题目

Problem

A bag contains a large number of coloured counters. Each counter is labelled A, B or C

$30\%$ of the counters are labelled A

$45\%$ of the counters are labelled B

The rest of the counters are labelled C

It is known that

$2\%$ of the counters labelled A are red

$4\%$ of the counters labelled B are red

$6\%$ of the counters labelled C are red

One counter is selected at random from the bag.

(a) Complete the tree diagram on the opposite page to illustrate this information.

Tree diagram

(2)

(b) Calculate the probability that the counter is labelled A and is not red.

(2)

(d) Given that the counter is red, find the probability that it is labelled C

(3)

解答

(a)

解法一

思路

展开

第一层标签概率为 $0.30,0.45,0.25$ 。第二层每个标签下红色概率分别为 $0.02,0.04,0.06$ ，不红就是补到 $1$ 。

答题过程

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The missing label probabilities are

\begin{align*} P(B)=0.45,\qquad P(C)=1-0.30-0.45=0.25. \end{align*}

The missing conditional probabilities are

\begin{gathered} P(\text{not red}\mid A)=0.98,\\[3mm] P(\text{red}\mid B)=0.04,\qquad P(\text{not red}\mid B)=0.96,\\[3mm] P(\text{red}\mid C)=0.06,\qquad P(\text{not red}\mid C)=0.94. \end{gathered}

(b)

解法一

思路

展开

沿树图的 A then not red 路线相乘。

答题过程

展开

\begin{align*} P(A\cap \text{not red})=0.30(0.98)=0.294. \end{align*}

(c)

解法一

思路

展开

红色可以来自 A、B、C 三条路线，把三条路线的概率相加。

答题过程

展开

\begin{align*} P(\text{red}) =&\,0.30(0.02)+0.45(0.04)+0.25(0.06)\\[3mm] =&\,0.006+0.018+0.015\\[3mm] =&\,0.039. \end{align*}

(d)

解法一

思路

展开

这是条件概率：分子是“C 且 red”，分母是全部 red 的概率。

答题过程

展开

\begin{align*} P(C\mid \text{red}) =&\,\frac{P(C\cap \text{red})}{P(\text{red})}\\[3mm] =&\,\frac{0.25(0.06)}{0.039}\\[3mm] =&\,\frac{0.015}{0.039}\\[3mm] =&\,\frac{5}{13}. \end{align*}

So the probability is

\begin{align*} 0.385 \end{align*}

to $3$ significant figures.

IAL 2023 June Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

(d)

解法一

思路

答题过程