Using the hypergeometric distribution, it is found that there is a 0.0316 = 3.16% probability that both Rose and Liz will be selected.
Hypergeometric distribution:
[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 both are selected is P(X = 2), hence:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 2) = h(2,20,4,2) = \frac{C_{2,2}C_{18,2}}{C_{20,4}} = 0.0316[/tex]
0.0316 = 3.16% probability that both Rose and Liz will be selected.
A similar problem is given at https://brainly.com/question/24826394
