Sample two individuals and find the probability of the following events

Blood can be classified according to ABO-type: $$A$$, $$B$$, $$AB$$ and $$O$$, but also according to Rh-type, $$P$$ (positive) and $$N$$ (negative). Suppose that every individual has one Rh-type and one ABO-type and that the two classifications groups are independent. The probabilities are: $$P(O)=0.45$$, $$P(A)=.40$$, $$P(B)=0.11$$, $$P(AB)=0.04$$, $$P(P)=0.84$$ and $$P(N)=0.16$$. Sample two individuals at random and find the probability that

a) both are $$AN$$,

b) one is $$OP$$, while the other is not $$OP$$,

c) at least one of them is $$OP$$,

d) one is $$P$$ and the other is not $$AB$$,

My Solution:

Suppose the groups are large and therefore the probabilities doesn't change for the second pick.

a) $$P(\textrm{both are }AN)=P(AN)P(AN)=(P(A)P(N))^2=(0.40\cdot0.16)^2=0.0041$$

b) $$P(\text{one is }OP,\text{ while the other is not }OP)=P(OP)P(OP^c)+P(OP^c)P(OP)=2P(OP)(1-P(OP))=2P(O)P(P)(1-P(O)P(P))=2\cdot0.45\cdot0.84(1-0.45\cdot0.84)=0.47$$

c) There are two ways to pick one with $$OP$$ and one way to pick two with $$OP$$. The three events are pairwise disjoint $$P(\textrm{at least one of them is }OP)=2P(OP)P(OP^c)+P(OP)^2=2P(O)P(P)(1-P(O)P(P))+(P(O)P(P))^2=2\cdot0.45\cdot0.84(1-0.45\cdot0.84)+(0.45\cdot0.84)^2=0.61$$

d) $$P(\text{one is }P\text{ and the other is not }AB)=P(P)P(AB^c)+P(AB^c)P(P)-P(\text{both are }AB^cP)=2(1-P(AB))P(P)-((1-P(AB))P(P))^2=2(1-0.04)0.84-((1-0.04)0.84)^2=0.96$$

I'm not sure my solutions are correct. My answer in d) does not match my textbooks, which says $$0.96×0.84 + 0.96×0.84×0.16 = 0.94$$.