Using the hypergeometric distribution, the probabilities are given as follows:
a) 0.034 = 3.4%.
b) 0.665 = 66.5%.
c) 0.3352 = 33.5%.
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:
For this problem, the parameters are given as follows:
N = 1100, k = 203, n = 2.
In item a, the probability 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(2,1100,2,203) = \frac{C_{203,2}C_{897,0}}{C_{1100,2}} = 0.034[/tex]
For item b, the probability is P(X = 0), 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 = 0) = h(0,1100,2,203) = \frac{C_{203,0}C_{897,2}}{C_{1100,2}} = 0.665[/tex]
For item c, the probability is:
P(X >= 0) = 1 - P(X = 0) = 1 - 0.665 = 0.335.
More can be learned about the hypergeometric distribution at https://brainly.com/question/24826394
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