Using the hypergeometric distribution, it is found that there is a 0.1515 = 15.15% probability that you pick at least 3 green balls.
The balls are chosen without replacement, hence the hypergeometric distribution is used.
The formula is:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
In this problem:
The probability that you pick at least 3 green balls is:
[tex]P(X \geq 3) = P(X = 3) + P(X = 4)[/tex]
Hence:
[tex]P(X = 3) = h(3,12,4,5) = \frac{C_{5,3}C_{7,1}}{C_{12,4}} = 0.1414[/tex]
[tex]P(X = 4) = h(4,12,4,5) = \frac{C_{5,4}C_{7,0}}{C_{12,4}} = 0.0101[/tex]
Then:
[tex]P(X \geq 3) = P(X = 3) + P(X = 4) = 0.1414 + 0.0101 = 0.1515[/tex]
0.1515 = 15.15% probability that you pick at least 3 green balls.
You can learn more about the hypergeometric distribution at https://brainly.com/question/4818951