if a woman is a carrier for a disease (Hh) and a man has a 50% chance of being a carrier for that disease, what is the probability that they have a child with that disease?
i've figured out that if the man IS a carrier, there is a 25% chance and if the man IS NOT a carrier, there is no chance. but what is the overall probability?
To determine the overall probability that the couple will have a child with the disease, you need to consider both the probability that the woman is a carrier (Hh) and the probability that the man is a carrier.
Let's break down the possibilities step by step:
1. The woman is a carrier (Hh) and the man is a carrier (50% chance): There is a 25% chance that the child will inherit the disease because the child would have a 50% chance of inheriting the disease allele from each parent (Hh).
2. The woman is a carrier (Hh) and the man is not a carrier (50% chance): In this case, the child has no chance of inheriting the disease because the child would receive the healthy allele (h) from the father.
Since these two cases are exclusive (meaning only one of them can happen at a time), you need to calculate their probabilities separately:
Probability of case 1: (Probability that the woman is a carrier) × (Probability that the man is a carrier) = 50% × 50% = 25%.
Probability of case 2: (Probability that the woman is a carrier) × (Probability that the man is not a carrier) = 50% × 50% = 25%.
Now, add the probabilities of both cases to get the overall probability:
Overall probability = Probability of case 1 + Probability of case 2 = 25% + 25% = 50%.
Therefore, the overall probability that they will have a child with the disease is 50%.