概率论!!求大佬解惑(。•́︿•̀。)？
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P(A)=0.4 , P(B∩~A)=0.2 , P(C∩~A∩轿桐~B)=0.1
To find :P(AUBUC)
solution :
P(B∩~A)=0.2
P(B) - P(A∩做袜B) =0.2 (1)
P(C∩闭胡坦~A∩~B)=0.1
P(C∩~(AUB) )=0.1
P(C) - P(C∩(AUB)) =0.1
P(C) -[ P(C) +P(AUB) -P(AUBUC) ] =0.1
-P(AUB) +P(AUBUC) =0.1
P(AUBUC) =0. 1 + P(AUB)
=0.1 + P(A)+P(B)-P(A∩B)
=0.1 +0.4 + 0.2
=0.7
To find :P(AUBUC)
solution :
P(B∩~A)=0.2
P(B) - P(A∩做袜B) =0.2 (1)
P(C∩闭胡坦~A∩~B)=0.1
P(C∩~(AUB) )=0.1
P(C) - P(C∩(AUB)) =0.1
P(C) -[ P(C) +P(AUB) -P(AUBUC) ] =0.1
-P(AUB) +P(AUBUC) =0.1
P(AUBUC) =0. 1 + P(AUB)
=0.1 + P(A)+P(B)-P(A∩B)
=0.1 +0.4 + 0.2
=0.7
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