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CIE 9709 2025 November Paper 52 Q4

A Level / CIE / S1

CIE 9709 2025 Nov Paper 52 Paper · Question 4

题目

Problem

Gio has a pack of 18 cards. Ivy has a pack of xx cards.

Each card has a picture of a bus or a car or a train. The number of cards with each picture in the two packs is shown in the table.

BusCarTrain
Gio’s pack6102
Ivy’s packx−12x-1293

One card is chosen at random from each pack. The probability that the two cards have pictures of buses on them is equal to twice the probability that the two cards have pictures of cars on them.

(a) Write down an equation in terms of xx and hence find the value of xx.

(4)

(b) Find the probability that the two cards have pictures of the same type of vehicle on them.

(3)
题目中文翻译

Gio 有一副 18 张的卡片，Ivy 有一副 xx 张的卡片。

每张卡片上都画有一辆公共汽车、一辆小汽车或一列火车。两副卡片中各类图案的张数如下表所示。

公共汽车小汽车火车
Gio 的卡片6102
Ivy 的卡片x−12x-1293

从每副卡片中各随机抽取一张。两张都画有公共汽车的概率等于两张都画有小汽车的概率的两倍。

(a) 写出一个关于 xx 的方程，并由此求出 xx 的值。

(4)

(b) 求两张卡片上画有同一种交通工具的概率。

(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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两副牌独立抽取，相同车型的联合概率为对应比例相乘。按“两辆巴士的概率为两辆轿车概率的两倍”列式，再消去共同分母。

答题过程

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The probabilities of two buses and two cars are respectively

P(BB)=618x−12x,P(CC)=10189x.\mathrm{P}(BB)=\frac6{18}\frac{x-12}{x}, \qquad \mathrm{P}(CC)=\frac{10}{18}\frac9x.

Thus the required equation is

618x−12x=210189x.\boxed{\frac6{18}\frac{x-12}{x} =2\frac{10}{18}\frac9x}.

Since xx is a positive number of cards, multiplying by 18x18x gives

6(x−12)= 180,x−12= 30,x= 42.\begin{align*} 6(x-12)=&\,180,\\[4mm] x-12=&\,30,\\[4mm] x=&\,\boxed{42}. \end{align*}

(b)

解法一

思路

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同车型分为巴士、轿车、火车三个互斥事件，使用(a)求得的42张牌计算。

答题过程

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With x=42x=42, Ivy has 30 bus cards, 9 car cards and 3 train cards. Therefore,

P(same vehicle)= 6183042+1018942+218342= 1042+542+1126= 2363.\begin{align*} \mathrm{P}(\text{same vehicle}) =&\,\frac6{18}\frac{30}{42} +\frac{10}{18}\frac9{42} +\frac2{18}\frac3{42}\\[4mm] =&\,\frac{10}{42}+\frac5{42}+\frac1{126}\\[4mm] =&\,\boxed{\frac{23}{63}}. \end{align*}