Question 799062
p(E) = 0.6
p(C) = 0.3
p(E&C) = 0.12

a) p(EorC) = 0.6+0.3 = 0.9
b) p(notE) = 1-0.6 = 0.4
c) p(C)*p(E') = p(E&C) 
p{E') = {{{ 0.12 / 0.3 = 0.4 }}}
d) not independent. 
p(E') < p(E) and p(E&C) < p(E)*p(C) = 0.18