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IAL 2024 Jan Q6

A Level / Edexcel / S1

IAL 2024 Jan Paper · Question 6

题目

Problem

The events AA and BB satisfy

P(A)=x,P(B)=y,P(A∪B)=0.65,P(B∣A)=0.3.P(A)=x,\qquad P(B)=y,\qquad P(A\cup B)=0.65,\qquad P(B\mid A)=0.3.

(a) Show that

14x+20y=13.14x+20y=13.
(3)

The events BB and CC are mutually exclusive such that

P(B∪C)=0.85,P(C)=12x+y.P(B\cup C)=0.85,\qquad P(C)=\frac12x+y.

(b) (i) Find a second equation in xx and yy.

(ii) Hence find the value of xx and the value of yy.

(4)

(c) Determine whether or not AA and BB are statistically independent. You must show your working clearly.

(2)

解答

(a)

解法一

思路

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由条件概率得

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

再代入加法公式

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

答题过程

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Since

P(B∣A)=0.3,\begin{align*} P(B\mid A)=0.3, \end{align*}

we have

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

So

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

Using

P(A∪B)=P(A)+P(B)−P(A∩B),\begin{align*} P(A\cup B)=P(A)+P(B)-P(A\cap B), \end{align*}

we get

0.65= x+y−0.3x0.65= 0.7x+y.\begin{align*} 0.65=&\,x+y-0.3x\\[3mm] 0.65=&\,0.7x+y. \end{align*}

Multiplying by 2020,

13=14x+20y.\begin{align*} 13=14x+20y. \end{align*}

Therefore

14x+20y=13.\begin{align*} 14x+20y=13. \end{align*}

(b)(i)

解法一

思路

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BB 和 CC 互斥，所以

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

答题过程

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Since BB and CC are mutually exclusive,

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

Therefore

0.85= y+(12x+y)0.85= 12x+2y.\begin{align*} 0.85=&\,y+\left(\frac12x+y\right)\\[3mm] 0.85=&\,\frac12x+2y. \end{align*}

(b)(ii)

解法一

思路

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联立

14x+20y=13\begin{align*} 14x+20y=13 \end{align*}

和

12x+2y=0.85.\begin{align*} \frac12x+2y=0.85. \end{align*}

答题过程

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From part (b)(i),

12x+2y=0.85.\begin{align*} \frac12x+2y=0.85. \end{align*}

Multiplying by 2020,

10x+40y=17.\begin{align*} 10x+40y=17. \end{align*}

Also,

14x+20y=13.\begin{align*} 14x+20y=13. \end{align*}

Double the second equation:

28x+40y=26.\begin{align*} 28x+40y=26. \end{align*}

Subtracting 10x+40y=1710x+40y=17 gives

18x= 9x= 0.5.\begin{align*} 18x=&\,9\\[3mm] x=&\,0.5. \end{align*}

Substituting into 12x+2y=0.85\frac12x+2y=0.85,

0.25+2y= 0.852y= 0.60y= 0.30.\begin{align*} 0.25+2y=&\,0.85\\[3mm] 2y=&\,0.60\\[3mm] y=&\,0.30. \end{align*}

Therefore

x=0.5,y=0.3.\begin{align*} x=0.5,\qquad y=0.3. \end{align*}

(c)

解法一

思路

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判断独立性可比较 P(B∣A)P(B\mid A) 和 P(B)P(B)。这里 P(B∣A)=0.3P(B\mid A)=0.3，而 P(B)=y=0.3P(B)=y=0.3，相等，所以独立。

答题过程

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We have

P(B∣A)=0.3\begin{align*} P(B\mid A)=0.3 \end{align*}

and

P(B)=y=0.3.\begin{align*} P(B)=y=0.3. \end{align*}

Since

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

the events AA and BB are statistically independent.