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S2 2021 June Q4

A Level / Edexcel / S2

IAL 2021 June Paper · Question 4

题目

Problem

A bag contains a large number of balls, each with one of the numbers 1, 2 or 5 written on it in the ratio 2:3:42:3:4 respectively.

A random sample of 3 balls is taken from the bag.

The random variable BB represents the range of the numbers written on the balls in the sample.

(i) Find P(B=4)P(B = 4)

(ii) Find the sampling distribution of BB.

(10)

(Total for Question 4 is 10 marks)

题目中文翻译

一个袋子里有很多球，每个球上写着 1、2 或 5 之一，比例分别为 2:3:42:3:4。

从袋中随机抽取 3 个球。

随机变量 BB 表示样本中所写数字的极差。

(i) 求 P(B=4)P(B=4)。

(ii) 求 BB 的抽样分布。

(10)

（第 4 题共 10 分）

解答

(i)

解法一

思路

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先把比例化成单次抽到各数字的概率。极差为 4 表示样本中必须同时出现 1 和 5；三个球的无序组合只有 (1,1,5)(1,1,5)、(1,2,5)(1,2,5)、(1,5,5)(1,5,5)。分别考虑排列数后相加。

答题过程

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Let AA be the number written on one selected ball. Then

P(A=1)= 29P(A=2)= 13P(A=5)= 49.\begin{align*} P(A=1)=&\,\frac29 \\[4mm] P(A=2)=&\,\frac13 \\[4mm] P(A=5)=&\,\frac49. \end{align*}

The possible unordered samples with range 4 are

(1,1,5),(1,2,5),(1,5,5).\begin{gathered} (1,1,5), \\[4mm] (1,2,5), \\[4mm] (1,5,5). \end{gathered}

Samples with two equal numbers have 3!/2!=33!/2!=3 possible orders; a sample with three different numbers has 3!=63!=6 possible orders. Their probabilities, including all possible orders, are

 P(1,1,5) =3(29)2(49) =16243,\begin{align*} &\,P(1,1,5) \\[4mm] &\,=3\Big(\frac29\Big)^2\Big(\frac49\Big) \\[4mm] &\,=\frac{16}{243}, \end{align*}  P(1,2,5) =6(29)(13)(49) =1681,\begin{align*} &\,P(1,2,5) \\[4mm] &\,=6\Big(\frac29\Big)\Big(\frac13\Big)\Big(\frac49\Big) \\[4mm] &\,=\frac{16}{81}, \end{align*}

and

 P(1,5,5) =3(29)(49)2 =32243.\begin{align*} &\,P(1,5,5) \\[4mm] &\,=3\Big(\frac29\Big)\Big(\frac49\Big)^2 \\[4mm] &\,=\frac{32}{243}. \end{align*}

Therefore,

 P(B=4) =16243+1681 +32243 =3281.\begin{align*} &\,P(B=4) \\[4mm] &\,=\frac{16}{243}+\frac{16}{81} \\[4mm] &\,\hspace{2pt}+\frac{32}{243} \\[4mm] &\,=\boxed{\frac{32}{81}}. \end{align*}

解法二

思路

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也可以把“同时出现 1 和 5”作为两个事件的交集，用容斥原理计算。总概率减去“没有 1”与“没有 5”，再加回被重复减去的“只有 2”。

答题过程

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The event B=4B=4 occurs when the sample contains at least one 1 and at least one 5. Hence, by inclusion-exclusion,

 P(B=4) =1−P(no 1) −P(no 5) +P(only 2s) =1−(79)3 −(59)3 +(13)3 =3281.\begin{align*} &\,P(B=4) \\[4mm] &\,=1-P(\text{no 1}) \\[4mm] &\,\hspace{2pt}-P(\text{no 5}) \\[4mm] &\,\hspace{4pt}+P(\text{only 2s}) \\[4mm] &\,=1-\Big(\frac79\Big)^3 \\[4mm] &\,\hspace{2pt}-\Big(\frac59\Big)^3 \\[4mm] &\,\hspace{4pt}+\Big(\frac13\Big)^3 \\[4mm] &\,=\boxed{\frac{32}{81}}. \end{align*}

(ii)

解法一

思路

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三个可抽到的数字为 1、2、5，所以极差只可能是 0、1、3、4。极差为 0 时三个数字相同；极差为 1 时只出现 1 和 2 且两者都要出现；极差为 3 时只出现 2 和 5 且两者都要出现；极差为 4 已由 (i) 求出。

答题过程

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For B=0B=0, all three numbers must be the same:

 P(B=0) =(29)3 +(13)3 +(49)3 =1181.\begin{align*} &\,P(B=0) \\[4mm] &\,=\Big(\frac29\Big)^3 \\[4mm] &\,\hspace{2pt}+\Big(\frac13\Big)^3 \\[4mm] &\,\hspace{4pt}+\Big(\frac49\Big)^3 \\[4mm] &\,=\frac{11}{81}. \end{align*}

For B=1B=1, both 1 and 2 must occur, but 5 must not occur:

 P(B=1) =3(29)2(13) +3(29)(13)2 =1081.\begin{align*} &\,P(B=1) \\[4mm] &\,=3\Big(\frac29\Big)^2\Big(\frac13\Big) \\[4mm] &\,\hspace{2pt}+3\Big(\frac29\Big)\Big(\frac13\Big)^2 \\[4mm] &\,=\frac{10}{81}. \end{align*}

For B=3B=3, both 2 and 5 must occur, but 1 must not occur:

 P(B=3) =3(13)2(49) +3(13)(49)2 =2881.\begin{align*} &\,P(B=3) \\[4mm] &\,=3\Big(\frac13\Big)^2\Big(\frac49\Big) \\[4mm] &\,\hspace{2pt}+3\Big(\frac13\Big)\Big(\frac49\Big)^2 \\[4mm] &\,=\frac{28}{81}. \end{align*}

Using part (i), P(B=4)=32/81P(B=4)=32/81.

Alternatively, once the probabilities for B=0B=0, B=1B=1 and B=4B=4 are known, the probability for B=3B=3 can be obtained by subtracting them from 11, since 00, 11, 33 and 44 are the only possible ranges:

 P(B=3)=1−1181 −1081−3281 =2881.\begin{align*} &\,P(B=3)=1-\frac{11}{81} \\[4mm] &\,\hspace{2pt}-\frac{10}{81}-\frac{32}{81} \\[4mm] &\,=\frac{28}{81}. \end{align*}

Therefore the sampling distribution is

bbP(B=b)P(B=b)
001181\dfrac{11}{81}
111081\dfrac{10}{81}
332881\dfrac{28}{81}
443281\dfrac{32}{81}

The probabilities sum to

11+10+28+3281=1.\frac{11+10+28+32}{81}=1.