题目

Problem

A club has 12 members of which 7 are men and 5 are women. A committee of 4 members is to be chosen from these 12 members.

Find the probability that a randomly chosen committee includes at least 2 men.

(4)

题目中文翻译

某俱乐部有 12 名成员，其中 7 名为男性，5 名为女性。现要从这 12 名成员中选出 4 人组成委员会。

求随机选出的委员会包含至少 2 名男性的概率。

(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

解法一

思路

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至少两位男生即2、3、4位，分别用组合数计数，再除以所有等可能的委员会数。

答题过程

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There are $\binom{12}{4}=495$ equally likely committees. The numbers with 2, 3 and 4 men are respectively

\begin{align*} \binom72\binom52=&\,210,\\[4mm] \binom73\binom51=&\,175,\\[4mm] \binom74\binom50=&\,35. \end{align*}

Hence

\mathrm{P}(\text{at least 2 men}) =\frac{210+175+35}{495} =\boxed{\frac{28}{33}}.

解法二

思路

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也可以取补集：不满足条件的委员会只可能有0或1位男生，所需计数更少。

答题过程

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The committees failing the condition contain either no men or exactly one man. Thus

\begin{align*} \mathrm{P}(\text{at least 2 men}) =&\,1-\frac{\binom70\binom54+\binom71\binom53} {\binom{12}{4}}\\[4mm] =&\,1-\frac{5+70}{495}\\[4mm] =&\,\boxed{\frac{28}{33}}. \end{align*}

解法三

思路

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另一条路线是想象按次序不放回抽4人。对每种男女构成，乘上其中男性可占的位置数，再将互斥情况相加。

答题过程

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Imagine selecting the four committee members in order, without replacement. The probabilities of 2, 3 and 4 men are

\begin{align*} P_2=&\,\binom42\frac7{12}\frac6{11}\frac5{10}\frac49 =\frac{5040}{11880},\\[4mm] P_3=&\,\binom43\frac7{12}\frac6{11}\frac5{10}\frac59 =\frac{4200}{11880},\\[4mm] P_4=&\,\frac7{12}\frac6{11}\frac5{10}\frac49 =\frac{840}{11880}. \end{align*}

Hence

P_2+P_3+P_4=\frac{10080}{11880}=\boxed{\frac{28}{33}}.

CIE 9709 2026 March Paper 52 Q1

题目

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解法一

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解法二

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解法三

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