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IAL 2025 May Q4

A Level / Edexcel / S1

IAL 2025 May Paper · Question 4

题目

Problem

Two events CC and DD are such that

P(C∪D)=0.59,P(D)=0.45,P(C∣D)=0.2.P(C\cup D)=0.59,\qquad P(D)=0.45,\qquad P(C\mid D)=0.2.

Find the value of

(a) P(C∩D)P(C\cap D)

(2)

(b) P(C)P(C)

(2)

解答

(a)

解法一

思路

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条件概率公式是

P(C∣D)=P(C∩D)P(D).\begin{align*} P(C\mid D)=\frac{P(C\cap D)}{P(D)}. \end{align*}

题目给了 P(C∣D)P(C\mid D) 和 P(D)P(D)，所以直接代入求交集概率。

答题过程

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Using conditional probability,

P(C∣D)= P(C∩D)P(D).\begin{align*} P(C\mid D) =&\,\frac{P(C\cap D)}{P(D)}. \end{align*}

Substitute the given values:

0.2= P(C∩D)0.45P(C∩D)= 0.2(0.45)= 0.09.\begin{align*} 0.2 =&\,\frac{P(C\cap D)}{0.45}\\[3mm] P(C\cap D) =&\,0.2(0.45)\\[3mm] =&\,0.09. \end{align*}

(b)

解法一

思路

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两个事件的并集公式是

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

把 (a) 的结果代入即可。

答题过程

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Using

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

we have

0.59= P(C)+0.45−0.09P(C)= 0.59−0.45+0.09= 0.23.\begin{align*} 0.59 =&\,P(C)+0.45-0.09\\[3mm] P(C) =&\,0.59-0.45+0.09\\[3mm] =&\,0.23. \end{align*}