Respuesta :
Answer:
the probability that exactly one of their four children will have that trait is 0.287%
Step-by-step explanation:
Given;
The probability that a child born to them have a particular trait is
P = 85% = 0.85
The probability that a child born to them will not have a particular trait is
P' = 1 - P = 1-0.85 = 0.15
The probability that exactly one of four children have the trait is:
P(1) = P × P' × P' × P'
P(1) = 0.85 × 0.15 × 0.15 × 0.15
P(1) = 0.00286875
P(1) = 0.287 %
the probability that exactly one of their four children will have that trait is 0.287%
Using the binomial distribution, it is found that there is a 0.011 = 1.1% probability that exactly one of their four children will have that trait.
For each children, there are only two possible outcomes, either they have the trait, or they do not. The probability of a children having the trait is independent of any other children, hence, the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 85% probability of a children having the trait, hence [tex]p = 0.85[/tex].
- They have four children, hence [tex]n = 4[/tex]
The probability is P(X = 1), hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{4,1}.(0.85)^{1}.(0.15)^{3} = 0.011[/tex]
0.011 = 1.1% probability that exactly one of their four children will have that trait.
A similar problem is given at https://brainly.com/question/24863377
