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

Comments:

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

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