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S2 2024 Jun Q4

A Level / Edexcel / S2

IAL 2024 June Paper · Question 4

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 D1D_1 and D2D_2

The random variable MM represents the mean of D1D_1 and D2D_2

(a) Show that P(M=4)=9245P(M=4)=\dfrac9{245}

(1)

(b) Find the sampling distribution of MM

(6)

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

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

(c) calculate the minimum value of nn such that P(T=0)<0.15P(T=0)<0.15

(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 D1D_1 和 D2D_2。

随机变量 MM 表示 D1D_1 和 D2D_2 的均值。

(a) 证明 P(M=4)=9245P(M=4)=\dfrac9{245}

(1)

(b) 求 MM 的抽样分布。

(6)

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

已知每组 2 个筹码在抽取后被放回，

(c) 计算最小的 nn 值，使得 P(T=0)<0.15P(T=0)<0.15

(3)

（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 50×210=1050\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

50×22+3+5=1050\times\frac{2}{2+3+5}=10

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

 P(M=4) =1050×949 =9245.\begin{align*} &\,P(M=4)\\[2mm] &\,=\frac{10}{50}\times\frac9{49}\\[2mm] &\,=\boxed{\frac9{245}}. \end{align*}

(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 MM 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, 15, 25.10,\ 15,\ 25.

The possible values of MM are

4, 5.5, 7, 8.5, 10.4,\ 5.5,\ 7,\ 8.5,\ 10.

Using sampling without replacement,

P(M=4)=9245,P(M=4)=\frac9{245},  P(M=5.5) =2(1050)(1549) =649,\begin{align*} &\,P(M=5.5)\\[2mm] &\,=2\Big(\frac{10}{50}\Big) \Big(\frac{15}{49}\Big)\\ &\,=\frac6{49}, \end{align*}  P(M=7) =2(1050)(2549) +(1550)(1449) =71245,\begin{align*} &\,P(M=7)\\[2mm] &\,=2\Big(\frac{10}{50}\Big) \Big(\frac{25}{49}\Big)\\ &\,\hspace{2pt}+\Big(\frac{15}{50}\Big) \Big(\frac{14}{49}\Big)\\ &\,=\frac{71}{245}, \end{align*}  P(M=8.5) =2(1550)(2549) =1549,\begin{align*} &\,P(M=8.5)\\[2mm] &\,=2\Big(\frac{15}{50}\Big) \Big(\frac{25}{49}\Big)\\ &\,=\frac{15}{49}, \end{align*}

and

 P(M=10) =(2550)(2449) =1249.\begin{align*} &\,P(M=10)\\[2mm] &\,=\Big(\frac{25}{50}\Big) \Big(\frac{24}{49}\Big)\\ &\,=\frac{12}{49}. \end{align*}

Hence the sampling distribution of MM is

mmP(M=m)P(M=m)
49245\dfrac9{245}
5.5649\dfrac6{49}
771245\dfrac{71}{245}
8.51549\dfrac{15}{49}
101249\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 9/2459/245。因此 TT 服从二项分布；将 P(T=0)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∼B⁡(n,9245).T\sim\operatorname{B}\Big(n,\frac9{245}\Big).

Therefore,

 P(T=0) =(1−9245)n =(236245)n.\begin{align*} &\,P(T=0)\\[2mm] &\,=\Big(1-\frac9{245}\Big)^n\\ &\,=\Big(\frac{236}{245}\Big)^n. \end{align*}

The required inequality is

(236245)n<0.15.\Big(\frac{236}{245}\Big)^n<0.15.

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

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

Hence the minimum integer value is

n=51.\boxed{n=51}.

The adjacent values confirm the minimum. With 50 sets,

 (236245)50 =0.1539203103… >0.15,\begin{align*} &\,\Big(\frac{236}{245}\Big)^{50}\\[2mm] &\,=0.1539203103\ldots\\ &\,>0.15, \end{align*}

whereas with 51 sets,

 (236245)51 =0.1482660948… <0.15.\begin{align*} &\,\Big(\frac{236}{245}\Big)^{51}\\[2mm] &\,=0.1482660948\ldots\\ &\,<0.15. \end{align*}