Answer: the following are required for IV-3 to have condition
Explanation:
-II-4 passes an X b chromosome to III-4 (probability = 1/2).
-If III-4 has the genotype X B X b (accounted for by the above probability), then she passes an X b chromosome to IV-3 (probability = 1/2).
-III-5 passes a Y chromosome to IV-3 (probability = 1/2).
All of these requirements are needed in sequence, so you apply the product rule here, too (1/2 x 1/2 x 1/2 = 1/8).
Once the individual probabilities are known, the sum and/or product rules can be used for various combinations (both conditions, either condition, etc.).
