Final answer:
The probability that Bob and Barb's next two children consist of a boy with blood type A+ followed by a girl with blood type O- is 3/128.
Step-by-step explanation:
Given that Bob has blood type A+ (genotype AO or AA) and Barb has blood type B+ (genotype BO or BB), their first child with blood type O- indicates that both parents must have one O allele, and at least one parent must have a negative Rh factor allele. Since the Rh+ allele is dominant over the Rh- allele, both Bob and Barb must be heterozygous for the Rh factor (Rh+Rh-).
For their next child to have a specific blood type and gender, we need to calculate the probability of each independently and then multiply them together. The probability of having a boy is 1/2, and the probability of having a girl is also 1/2. The probability the child has blood type A+ is the probability that they inherit an A allele from Bob (1/2 if Bob is AO and 1 if Bob is AA) and an O allele from Barb (1/2), as well as inheriting the Rh+ allele from either parent when both are Rh+Rh-.
Similarly, the probability the child has blood type O- is calculated by inheriting an O allele from each parent (1/4) and the Rh- allele from both (1/4).
The overall probability for a boy with blood type A+ is (1/2 * 1/2 * 3/4) = 3/16. The overall probability for a girl with blood type O- is (1/2 * 1/4) = 1/8. Finally, the probability that these two events happen in sequence for the next two children is the product of their independent probabilities (3/16 * 1/8) = 3/128.