Using the hypergeometric distribution, it is found that there is a 0.7568 = 75.68% probability that neither can wiggle his or her ears.
The people are chosen from the sample without replacement, which is why the hypergeometric distribution is used to solve this question.
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 neither can wiggle his or her ears is P(X = 0), thus:
[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 = x) = h(0,1000,2,130) = \frac{C_{130,0}C_{870,2}}{C_{1000,2}} = 0.7568[/tex]
0.7568 = 75.68% probability that neither can wiggle his or her ears.
A similar problem is given at https://brainly.com/question/24826394