Respuesta :
Answer:
[tex]P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}[/tex]
For this case N = 7, M = 3, n =3, k=1
[tex]P(X=1)= \frac{(3C1)(7-3 C 3-1)}{7C3} = \frac{3*6}{35}= \frac{18}{35}= 0.514[/tex]
Step-by-step explanation:
Previous concepts
The hypergeometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:
[tex]P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}[/tex]
Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes
The expected value and variance for this distribution are given by:
[tex]E(X)= n\frac{M}{N}[/tex]
[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}[/tex]
For this case N = 7, M = 3, n =3, k=1
Solution to the problem
For this case we want this probability:
[tex] P(X=1)[/tex] and if we use the probability mass function we got:
[tex]P(X=1)= \frac{(3C1)(7-3 C 3-1)}{7C3} = \frac{3*6}{35}= \frac{18}{35}= 0.514[/tex]
