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S1 2021 June Q1

A Level / Edexcel / S1

IAL 2021 June Paper · Question 1

题目

Problem

There are 7 red counters, 3 blue counters and 2 yellow counters in a bag. Gina selects a counter at random from the bag and keeps it. If the counter is yellow she does not select any more counters. If the counter is not yellow she randomly selects a second counter from the bag.

(a) Complete the tree diagram.

(2)

Given that Gina has selected a yellow counter,

(b) find the probability that she has 2 counters.

(3)
题目中文翻译

一个袋子里有 7 个红色计数片、3 个蓝色计数片和 2 个黄色计数片。Gina 从袋中随机取出一个计数片并保留它。如果取出的是黄色计数片，她不再取出其他计数片；如果取出的不是黄色计数片，她再从袋中随机取出第二个计数片。

(a) 补全树状图。

已知 Gina 取到了一个黄色计数片，

(b) 求她取到 2 个计数片的概率。

解答

(a)

解法一

思路

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第一抽后如果不是 yellow，袋中会剩下 11 个 counters。第二抽的分母因此都是 11，并且要根据第一抽拿走了哪种颜色来调整分子。

答题过程

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For the first counter,

P(R)=712,P(B)=312,P(Y)=212.\begin{align*} P(R)=\frac{7}{12},\quad P(B)=\frac{3}{12},\quad P(Y)=\frac{2}{12}. \end{align*}

If the first counter is red, then the remaining counters are 6R,3B,2Y6R,3B,2Y, so

P(R∣R first)=611,P(B∣R first)=311,P(Y∣R first)=211.P(R\mid R\text{ first})=\frac{6}{11},\quad P(B\mid R\text{ first})=\frac{3}{11},\quad P(Y\mid R\text{ first})=\frac{2}{11}.

If the first counter is blue, then the remaining counters are 7R,2B,2Y7R,2B,2Y, so

P(R∣B first)=711,P(B∣B first)=211,P(Y∣B first)=211.P(R\mid B\text{ first})=\frac{7}{11},\quad P(B\mid B\text{ first})=\frac{2}{11},\quad P(Y\mid B\text{ first})=\frac{2}{11}.

The completed tree diagram is shown below.

(b)

解法一

思路

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题目已经告诉我们“Gina selected a yellow counter”，所以这是条件概率。分母是所有出现 yellow 的情况；分子是“有两个 counters 且其中第二个是 yellow”的情况。

答题过程

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The probability that Gina selects a yellow counter is

P(Y)= 212+712⋅211+312⋅211= 22132+14132+6132= 42132.\begin{align*} P(Y) =&\,\frac{2}{12} +\frac{7}{12}\cdot\frac{2}{11} +\frac{3}{12}\cdot\frac{2}{11}\\ =&\,\frac{22}{132}+\frac{14}{132}+\frac{6}{132}\\ =&\,\frac{42}{132}. \end{align*}

The probability that Gina selects a yellow counter and has 2 counters is

P(Y and 2 counters)= 712⋅211+312⋅211= 14132+6132= 20132.\begin{align*} P(Y\text{ and 2 counters}) =&\,\frac{7}{12}\cdot\frac{2}{11} +\frac{3}{12}\cdot\frac{2}{11}\\ =&\,\frac{14}{132}+\frac{6}{132}\\ =&\,\frac{20}{132}. \end{align*}

Therefore

P(2 counters∣Y)=2013242132=2042=1021.P(\text{2 counters}\mid Y) =\frac{\frac{20}{132}}{\frac{42}{132}} =\frac{20}{42} =\frac{10}{21}.

解法二

思路

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也可以先用补集求“至少取到一个黄色计数片”的概率。排除两次均不是黄色的路径，再与“第二次取到黄色”的概率组成条件概率。

答题过程

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The probability of selecting no yellow counter is

P(no yellow)= 712⋅911+312⋅911= 90132.\begin{align*} P(\text{no yellow}) =&\,\frac{7}{12}\cdot\frac{9}{11} +\frac{3}{12}\cdot\frac{9}{11}\\ =&\,\frac{90}{132}. \end{align*}

Hence

P(Y)= 1−90132= 42132.\begin{align*} P(Y) =&\,1-\frac{90}{132}\\ =&\,\frac{42}{132}. \end{align*}

Also,

P(Y and 2 counters)= 712⋅211+312⋅211= 20132.\begin{align*} P(Y\text{ and 2 counters}) =&\,\frac{7}{12}\cdot\frac{2}{11} +\frac{3}{12}\cdot\frac{2}{11}\\ =&\,\frac{20}{132}. \end{align*}

Therefore

P(2 counters∣Y)= 20/13242/132= 1021.\begin{align*} P(\text{2 counters}\mid Y) =&\,\frac{20/132}{42/132}\\ =&\,\frac{10}{21}. \end{align*}