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# 概率论 - PowerPoint PPT Presentation

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## PowerPoint Slideshow about ' 概率论' - erwin

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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？

A表示电路断电,

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

=1-0.168=0.832

1

2

1

2

(或系统)的可靠性.

A2

A1

B2

B1

S1:

A2

A1

B1

B2

S2:

P(B0)=

=(1-0.6)5

=0.45

P(B1)=

=5×0.6×(1-0.6)4