Answer
a) No
b) Not independent
c) 0.8
Work Step by Step
a) Since $P(A \cap B) = 0.2 \ne 0$, A and B are not mutually exclusive
b) $P(A) . P(B) = 0.6 \times 0.5 = 0.3$
$P(A \cap B) = 0.2 $
$P(A \cap B) \ne P(A) . P(B)$ proving that A and B are not independent.
c) $P(A^C \cup B^C)=P((A \cap B)^c)=1 - P(A \cap B) =1 - 0.2 = 0.8$
