By their mutual exclusivity,
[tex]P(A\cup B\cup C)=P(A)+P(B)+P(C)=0.96[/tex]
[tex]P(A\cap B\cap C)=0[/tex]
[tex]P(A\cap B)=0[/tex]
For the last probability, first distribute the intersection:
[tex](A\cup B)\cap C=(A\cap C)\cup(B\cap C)[/tex]
Recall that for two event [tex]X,Y[/tex],
[tex]P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)[/tex]
so that
[tex]P((A\cap C)\cup(B\cap C))=P(A\cap C)+P(B\cap C)-P((A\cap C)\cap(B\cap C))[/tex]
[tex]P((A\cap C)\cup(B\cap C))=P(A\cap C)+P(B\cap C)-P(A\cap B\cap C)=0[/tex]
