To determine the chances of a male child being color blind when both parents are normal, we need to analyze the genetics behind color blindness, which is a sex-linked recessive disorder.
### Step-by-Step Solution:
1. **Understanding Color Blindness**:
- Color blindness is primarily caused by a defect in the genes located on the X chromosome. It is a recessive trait, meaning that a person must have two copies of the defective gene to express the trait (in females) or one copy (in males).
2. **Genetic Makeup of Parents**:
- In this scenario, both parents are said to be normal. However, it is important to consider that the mother could be a carrier of the color blindness gene.
- The father, being male, has one X chromosome and one Y chromosome (XY). Since he is normal, his X chromosome does not carry the color blindness gene (denoted as X^N).
- The mother, being female, has two X chromosomes (XX). If she is normal, she could either have two normal X chromosomes (X^N X^N) or one normal and one defective X chromosome (X^N X^c), where X^c represents the color blindness allele.
3. **Possible Genotypes of the Mother**:
- If the mother is normal and not a carrier: X^N X^N (no chance of color blindness).
- If the mother is a carrier: X^N X^c (there is a chance of passing the color blindness gene to her offspring).
4. **Inheritance Pattern**:
- Males inherit their X chromosome from their mother and their Y chromosome from their father. Therefore, a male child will be affected by color blindness if he inherits the X^c chromosome from his mother.
- If the mother is a carrier (X^N X^c), there is a 50% chance that she will pass the X^c chromosome to her son.
5. **Conclusion**:
- If both parents are normal and the mother is not a carrier, the child cannot be color blind. However, if the mother is a carrier, there is a 50% chance that the male child will be color blind.
- Therefore, the chances of the male child being color blind is **50%** if the mother is a carrier.
### Final Answer:
The chances of the male child being color blind, assuming the mother is a carrier, is **50%**.
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