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S1 2024 Oct Q7

A Level / Edexcel / S1

IAL 2024 Oct Paper · Question 7

题目

Problem

A box contains only red counters and black counters. There are nn red counters and n+1n+1 black counters.

Two counters are selected at random, one at a time without replacement, from the box.

(a) Complete the tree diagram for this information. Give your probabilities in terms of nn where necessary.

(3)

(b) Show that the probability that the two counters selected are different colours is

n+12n+1.\frac{n+1}{2n+1}.
(2)

The probability that the two counters selected are different colours is 2549\dfrac{25}{49}.

(c) Find the total number of counters in the box before any counters were selected.

(2)

Given that the two counters selected are different colours,

(d) find the probability that the 11st counter is black. You must show your working.

(2)
题目中文翻译

一个盒子里只有红色和黑色计数片，其中有 nn 个红色计数片和 n+1n+1 个黑色计数片。

从盒子里逐个随机抽取两个计数片，抽后不放回。

(a) 完成表示这些信息的树状图；必要时用 nn 表示概率。

(b) 证明抽出的两个计数片颜色不同的概率为

n+12n+1.\frac{n+1}{2n+1}.

已知抽出的两个计数片颜色不同的概率为 2549\dfrac{25}{49}。

(c) 求抽取前盒子中计数片的总数。

已知抽出的两个计数片颜色不同，

(d) 求第一个抽出的计数片为黑色的概率。你必须写出计算过程。

解答

(a)

解法一

思路

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一开始共有

n+(n+1)=2n+1\begin{align*} n+(n+1)=2n+1 \end{align*}

个 counters。第一次抽完后，总数变成 2n2n，并且对应颜色数量会减少 11。

答题过程

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For the first counter,

P(R)= n2n+1,P(B)= n+12n+1.\begin{align*} P(R)=&\,\frac{n}{2n+1},\\ P(B)=&\,\frac{n+1}{2n+1}. \end{align*}

If the first counter is red, then there are n−1n-1 red counters and n+1n+1 black counters left, out of 2n2n counters. So

P(R∣R)= n−12n,P(B∣R)= n+12n.\begin{align*} P(R\mid R)=&\,\frac{n-1}{2n},\\ P(B\mid R)=&\,\frac{n+1}{2n}. \end{align*}

If the first counter is black, then there are nn red counters and nn black counters left, out of 2n2n counters. So

P(R∣B)= n2n,P(B∣B)= n2n.\begin{align*} P(R\mid B)=&\,\frac{n}{2n},\\ P(B\mid B)=&\,\frac{n}{2n}. \end{align*}

(b)

解法一

思路

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颜色不同有两种顺序：先红后黑，或先黑后红。分别沿树图相乘，再相加。

答题过程

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The counters are different colours if the outcomes are RBRB or BRBR.

Therefore

P(RB)+P(BR)= n2n+1⋅n+12n +n+12n+1⋅n2n= n(n+1)+n(n+1)2n(2n+1)= 2n(n+1)2n(2n+1)= n+12n+1.\begin{align*} P(RB)+P(BR) =&\,\frac{n}{2n+1}\cdot\frac{n+1}{2n}\\ &\,\hspace{2pt}+\frac{n+1}{2n+1}\cdot\frac{n}{2n}\\ =&\,\frac{n(n+1)+n(n+1)}{2n(2n+1)}\\ =&\,\frac{2n(n+1)}{2n(2n+1)}\\ =&\,\frac{n+1}{2n+1}. \end{align*}

(c)

解法一

思路

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把上一小题的表达式等于 2549\dfrac{25}{49}，解出 nn。题目问原来盒子中 counters 的总数，也就是 2n+12n+1。

答题过程

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We have

n+12n+1=2549.\begin{align*} \frac{n+1}{2n+1}=\frac{25}{49}. \end{align*}

Cross-multiplying,

49(n+1)= 25(2n+1)49n+49= 50n+25n= 24.\begin{align*} 49(n+1)=&\,25(2n+1)\\ 49n+49=&\,50n+25\\ n=&\,24. \end{align*}

The total number of counters is

2n+1=2(24)+1=49.\begin{align*} 2n+1=2(24)+1=49. \end{align*}

解法二

思路

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由于 gcd⁡(n+1,2n+1)=1\gcd(n+1,2n+1)=1，左边始终是最简分数。因此可直接与最简分数 2549\dfrac{25}{49} 对应分子与分母。

答题过程

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Since

gcd⁡(n+1,2n+1)=1,\begin{align*} \gcd(n+1,2n+1)=1, \end{align*}

the fraction on the left is in its simplest form. As 2549\dfrac{25}{49} is also in its simplest form,

n+1= 25,2n+1= 49.\begin{align*} n+1=&\,25,\\ 2n+1=&\,49. \end{align*}

So n=24n=24 and the total number of counters is 4949.

(d)

解法一

思路

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已知两次抽到不同颜色。要求“第一次是黑色”的条件概率，就是

P(BR)P(different colours).\begin{align*} \frac{P(BR)}{P(\text{different colours})}. \end{align*}

代入 n=24n=24 后算即可。

答题过程

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Using n=24n=24,

P(BR)=2549⋅2448.\begin{align*} P(BR)=\frac{25}{49}\cdot\frac{24}{48}. \end{align*}

Also,

P(different colours)=2549.\begin{align*} P(\text{different colours})=\frac{25}{49}. \end{align*}

Therefore

 P(first black∣different)= P(BR)÷P(different)= (2549⋅2448)÷2549= 2448= 12.\begin{align*} &\,P(\text{first black}\mid\text{different})\\ =&\,P(BR)\div P(\text{different})\\ =&\,\Big(\frac{25}{49}\cdot\frac{24}{48}\Big)\div\frac{25}{49}\\ =&\,\frac{24}{48}\\ =&\,\frac12. \end{align*}