Skip to content
CalcGospel 國際數學圖譜
返回

IAL 2021 June Q2

A Level / Edexcel / S1

IAL 2021 June Paper · Question 2

题目

Problem

In the Venn diagram below, AA, BB and CC are events and pp, qq, rr and ss are probabilities. The events AA and CC are independent and P(A)=0.65P(A)=0.65

(a) State which two of the events AA, BB and CC are mutually exclusive.

(1)

(b) Find the value of rr and the value of ss.

(5)

The events (AC)(A\cap C') and (BC)(B\cup C) are also independent.

(c) Find the exact value of pp and the exact value of qq. Give your answers as fractions.

(6)

解答

(a)

解法一

思路

展开

互斥表示两个事件没有重叠区域。从图中看,BBCC 没有共同部分。

答题过程

展开

BB and CC are mutually exclusive.

(b)

解法一

思路

展开

图中 ACA\cap C 的概率是 0.130.13。由于 AACC 独立,P(AC)=P(A)P(C)P(A\cap C)=P(A)P(C),可以先求出 P(C)P(C),再求 rr。最后用全部概率加起来等于 1 求 ss

答题过程

展开

Since AA and CC are independent,

P(AC)=P(A)P(C).\begin{align*} P(A\cap C)=P(A)P(C). \end{align*}

Hence

0.13=0.65P(C),\begin{align*} 0.13=0.65P(C), \end{align*}

so

P(C)=0.130.65=0.2.\begin{align*} P(C)=\frac{0.13}{0.65}=0.2. \end{align*}

From the diagram,

P(C)=r+0.13,\begin{align*} P(C)=r+0.13, \end{align*}

so

r=0.20.13=0.07.\begin{align*} r=0.2-0.13=0.07. \end{align*}

Also,

P(A)+r+s=1.\begin{align*} P(A)+r+s=1. \end{align*}

Therefore

0.65+0.07+s=1,\begin{align*} 0.65+0.07+s=1, \end{align*}

and

s=0.28.\begin{align*} s=0.28. \end{align*}

(c)

解法一

思路

展开

先把两个独立事件的概率写出来。ACA\cap C' 就是 AA 中不在 CC 的部分,即 p+qp+q。由于 P(A)=0.65P(A)=0.65,而 AC=0.13A\cap C=0.13,所以 p+q=0.52p+q=0.52。然后用独立性建立方程。

答题过程

展开

Since

P(A)=p+q+0.13=0.65,\begin{align*} P(A)=p+q+0.13=0.65, \end{align*}

we have

p+q=0.52.\begin{align*} p+q=0.52. \end{align*}

Also,

P(BC)=q+r+0.13=q+0.07+0.13=q+0.2.\begin{align*} P(B\cup C)=q+r+0.13=q+0.07+0.13=q+0.2. \end{align*}

The intersection of (AC)(A\cap C') and (BC)(B\cup C) is the region qq, so

P((AC)(BC))=q.\begin{align*} P\bigl((A\cap C')\cap(B\cup C)\bigr)=q. \end{align*}

Using independence,

P(AC)P(BC)=q.\begin{align*} P(A\cap C')P(B\cup C)=q. \end{align*}

Therefore

0.52(q+0.2)=q.\begin{align*} 0.52(q+0.2)=q. \end{align*}

So

0.52q+0.104=q,\begin{align*} 0.52q+0.104=q, \end{align*}

and

0.48q=0.104.\begin{align*} 0.48q=0.104. \end{align*}

Hence

q=0.1040.48=1360.\begin{align*} q=\frac{0.104}{0.48}=\frac{13}{60}. \end{align*}

Since p+q=0.52=1325p+q=0.52=\frac{13}{25},

p=13251360=15630065300=91300.p=\frac{13}{25}-\frac{13}{60} =\frac{156}{300}-\frac{65}{300} =\frac{91}{300}.