Extra exercise 4
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Transmission of ABO blood types. 
Let X denote the allele received from father: values A,B.
and let Y denote the allele received from mother: A,B.


(a) The possible allele combinations and blood types of children are:
    AA - blood type A
    AB - blood type AB
    BB - blood type B
We compute the probabilities using the multiplication rule (because
of the independence of the alleles received from the two parents, 
denoted by X and Y):

P(AA) = P(X=A,Y=A) = P(X=A)*P(Y=A) = 0.5*0.5 = 0.25
P(BB) = P(X=B,Y=B) = P(X=B)*P(Y=B) = 0.5*0.5 = 0.25
P(AB) = P(X=A,Y=B or X=B,Y=A) = P(X=A,Y=B) + P(X=B,Y=A)= 0.25 + 0.25 = 0.5


(b) For this setting the possible allele combinations and blood types 
of children are:
    AA - blood type A
    AO - blood type A
    AB - blood type AB
    BO - blood type B
Each of these combinations have probability 0.25, thus blood type A has
probability 0.5 and the two others 0.25.

Q1: n=2 children, let X1 and X2 denote their blood types
P(X1=A,X2=A) = P(X1=A)*P(X2=A) = 0.5*0.5 = 0.25
because the children's blood types are independent

Q2: n=2 children, let again X1 and X2 denote their blood types
P(X1=X2) = P(X1=X2=A or X1=X2=AB or X1=X2=B)
         = P(X1=A,X2=A) + P(X1=AB,X2=AB) + P(X1=B,X2=B)
         = 0.5*0.5 + 0.25*0.25 + 0.25*0.25 = 0.375


(c) In (a), the probabilities are actually from a binomial distribution 
with n=2 and p=0.5, corresponding to counting the number of A alleles (values
2, 0, and 1, respectively). If this is not obvious for you,
try computing them from the binomial distribution (use Table C or Minitab).

In (b), the probability in Q1 is actually from a binomial distribution 
with n=2 and p=0.5. If this is not obvious for you, try computing it from 
the binomial distribution (use Table 1 from Stevens or Minitab).

In contrast, the probability in Q2 is NOT from a binomial distribution.