Statistics excercices

[Tanne07]

Hello. How I can solve those problems?

a) (A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)

(b) P(A∪B)≥1−(P(A¯)+P(B¯))P(A∪B)≥1−(P(A¯)+P(B¯))

(c) P(A∩B)≥1−(P(A¯)+P(B¯))

 

[Deleted member 4993]

Hello. How I can solve those problems?

a) (A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)

(b) P(A∪B)≥1−(P(A¯)+P(B¯))P(A∪B)≥1−(P(A¯)+P(B¯))

(c) P(A∩B)≥1−(P(A¯)+P(B¯))

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[pka]

Hello. How I can solve those problems?

a) (A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)

(b) P(A∪B)≥1−(P(A¯)+P(B¯))P(A∪B)≥1−(P(A¯)+P(B¯))

(c) P(A∩B)≥1−(P(A¯)+P(B¯))

To Tanne07, where did you get these questions? I think that you have several mistakes in what you copied.
Are you using \(\displaystyle A^{\_}\) for complement? If so, please use \(\displaystyle A^c\text{ or }\overline{A}\) the tex code is [ tex]A^c\text{ or }\overline{A}[/tex] without the first space.
Nonetheless, please check the problems for mistakes.