题目

Problem

For her morning break, Jules has a drink. She has tea, juice or water, with probabilities 0.2, 0.35 and 0.45 respectively.

When she has tea, the probability that she has a cake is 0.25.

When she has juice, the probability that she has a cake is 0.4.

When she has water, the probability that she has a cake is $x$ .

On a randomly chosen day, the probability that Jules has a cake is 0.505.

(a) Find the value of $x$ .

(2)

(b) Find the probability that Jules has juice given that she does not have a cake.

(3)

题目中文翻译

Jules 在上午休息时会喝一种饮品。她喝茶、果汁或水的概率分别为 0.2、0.35 和 0.45。

当她喝茶时，她吃蛋糕的概率为 0.25。

当她喝果汁时，她吃蛋糕的概率为 0.4。

当她喝水时，她吃蛋糕的概率为 $x$ 。

随机选取一天，Jules 吃蛋糕的概率为 0.505。

(a) 求 $x$ 的值。

(2)

(b) 已知 Jules 没有吃蛋糕，求她喝果汁的概率。

(3)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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分别计算饮茶、果汁和水且吃蛋糕的概率，三个互斥分支相加等于吃蛋糕的总概率。

答题过程

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Let $C$ denote having a cake. By conditioning on the drink,

\begin{align*} \mathrm{P}(C)=&\,0.2(0.25)+0.35(0.4)+0.45x\\[4mm] 0.505=&\,0.05+0.14+0.45x,\\[4mm] 0.45x=&\,0.315,\\[4mm] x=&\,\boxed{0.7}. \end{align*}

(b)

解法一

思路

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分子是喝果汁且不吃蛋糕，分母是所有不吃蛋糕的概率。果汁分支中不吃蛋糕的条件概率为0.6。

答题过程

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Let $J$ denote having juice. Then

\mathrm{P}(J\cap C')=0.35(1-0.4)=0.21,

and $\mathrm{P}(C')=1-0.505=0.495$ . Therefore,

\mathrm{P}(J\mid C')=\frac{0.21}{0.495} =\boxed{\frac{14}{33}}.

CIE 9709 2026 June Paper 55 Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程