Answer:
the probability of exactly 5 out of the next 10 children born to couples where one parent has Huntington's disease having the disease is 0.246, or about 24.6%.
Explanation:
The probability of a child being born with Huntington's disease when one parent is a carrier is 50%, since it is an autosomal dominant disease. To calculate the probability of exactly 5 out of the next 10 children born to such couples having the disease, we can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
P(X = k) is the probability of getting k successes (in this case, 5 children with the disease)
n is the total number of trials (in this case, 10 children born)
k is the number of successes we want (in this case, 5 children with the disease)
p is the probability of success (in this case, 0.5, since the disease has a 50% chance of being inherited)
(1-p) is the probability of failure (in this case, also 0.5)
Using this formula, we get:
P(X = 5) = (10 choose 5) * 0.5^5 * 0.5^(10-5)
P(X = 5) = (252) * 0.5^10
P(X = 5) = 0.246
Therefore, the probability of exactly 5 out of the next 10 children born to couples where one parent has Huntington's disease having the disease is 0.246, or about 24.6%.