Respuesta :
Answer:
0.384 = 38.4% probability that exactly two of their three children will have that trait.
Step-by-step explanation:
For each children, there are only two possible outcomes. Either they have the trait, or they do not. The probability of a child having the trait is independent of any other child. This means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 0.8.
This means that [tex]p = 0.8[/tex]
Three children:
This means that [tex]n = 3[/tex]
What is the probability that exactly two of their three children will have that trait?
This is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{3,2}.(0.8)^{2}.(0.2)^{1} = 0.384[/tex]
0.384 = 38.4% probability that exactly two of their three children will have that trait.
