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CIE 9709 2025 March Paper 52 Q4

A Level / CIE / S1

CIE 9709 2025 March Paper 52 Paper · Question 4

题目

Problem

Eddie has 16 toy cars, of which 8 are white, 5 are black and 3 are silver. He places all the cars in a bag and selects three of them at random, without replacement.

(a) Find the probability that all three cars are the same colour.

(3)

(b) Find the probability that, when the 3 cars are selected, at least one car is white and at least one car is black.

(4)
题目中文翻译

Eddie 有 16 辆玩具车,其中 8 辆是白色的,5 辆是黑色的,3 辆是银色的。他将所有玩具车放入一个袋子,从中随机不放回地抽取 3 辆。

(a) 求这 3 辆车颜色全部相同的概率。

(3)

(b) 求所抽取的 3 辆车中至少有一辆白色车且至少有一辆黑色车的概率。

(4)
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

展开

同色可以是全白、全黑、全银三种。分别从各颜色中选3,再除以全部3车选法。

答题过程

展开

There are (163)=560\binom{16}{3}=560 equally likely selections. Hence

P(all same colour)= (83)+(53)+(33)(163)= 56+10+1560= 67560.\begin{align*} \mathrm{P}(\text{all same colour}) =&\,\frac{\binom83+\binom53+\binom33}{\binom{16}{3}}\\[4mm] =&\,\frac{56+10+1}{560}\\[4mm] =&\,\boxed{\frac{67}{560}}. \end{align*}

解法二

思路

展开

也可以按不放回抽取次序,分别算白白白、黑黑黑、银银银三条路径。

答题过程

展开 P= 816715614+516415314 +316215114= 336+60+63360= 67560.\begin{align*} \mathrm{P}=&\,\frac8{16}\frac7{15}\frac6{14} +\frac5{16}\frac4{15}\frac3{14}\\[4mm] &\,\hspace{2pt}+\frac3{16}\frac2{15}\frac1{14}\\[4mm] =&\,\frac{336+60+6}{3360}\\[4mm] =&\,\boxed{\frac{67}{560}}. \end{align*}

(b)

解法一

思路

展开

三车中至少一白一黑,颜色构成只有两白一黑、一白两黑、一白一黑一银。按三类组合数相加。

答题过程

展开

The permitted colour compositions are WWB, WBB and WBS. Thus

P= (82)(51)+(81)(52)(163) +(81)(51)(31)(163)= 140+80+120560= 1728.\begin{align*} \mathrm{P} =&\,\frac{\binom82\binom51+\binom81\binom52}{\binom{16}{3}}\\[4mm] &\,\hspace{2pt}+\frac{\binom81\binom51\binom31}{\binom{16}{3}}\\[4mm] =&\,\frac{140+80+120}{560}\\[4mm] =&\,\boxed{\frac{17}{28}}. \end{align*}

解法二

思路

展开

另一条路线按有序抽取,前两种构成各有3种颜色次序,三色各一则有6种。

答题过程

展开 P= 3816715514+3816515414 +6816515314= 840+480+7203360= 1728.\begin{align*} \mathrm{P} =&\,3\frac8{16}\frac7{15}\frac5{14} +3\frac8{16}\frac5{15}\frac4{14}\\[4mm] &\,\hspace{2pt}+6\frac8{16}\frac5{15}\frac3{14}\\[4mm] =&\,\frac{840+480+720}{3360}\\[4mm] =&\,\boxed{\frac{17}{28}}. \end{align*}