Respuesta :
Answer:
31.25% probability that the Blair family had at least 3 girls
Step-by-step explanation:
For each children, there are only two possible outcomes. Either it was a girl, or it was not. The probability of a child being a girl is independent of any other children. So we use the binomial probability distribution 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.
They had 4 children.
This means that [tex]n = 4[/tex]
The probability of a child being a girl is 0.5
This means that [tex]p = 0.5[/tex]
Probability of at least 3 children:
[tex]P(X \geq 3) = P(X = 3) + P(X = 4)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{4,3}.(0.5)^{3}.(0.5)^{1} = 0.25[/tex]
[tex]P(X = 4) = C_{4,4}.(0.5)^{4}.(0.5)^{0} = 0.0625[/tex]
[tex]P(X \geq 3) = P(X = 3) + P(X = 4) = 0.25 + 0.0625 = 0.3125[/tex]
31.25% probability that the Blair family had at least 3 girls
