58 ELEMENTARY PROBABILITY Ch. 3 



time in the long run of many such matings ; hence the individual pair- 

 ings are considered to be single events. It follows that the probability 

 that any specified future child of these parents will have type B 

 blood is P(B) = 2/4 = 1/2. Similarly, P(A) = 1/2. There is no 

 possibility that these parents will produce a child of either type 

 or type AB; hence P(AB) = P(O) = 0/4 = 0. 



In view of the fact that the inheritance which a child receives from 

 its father is an independent event with respect to its inheritance 

 from its mother, formula 3.16 can be applied. This gives the follow- 

 ing simple solution for the probability that a child with blood type 

 B will be produced: 



P(B from father) = 1/2; P(0 from mother) = 1; therefore, P(B 

 from father and from mother) = (1/2) (1) = 1/2. 



Since that is the only way a child with B-type blood can be produced 

 by these parents, that is the solution to the problem. 



If one parent is type and the other type A, something must be 

 known or assumed regarding the specific type A, that is, A/O or A/A. 

 If one parent is A/O and the other is type O/O, P(A) = P(O) = 1/2. 

 No other type is possible. But if the parent with type A blood is 

 actually A/A, all children will be A/O. 



If it is known only that the parents are of types A and 0, respec- 

 tively, and if it assumed that type A is equally frequently A/O and 

 A/A, the probability that any particular future child will be type 

 A is 



P(type A child) = P(A/0 parent) -P[ A child/ (A/O parent)] 



+ P(A/A parent) -P[A child/ (A/A parent)] 



= (l/2)-(l/2) + (l/2).(l) = 3/4 



by the probability laws of (3.15) and (3.17). 



Table 3.12 was derived by the above methods under the assump- 

 tion that type A is equally frequently A/A and A/O; and similarly 

 for type B. (Actually this assumption is unrealistic, but it is con- 

 venient here.) The reader should verify several of the probabilities 

 in this table, noting particularly where the assumption regarding 

 the relative frequency of A/A and A/O among type A parents (or 

 likewise for type B) affects the calculations. 



