Respuesta :

LammettHash
If [tex]A[/tex] and [tex]B[/tex] are mutually exclusive, then

[tex]\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)=0.8+0.1=0.9[/tex]

Then [tex]\mathbb P((A\cup B)')=1-\mathbb P(A\cup B)=0.1[/tex].

Otras preguntas