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)

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,

\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 $6R,3B,2Y$ , so

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 $7R,2B,2Y$ , so

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}.

(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

\begin{aligned} 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{aligned}

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

\begin{aligned} 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{aligned}

Therefore

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

IAL 2021 June Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程