Answer:
The probability is 0.74719
Step-by-step explanation:
Let's start defining the random variable X.
X : ''Number of children with hyperlipidemia out of 12 children''
X can be modeled as a binomial random variable.
X ~ Bi (n,p)
Where n is the sample size and p is the ''success probability''.
We defining as a success to find a child that has hyperlipidemia.
The probability function for X is :
[tex]P(X=x)=(nCx).p^{x}.(1-p)^{n-x}[/tex]
Where nCx is the combinatorial number define as :
[tex]nCx=\frac{n!}{x!(n-x)!}[/tex]
We are looking for [tex]P(X\geq 3)[/tex]
[tex]P(X\geq 3)=1-P(X\leq 2)[/tex]
[tex]P(X\geq 3)=1-[P(X=0)+P(X=1)+P(X=2)][/tex]
[tex]P(X\geq 3)=1-[(12C0)0.3^{0}0.7^{12}+(12C1)0.3^{1}0.7^{11}+(12C2)0.3^{2}0.7^{10}][/tex]
[tex]P(X\geq 3)=1-(0.7^{12}+0.07118+0.16779)=1-0.25281=0.74719[/tex]
There is a probability of 0.74719 that at least 3 children are hyperlipidemic.
