CalcGospel 國際數學圖譜
返回

IAL 2022 June Q4

A Level / Edexcel / S1

IAL 2022 June Paper · Question 4

题目

Problem

The events HH and WW are such that

P(H)=38P(H∪W)=34P(H)=\frac{3}{8}\qquad P(H\cup W)=\frac{3}{4}

Given that HH and WW are independent,

(a) show that P(W)=35P(W)=\dfrac{3}{5}

(4)

The event NN is such that

P(N)=115P(H∩N)=P(N)P(N)=\frac{1}{15}\qquad P(H\cap N)=P(N)

(b) Find P(N′∣H)P(N'\mid H)

(2)

Given that WW and NN are mutually exclusive,

(c) draw a Venn diagram to represent the events HH, WW and NN giving the exact probabilities of each region in the Venn diagram.

(5)

解答

(a)

解法一

思路

展开

用加法公式，并利用 independent 得 P(H∩W)=P(H)P(W)P(H\cap W)=P(H)P(W)。

答题过程

展开 P(H∪W)=P(H)+P(W)−P(H∩W).\begin{align*} P(H\cup W)=P(H)+P(W)-P(H\cap W). \end{align*}

Since HH and WW are independent,

P(H∩W)=P(H)P(W)=38P(W).\begin{align*} P(H\cap W)=P(H)P(W)=\frac{3}{8}P(W). \end{align*}

Therefore

34= 38+P(W)−38P(W)38= 58P(W)P(W)= 35.\begin{align*} \frac{3}{4} =&\,\frac{3}{8}+P(W)-\frac{3}{8}P(W)\\[3mm] \frac{3}{8} =&\,\frac{5}{8}P(W)\\[3mm] P(W)=&\,\frac{3}{5}. \end{align*}

(b)

解法一

思路

展开

P(H∩N)=P(N)P(H\cap N)=P(N) 表示 NN 完全在 HH 内。因此 P(N′∩H)=P(H)−P(N)P(N'\cap H)=P(H)-P(N)。

答题过程

展开 P(N′∣H)= P(N′∩H)P(H)= P(H)−P(N)P(H)= 38−11538= 3745.\begin{align*} P(N'\mid H) =&\,\frac{P(N'\cap H)}{P(H)}\\[3mm] =&\,\frac{P(H)-P(N)}{P(H)}\\[3mm] =&\,\frac{\frac{3}{8}-\frac{1}{15}}{\frac{3}{8}}\\[3mm] =&\,\frac{37}{45}. \end{align*}

(c)

解法一

思路

展开

先求 H∩W=38⋅35=940H\cap W=\dfrac38\cdot\dfrac35=\dfrac{9}{40}。又因为 NN 在 HH 内且 WW 与 NN 互斥，所以 NN 放在 HH 中但不与 WW 重叠。

答题过程

展开

We have

P(H∩W)=38⋅35=940.\begin{align*} P(H\cap W)=\frac{3}{8}\cdot\frac{3}{5}=\frac{9}{40}. \end{align*}

Since P(H∩N)=P(N)P(H\cap N)=P(N),

N⊂H.\begin{align*} N\subset H. \end{align*}

Also,

P(N)=115.\begin{align*} P(N)=\frac{1}{15}. \end{align*}

The HH only region is

38−940−115=112.\frac{3}{8}-\frac{9}{40}-\frac{1}{15} =\frac{1}{12}.

The WW only region is

35−940=38.\frac{3}{5}-\frac{9}{40} =\frac{3}{8}.

The outside region is

1−(112+115+940+38)= 14.\begin{align*} 1&-\left(\frac{1}{12}+\frac{1}{15} +\frac{9}{40}+\frac{3}{8}\right)\\[3mm] =&\,\frac{1}{4}. \end{align*}

So the regions are

H only=112,N=115,H∩W=940,W only=38,outside=14,\begin{gathered} H\text{ only}=\frac{1}{12},\quad N=\frac{1}{15},\quad H\cap W=\frac{9}{40},\\[3mm] W\text{ only}=\frac{3}{8},\quad \text{outside}=\frac{1}{4}, \end{gathered}

with W∩N=0W\cap N=0.