We can divide the sample space into 4 regions:
① part-time students only
② distance learning students only
③ D and E students
④ other students
①+②+③+④=100%
Since E=30%=①+③, ③=25% of 30%=7.5% therefore ①=22.5%.
Also D=15%=②+③, so ②=7.5%. (Although not needed, ④=100-(22.5+7.5+7.5)=100-37.5=62.5%.)
(A) P(D AND E)=③=7.5%.
(B) P(E | D)=50% because given D=15%, E (=7.5%) is 50% of 15%.
(C) P(D OR E)=P(D)+P(E)-P(D AND E)=15+30-7.5=37.5%.
(D) The events D and E are independent because the number of part-time students could, for example, be zero, without affecting the number of distance learning students. However, the percentages are not independent.
(E) If the events were mutually exclusive there could be no intersection (D AND E), in contradiction to the given information.