111k views
4 votes
if danielle and miguel were to have 5 children, what is the probability that one would develop hd and 4 would not?

User Glocore
by
8.3k points

1 Answer

0 votes
To calculate the probability, we need to know the probability of a child developing Huntington's disease (HD). HD is an inherited disorder caused by a genetic mutation, so the probability of a child developing HD depends on the parents' genetic makeup and whether they are carriers of the mutation.

If we assume that both Danielle and Miguel are carriers of the HD mutation, which follows an autosomal dominant inheritance pattern, there is a 50% chance that each of their children will inherit the mutation and develop HD.

The probability of one child developing HD and four children not developing HD can be calculated using the binomial probability formula:

P(X=k) = (nCk) * p^k * q^(n-k)

Where:
P(X=k) is the probability of getting k successes (in this case, one child developing HD)
n is the total number of children (5)
k is the number of successes (1)
p is the probability of success (0.5, since there is a 50% chance of inheriting the HD mutation)
q is the probability of failure (1-p = 0.5)

Plugging in the values, we get:

P(X=1) = (5C1) * (0.5)^1 * (0.5)^(5-1)
= 5 * 0.5 * 0.5^4
= 5 * 0.5 * 0.0625
= 0.15625

Therefore, the probability that one child would develop HD and four children would not, given that both Danielle and Miguel are carriers of the HD mutation, is 0.15625, or 15.625%.
User MTCoster
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories