题目

Problem

Three events $A$ , $B$ and $C$ are such that

P(A)=0.1\qquad P(B\mid A)=0.3\qquad P(A\cup B)=0.25\qquad P(C)=0.5

Given that $A$ and $C$ are mutually exclusive

(a) find $P(A\cup C)$

(1)

(b) Show that $P(B)=0.18$

(3)

Given also that $B$ and $C$ are independent,

(5)

解答

(a)

解法一

思路

展开

$A$ 和 $C$ mutually exclusive，所以交集为 $0$ ，联合概率直接相加。

答题过程

展开

\begin{align*} P(A\cup C)=P(A)+P(C)=0.1+0.5=0.6. \end{align*}

(b)

解法一

思路

展开

先由条件概率求 $P(A\cap B)$ ，再代入加法公式求 $P(B)$ 。

答题过程

展开

Using

\begin{align*} P(B\mid A)=\frac{P(A\cap B)}{P(A)}, \end{align*}

we get

\begin{align*} P(A\cap B)=0.3(0.1)=0.03. \end{align*}

Now

\begin{align*} P(A\cup B)=P(A)+P(B)-P(A\cap B). \end{align*}

\begin{align*} 0.25=&\,0.1+P(B)-0.03\\[3mm] P(B)=&\,0.18. \end{align*}

(c)

解法一

思路

展开

先放 $A\cap B=0.03$ 。因为 $A$ 与 $C$ 互斥，所以 $A\cap C=0$ ，三者交集也为 $0$ 。再用 $B$ 与 $C$ 独立求 $B\cap C$ 。

答题过程

展开

Since $B$ and $C$ are independent,

\begin{align*} P(B\cap C)=P(B)P(C)=0.18(0.5)=0.09. \end{align*}

Also,

\begin{align*} P(A\cap B)=0.03. \end{align*}

Since $A$ and $C$ are mutually exclusive,

\begin{align*} P(A\cap C)=0,\qquad P(A\cap B\cap C)=0. \end{align*}

The $A$ only region is

\begin{align*} 0.10-0.03=0.07. \end{align*}

The $B$ only region is

\begin{align*} 0.18-0.03-0.09=0.06. \end{align*}

The $C$ only region is

\begin{align*} 0.50-0.09=0.41. \end{align*}

The outside region is

\begin{align*} 1&-(0.07+0.03+0.06+0.09+0.41)\\[3mm] =&\,0.34. \end{align*}

So the Venn diagram should contain the regions

\begin{gathered} A\text{ only}=0.07,\quad A\cap B=0.03,\quad B\text{ only}=0.06,\\[3mm] B\cap C=0.09,\quad C\text{ only}=0.41,\quad \text{outside}=0.34, \end{gathered}

with the $A\cap C$ regions equal to $0$ .

IAL 2023 June Q6

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