中奖与抽奖次序无关

前言

典例剖析

【人教 2019A 版教材\(P_{262}\) 页习题10.3 第 6 题改编】在一个袋子中放 \(6\) 个白球，\(4\) 个红球，摇匀后随机摸球 \(3\) 次，采用放回和不放回两种方式摸球 . 设事件 \(A_{i}\)=“第 \(i\) 次摸到红球”，\(i=1,2,3\) .

(1). 在两种摸球方式下分别计算事件 \(A_{1}\)， \(A_{2}\)， \(A_{3}\) 发生的概率的大小关系 .

解：有放回的情况下：

计算\(P(A_1)\)时， \(n(\Omega_{1})\)\(=\)\(C_{10}^1\)\(=\)\(10\)，满足有限等可能性，\(n(A_1)\)\(=\)\(C_{4}^1\)\(=\)\(4\)，故 \(P(A_1)\)\(=\)\(\cfrac{4}{10}=0.4\)；

采用古典概型计算 \(P(A_2)\) 时， \(n(\Omega_{2})\)\(=\)\(C_{10}^1\)\(\times\)\(C_{10}^1\)\(=\)\(100\)，满足有限等可能性，\(n(A_2)\)\(=\)\(C_{6}^1\)\(\times\)\(C_{4}^1\)\(+\)\(C_{4}^1\)\(\times\)\(C_{4}^1\)\(=\)\(40\)[包括第一次取到白球第二次取到红球和包括第一次取到红球第二次取到红球两种情况]，故 \(P(A_2)\)\(=\)\(\cfrac{40}{100}\)\(=\)\(0.4\)；

简化计算模型，采用相互独立事件计算 \(P(A_2)\) ，令前半次取到白球为 \(A\)，后半次取到红球为 \(B\)，则 \(P(A)\)\(=\)\(\cfrac{6}{10}\)， \(P(B)\)\(=\)\(\cfrac{4}{10}\)， 前半次取到红球为\(P(\bar{A})\)\(=\)\(\cfrac{4}{10}\)， 后半次取到白球为 \(P(\bar{B})\)\(=\)\(\cfrac{6}{10}\)，则 \(A_2=AB\cup\bar{A}B\)，且 \(A\)、\(B\)相互独立[原因：\(P(AB)\)\(=\)\(\cfrac{24}{100}\)\(=\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{4}{10}\)]， \(\bar{A}\)、\(B\)相互独立，\(AB\) 和 \(\bar{A}B\)彼此互斥，

故 \(P(A_2)\)\(=\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{4}{10}\)\(+\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{4}{10}\)\(=\)\(\cfrac{2}{5}\)\(=\)\(0.4\)，

计算\(P(A_3)\)时， \(n(\Omega_{3})\)\(=\)\(C_{10}^1\)\(\times\)\(C_{10}^1\)\(\times\)\(C_{10}^1\)\(=\)\(1000\)，满足有限等可能性，\(n(A_3)\)\(=\)\(C_{6}^1\)\(\times\)\(C_{6}^1\)\(\times\)\(C_{4}^1\)\(+\)\(C_{6}^1\)\(\times\)\(C_{4}^1\)\(\times\)\(C_{4}^1\)\(+\)\(C_{4}^1\)\(\times\)\(C_{6}^1\)\(\times\)\(C_{4}^1\)\(+\)\(C_{4}^1\)\(\times\)\(C_{4}^1\)\(\times\)\(C_{4}^1\)\(=\)\(400\)[包括一白二白三红，一白二红三红，一红二白三红，一红二红三红共四种情况]，故 \(P(A_3)\)\(=\)\(\cfrac{400}{1000}\)\(=\)\(0.4\)；

简化计算模型，故 \(P(A_3)\)\(=\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{4}{10}\)\(+\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{4}{10}\)\(+\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{4}{10}\)\(+\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{4}{10}\)\(=\)\(0.4\)，

无放回的情况下：

\(P(A_1)\)\(=\)\(\cfrac{4}{10}=0.4\)；

\(P(A_2)\)\(=\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{4}{9}\)\(+\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{3}{9}\)\(=\)\(\cfrac{2}{5}\)\(=\)\(0.4\)，

\(P(A_3)\)\(=\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{5}{9}\)\(\times\)\(\cfrac{4}{8}\)\(+\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{4}{9}\)\(\times\)\(\cfrac{3}{8}\)\(+\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{6}{9}\)\(\times\)\(\cfrac{3}{8}\)\(+\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{3}{9}\)\(\times\)\(\cfrac{2}{8}\)\(=\)\(0.4\)，