题目

Problem

The Venn diagram shows the probability of a randomly selected student from a school being in the sets $L$ , $B$ and $C$ , where

$L$ represents the event that the student has instrumental music lessons

$B$ represents the event that the student plays in the school band

$C$ represents the event that the student sings in the school choir

$p$ , $q$ , $r$ and $s$ are probabilities.

(a) Select a pair of mutually exclusive events from $L$ , $B$ and $C$ .

(1)

Given that $P(L)=0.4$ , $P(B)=0.13$ , $P(C)=0.3$ and the events $L$ and $C$ are independent,

(b) find the value of $p$ ,

(2)

(3)

A student is selected at random from those who play in the school band or sing in the school choir.

(d) Find the exact probability that this student has instrumental music lessons.

(3)

解答

(a)

解法一

思路

展开

从图中看，band 和 choir 没有重叠区域。

答题过程

展开

$B$ and $C$ are mutually exclusive.

(b)

解法一

思路

展开

独立事件满足 $P(L\cap C)=P(L)P(C)$ ，而图中 $L\cap C$ 就是 $p$ 。

答题过程

展开

Since $L$ and $C$ are independent,

\begin{align*} p=P(L\cap C)=P(L)P(C)=0.4(0.3)=0.12. \end{align*}

(c)

解法一

思路

展开

分别用 $P(L)$ 、 $P(C)$ 和总概率为 1 来求剩下区域。

答题过程

展开

Since $P(L)=0.4$ ,

\begin{align*} q+0.13+p=0.4. \end{align*}

\begin{align*} q=0.4-0.13-0.12=0.15. \end{align*}

Since $P(C)=0.3$ ,

\begin{align*} r+p=0.3, \end{align*}

\begin{align*} r=0.3-0.12=0.18. \end{align*}

Finally,

\begin{align*} s=1-(0.15+0.13+0.12+0.18)=0.42. \end{align*}

(d)

解法一

思路

展开

条件是 $B\cup C$ ，分母是 band 或 choir 的总概率。分子是同时在 $L$ 且属于 $B\cup C$ ，即 $0.13+p$ 。

答题过程

展开

\begin{align*} P(L\mid B\cup C)=\frac{P(L\cap(B\cup C))}{P(B\cup C)}. \end{align*}

The numerator is

\begin{align*} P(L\cap(B\cup C))=0.13+0.12=0.25. \end{align*}

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

\begin{align*} P(B\cup C)=0.13+0.30=0.43. \end{align*}

Therefore

\begin{align*} P(L\mid B\cup C)=\frac{0.25}{0.43}=\frac{25}{43}. \end{align*}

IAL 2019 Jan 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

(c)

解法一

思路

答题过程

(d)

解法一

思路

答题过程