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Sagot :
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:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- 1000 people means that [tex]N = 1000[/tex]
- 130 can wiggle their ears, thus [tex]k = 130[/tex]
- Two are selected, thus [tex]n = 2[/tex].
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
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