Respuesta :
Using the hypergeometric distribution, it is found that:
a) The probability distribution is:
[tex]P(X = 0) = 0.357[/tex]
[tex]P(X = 1) = 0.536[/tex]
[tex]P(X = 2) = 0.107[/tex]
b) The expected value is of 0.75.
The cameras are chosen without replacement, which means that 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:
- Contains 8 cameras, thus [tex]N = 8[/tex].
- 3 are defective, thus [tex]k = 3[/tex].
- Sample of 2, thus [tex]n = 2[/tex].
Item a:
The distribution is the probability of each outcome, 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 = 0) = h(0,8,2,3) = \frac{C_{3,0}C_{5,2}}{C_{8,2}} = 0.357[/tex]
[tex]P(X = 1) = h(1,8,2,3) = \frac{C_{3,1}C_{5,1}}{C_{8,2}} = 0.536[/tex]
[tex]P(X = 2) = h(2,8,2,3) = \frac{C_{3,2}C_{5,0}}{C_{8,2}} = 0.107[/tex]
Item b:
The expected value for the hypergeometric distribution is:
[tex]E(X) = \frac{nk}{N}[/tex]
Then:
[tex]E(X) = \frac{2(3)}{8} = 0.75[/tex]
The expected value is of 0.75.
A similar problem is given at https://brainly.com/question/24008577
