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§1.6独立性

P(AB)=P(A)P(B)

<=>P(B|A)=P(B)

① 由定义判断,是否满足公式;

② 由问题的性质从直观上去判断.

=“学生视力有缺陷”，

=“学生听力有缺陷”

=“学生听力与视力都有缺陷”，

1）事件

（2）如果已知一学生视力有缺陷，那么他听力也

,由定义知

（3）如果已知一学生听力有缺陷，那么他视力也有缺陷的概率是多少？

2随机投掷编号为 1 与 2 的两个骰子事件A 表示1号骰子向上一面出现奇数,B 表示2号骰子向上一面出现奇数,C 表示两骰子出现的点数之和为奇数.

A, B, C相互独立

3三个元件串联的电路中,每个元件发生断电的概率依次为0.3,0.4,0.6,且各元件是否断电相互独立,求电路断电的概率是多少?

A表示电路断电,

A1,A2,A3相互独立,A= A1+A2+A3,

P(A)=P(A1+A2+A3)=

=1-0.168=0.832

5设每个人的血清中含肝炎病毒的概率为0.4%, 求来自不同地区的100个人的血清混合

1

2

1

2

(或系统)的可靠性.

A2

A1

B2

B1

S1:

A2

A1

B1

B2

S2:

7 某射手在相同条件下独立地进行5次射击,每次击中目标的概率是0.6,求：各个击中目标次数的概率.

P(B0)=

=(1-0.6)5

=0.45

P(B1)=

=5×0.6×(1-0.6)4