<0.15.

Taking logarithms, and noting that $\log(236/245)<0$ ,

\begin{align*} n>&\,\frac{\log(0.15)}{\log(236/245)}\\ >&\,50.689\ldots \end{align*}

Hence the minimum integer value is

\boxed{n=51}.

IAL 2024 Jun S2 Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程