中獎與抽獎次序無關

前言

典例剖析

【人教 2019A 版教材\(P_{262}\) 頁習題10.3 第 6 題改編】在一個袋子中放 \(6\) 個白球，\(4\) 個紅球，搖勻後隨機摸球 \(3\) 次，採用放回和不放回兩種方式摸球 . 設事件 \(A_{i}\)=“第 \(i\) 次摸到紅球”，\(i=1,2,3\) .

(1). 在兩種摸球方式下分別計算事件 \(A_{1}\)， \(A_{2}\)， \(A_{3}\) 發生的機率的大小關係 .

解：有放回的情況下：

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

採用古典概型計算 \(P(A_2)\) 時， \(n(\Omega_{2})\)\(=\)\(C_{10}^1\)\(\times\)\(C_{10}^1\)\(=\)\(100\)，滿足有限等可能性，\(n(A_2)\)\(=\)\(C_{6}^1\)\(\times\)\(C_{4}^1\)\(+\)\(C_{4}^1\)\(\times\)\(C_{4}^1\)\(=\)\(40\)[包括第一次取到白球第二次取到紅球和包括第一次取到紅球第二次取到紅球兩種情況]，故 \(P(A_2)\)\(=\)\(\cfrac{40}{100}\)\(=\)\(0.4\)；

簡化計算模型，採用相互獨立事件計算 \(P(A_2)\) ，令前半次取到白球為 \(A\)，後半次取到紅球為 \(B\)，則 \(P(A)\)\(=\)\(\cfrac{6}{10}\)， \(P(B)\)\(=\)\(\cfrac{4}{10}\)， 前半次取到紅球為\(P(\bar{A})\)\(=\)\(\cfrac{4}{10}\)， 後半次取到白球為 \(P(\bar{B})\)\(=\)\(\cfrac{6}{10}\)，則 \(A_2=AB\cup\bar{A}B\)，且 \(A\)、\(B\)相互獨立[原因：\(P(AB)\)\(=\)\(\cfrac{24}{100}\)\(=\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{4}{10}\)]， \(\bar{A}\)、\(B\)相互獨立，\(AB\) 和 \(\bar{A}B\)彼此互斥，

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

計算\(P(A_3)\)時， \(n(\Omega_{3})\)\(=\)\(C_{10}^1\)\(\times\)\(C_{10}^1\)\(\times\)\(C_{10}^1\)\(=\)\(1000\)，滿足有限等可能性，\(n(A_3)\)\(=\)\(C_{6}^1\)\(\times\)\(C_{6}^1\)\(\times\)\(C_{4}^1\)\(+\)\(C_{6}^1\)\(\times\)\(C_{4}^1\)\(\times\)\(C_{4}^1\)\(+\)\(C_{4}^1\)\(\times\)\(C_{6}^1\)\(\times\)\(C_{4}^1\)\(+\)\(C_{4}^1\)\(\times\)\(C_{4}^1\)\(\times\)\(C_{4}^1\)\(=\)\(400\)[包括一白二白三紅，一白二紅三紅，一紅二白三紅，一紅二紅三紅共四種情況]，故 \(P(A_3)\)\(=\)\(\cfrac{400}{1000}\)\(=\)\(0.4\)；

簡化計算模型，故 \(P(A_3)\)\(=\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{4}{10}\)\(+\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{4}{10}\)\(+\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{6}{10}\)\(\times\)\(\cfrac{4}{10}\)\(+\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{4}{10}\)\(\times\)\(\cfrac{4}{10}\)\(=\)\(0.4\)，

無放回的情況下：

\(P(A_1)\)\(=\)\(\cfrac{4}{10}=0.4\)；

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

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