题目内容
【题目】一个包装箱内有6件产品,其中4件正品,2件次品.现随机抽出两件产品,
(1)求恰好有一件次品的概率.
(2)求都是正品的概率.
(3)求抽到次品的概率.
【答案】(1)[Failed to download image : http://192.168.0.10:8086/QBM/2015/11/19/1572299785969664/1572299791925248/ANSWER/5ec6b369f48a4d56924ddc4afc74f3a8.png];(2)[Failed to download image : http://192.168.0.10:8086/QBM/2015/11/19/1572299785969664/1572299791925248/ANSWER/758efc4d68944a0ba5a04bdc579ab1b8.png];(3)[Failed to download image : http://192.168.0.10:8086/QBM/2015/11/19/1572299785969664/1572299791925248/ANSWER/a0051ac52b264572bea815047a738b3b.png];
【解析】
试题本题中三个小题考察的都是古典概型概率,求解时需找到所有基本事件总数和满足题意要求的基本事件的个数,求其比值即可,在求解时当情况比较多可首先考虑其对立事件
试题解析:将六件产品编号,ABCD(正品),ef(次品),从6件产品中选2件,其包含的基本事件为:(AB)(AC)(AD)(Ae)(Af)(BC)(BD)(Be)(Bf)(CD)(Ce)(Cf)(De)(Df)(ef).共有15种, 2分
(1)设恰好有一件次品为事件A,事件A中基本事件数为:8
则P(A)=[Failed to download image : http://192.168.0.10:8086/QBM/2015/7/10/1572177482039296/1572177487126528/EXPLANATION/3ef7009a7764464a83f3c2cbb9347f70.png]6分
(2)设都是正品为事件B,事件B中基本事件数为:6
则P(B)=[Failed to download image : http://192.168.0.10:8086/QBM/2015/7/10/1572177482039296/1572177487126528/EXPLANATION/63cee1ad927b4223b1c7cdc414063892.png]10分
(3)设抽到次品为事件C,事件C与事件B是对立事件,
则P(C)=1-P(B)=1-14分