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IAL 2025 Oct S2 A Q6

A Level / Edexcel / S2

IAL 2025 Oct A Paper · Question 6

题目

Problem

A bag contains a large number of coins. It contains only 1p, 5p and 10p coins.

The fraction of 1p coins in the bag is qq, the fraction of 5p coins in the bag is rr and the fraction of 10p coins in the bag is ss.

Two coins are selected at random from the bag and the coin with the highest value is recorded. Let MM represent the value of the highest coin.

The sampling distribution of MM is given below

mm1510
P(M=m)P(M = m)125\dfrac{1}{25}1380\dfrac{13}{80}319400\dfrac{319}{400}

(a) List all the possible samples of two coins which may be selected.

(2)

(b) Find the value of qq, the value of rr and the value of ss.

(7)

(Total for Question 6 is 9 marks)

题目中文翻译

一个袋子中装有大量硬币。只有 1p、5p 和 10p 硬币。

袋子中 1p 硬币的比例为 qq，5p 硬币的比例为 rr，10p 硬币的比例为 ss。

从袋子中随机选择两枚硬币，记录面值较大的硬币。令 MM 表示较大硬币的面值。

MM 的抽样分布如下所示

mm1510
P(M=m)P(M = m)125\dfrac{1}{25}1380\dfrac{13}{80}319400\dfrac{319}{400}

(a) 列出可能被选中的两枚硬币的所有可能样本。

(b) 求 qq 的值、rr 的值和 ss 的值。

（第 6 题共 9 分）

解答

(a)

解法一

思路

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每次抽取的面值可以是 1、5 或 10。系统列出两枚硬币的六种面值组合；对于两枚面值不同的组合，还要注明抽取时可以出现两种次序。

答题过程

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The possible samples are

(1,1), (1,5) [×2], (5,5), (1,10) [×2], (5,10) [×2], (10,10),\boxed{(1,1),\ (1,5)\ [\times2],\ (5,5),\ (1,10)\ [\times2],\ (5,10)\ [\times2],\ (10,10)},

where [×2][\times2] indicates that the two different coins may be selected in either order.

(b)

解法一

思路

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按记录到的最大面值依次建式。最大值为 1 时，两枚都必须是 1p，可先求 qq；最大值为 5 时，可能是 (1,5)(1,5)、(5,1)(5,1) 或 (5,5)(5,5)，由此求 rr；最后利用三类硬币的比例和为 1 求 ss。

答题过程

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For M=1M=1, both selected coins must be 1p coins. Thus,

q2=125.q^2=\frac1{25}.

Since qq is a proportion,

q=15.q=\boxed{\frac15}.

For M=5M=5, the possible ordered samples are (1,5)(1,5), (5,1)(5,1) and (5,5)(5,5). Hence,

2qr+r2=1380.2qr+r^2=\frac{13}{80}.

Substituting q=1/5q=1/5 gives

r2+25r−1380= 080r2+32r−13= 0(4r−1)(20r+13)= 0.\begin{align*} r^2+\frac25r-\frac{13}{80}=&\,0\\ 80r^2+32r-13=&\,0\\ (4r-1)(20r+13)=&\,0. \end{align*}

Since rr is a proportion, the negative root is rejected, so

r=14.r=\boxed{\frac14}.

Finally,

s= 1−q−r= 1−15−14= 1120.\begin{align*} s =&\,1-q-r\\ =&\,1-\frac15-\frac14\\ =&\,\boxed{\frac{11}{20}}. \end{align*}

解法二

思路

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在求得 qq 和 rr 后，也可直接使用题目给出的 P(M=10)P(M=10) 来求 ss。最大值为 10 的样本包括一枚或两枚 10p 硬币，因此概率为 2qs+2rs+s22qs+2rs+s^2。

答题过程

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Using q=1/5q=1/5 and r=1/4r=1/4,

2qs+2rs+s2=319400.2qs+2rs+s^2=\frac{319}{400}.

Therefore,

s2+2(15+14)s−319400= 0400s2+360s−319= 0(20s−11)(20s+29)= 0.\begin{align*} s^2+2\left(\frac15+\frac14\right)s-\frac{319}{400}=&\,0\\ 400s^2+360s-319=&\,0\\ (20s-11)(20s+29)=&\,0. \end{align*}

Since ss is a proportion, the negative root is rejected. Hence,

s=1120.s=\boxed{\frac{11}{20}}.