3.5. Additive and Multiplicative Rules http://www.ck12.org
P(A∪B) =P( 1 )+P( 2 )+P( 4 )+P( 6 )
Since
P(A) =P( 2 )+P( 4 )+P( 6 ) = 0. 4
P(B) =P( 1 )+P( 2 ) = 0. 3
P(A∩B) =P( 2 ) = 0. 1
If we add the probabilities ofP(A)andP(B), we get
P(A)+P(B) =P( 2 )+P( 4 )+P( 6 )+P( 1 )+P( 2 )
But since
P(A∪B) =P( 1 )+P( 2 )+P( 4 )+P( 6 )
Substituting, yields
P(A)+P(B) =P(A∪B)+P( 2 )
However,P( 2 ) =P(A∩B), thus
P(A)+P(B) =P(A∪B)+P(A∩B)
Or,