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IAL 2021 June S2 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 : 4 respectively.

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

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

(i) Find P(B = 4)

(ii) Find the sampling distribution of B.

(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 : 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。

（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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For one selected ball,

P(A=1)=29,P(A=2)=13,P(A=5)=49.P(A=1)=\frac29, \qquad P(A=2)=\frac13, \qquad P(A=5)=\frac49.

The possible unordered samples with range 4 are

(1,1,5),(1,2,5),(1,5,5).(1,1,5), \qquad (1,2,5), \qquad (1,5,5).

Their probabilities, including all possible orders, are

P(1,1,5)=3(29)2(49)=16243,P(1,1,5)=3\left(\frac29\right)^2\left(\frac49\right) =\frac{16}{243}, P(1,2,5)=6(29)(13)(49)=1681,P(1,2,5)=6\left(\frac29\right) \left(\frac13\right)\left(\frac49\right) =\frac{16}{81},

and

P(1,5,5)=3(29)(49)2=32243.P(1,5,5)=3\left(\frac29\right) \left(\frac49\right)^2 =\frac{32}{243}.

Therefore,

P(B=4)= 16243+1681+32243= 3281.\begin{align*} P(B=4) =&\,\frac{16}{243}+\frac{16}{81}+\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) =&\,1-P(\text{no 1})-P(\text{no 5}) \\[4mm] &\,\hspace{2pt}+P(\text{only 2s}) \\[4mm] =&\,1-\left(\frac79\right)^3 -\left(\frac59\right)^3+\left(\frac13\right)^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) =&\,\left(\frac29\right)^3 +\left(\frac13\right)^3 +\left(\frac49\right)^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) =&\,3\left(\frac29\right)^2\left(\frac13\right) \\[4mm] &\,\hspace{2pt}+3\left(\frac29\right) \left(\frac13\right)^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) =&\,3\left(\frac13\right)^2\left(\frac49\right) \\[4mm] &\,\hspace{2pt}+3\left(\frac13\right) \left(\frac49\right)^2 \\[4mm] =&\,\frac{28}{81}. \end{align*}

Using part (i), P(B=4)=32/81P(B=4)=32/81. Therefore the sampling distribution is

b0134P(B=b)1181108128813281\begin{array}{c|cccc} b & 0 & 1 & 3 & 4 \\ \hline P(B=b) & \dfrac{11}{81} & \dfrac{10}{81} & \dfrac{28}{81} & \dfrac{32}{81} \end{array}

and the probabilities sum to

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