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IAL 2023 Jan Q4

A Level / Edexcel / S1

IAL 2023 Jan Paper · Question 4

题目

Problem

(i) In the Venn diagram below, AA and BB represent events and pp, qq, rr and ss are probabilities.

Venn diagram
P(A)=725P(B)=15P((A∩B′)∪(A′∩B))=825P(A)=\frac{7}{25}\qquad P(B)=\frac{1}{5}\qquad P\left((A\cap B')\cup(A'\cap B)\right)=\frac{8}{25}

(a) Use algebra to show that 2p+2q+2r=452p+2q+2r=\dfrac{4}{5}

(4)

(b) Find the value of pp, the value of qq, the value of rr and the value of ss

(5)

(ii) Two events, CC and DD, are such that

P(C)=xx+5P(D)=5xP(C)=\frac{x}{x+5}\qquad P(D)=\frac{5}{x}

where xx is a positive constant.

By considering P(C)+P(D)P(C)+P(D) show that CC and DD cannot be mutually exclusive.

(4)

解答

(i)(a)

解法一

思路

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从 Venn 图可得 p+q=P(A)p+q=P(A)，q+r=P(B)q+r=P(B)，而“只在其中一个事件中”是 p+rp+r。把三式相加就会出现 2p+2q+2r2p+2q+2r。

答题过程

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From the Venn diagram,

p+q=725.\begin{align*} p+q=\frac{7}{25}. \end{align*}

Also,

q+r=15.\begin{align*} q+r=\frac{1}{5}. \end{align*}

And

p+r=825.\begin{align*} p+r=\frac{8}{25}. \end{align*}

Adding these three equations,

(p+q)+(q+r)+(p+r)= 725+15+8252p+2q+2r= 725+525+825= 2025= 45.\begin{align*} (p+q)+(q+r)+(p+r) =&\,\frac{7}{25}+\frac{1}{5}+\frac{8}{25}\\[3mm] 2p+2q+2r =&\,\frac{7}{25}+\frac{5}{25}+\frac{8}{25}\\[3mm] =&\,\frac{20}{25}\\[3mm] =&\,\frac{4}{5}. \end{align*}

(i)(b)

解法一

思路

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有三条关于 p,q,rp,q,r 的方程，先由上一小题得到 p+q+r=25p+q+r=\dfrac{2}{5}，再逐个相减。

答题过程

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From part (a),

p+q+r=25.\begin{align*} p+q+r=\frac{2}{5}. \end{align*}

Since

q+r=15,\begin{align*} q+r=\frac{1}{5}, \end{align*}

we have

p=25−15=15.\begin{align*} p=\frac{2}{5}-\frac{1}{5}=\frac{1}{5}. \end{align*}

Since

p+r=825,\begin{align*} p+r=\frac{8}{25}, \end{align*}

we have

r=825−15=325.\begin{align*} r=\frac{8}{25}-\frac{1}{5}=\frac{3}{25}. \end{align*}

Then

q=15−325=225.\begin{align*} q=\frac{1}{5}-\frac{3}{25}=\frac{2}{25}. \end{align*}

Finally,

s=1−(p+q+r)=1−25=35.\begin{align*} s=1-(p+q+r)=1-\frac{2}{5}=\frac{3}{5}. \end{align*}

(ii)

解法一

思路

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如果两个事件互斥，则 P(C∪D)=P(C)+P(D)P(C\cup D)=P(C)+P(D)，而联合概率不能超过 11。所以只要证明 P(C)+P(D)>1P(C)+P(D)>1，就可说明它们不可能互斥。

答题过程

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Consider

P(C)+P(D)= xx+5+5x= x2+5(x+5)x(x+5)= x2+5x+25x2+5x= 1+25x2+5x.\begin{align*} P(C)+P(D) =&\,\frac{x}{x+5}+\frac{5}{x}\\[3mm] =&\,\frac{x^2+5(x+5)}{x(x+5)}\\[3mm] =&\,\frac{x^2+5x+25}{x^2+5x}\\[3mm] =&\,1+\frac{25}{x^2+5x}. \end{align*}

Since x>0x>0,

25x2+5x>0.\begin{align*} \frac{25}{x^2+5x}>0. \end{align*}

Therefore

P(C)+P(D)>1.\begin{align*} P(C)+P(D)>1. \end{align*}

If CC and DD were mutually exclusive, then P(C∪D)=P(C)+P(D)>1P(C\cup D)=P(C)+P(D)>1, which is impossible. Hence CC and DD cannot be mutually exclusive.