Respuesta :
Answer:
[tex]X \sim Hyp (N = 810, M=102, n=4)[/tex]
[tex] P(X=0) = \frac{(102C0)(810-102 C 4-0)}{810C4}[/tex]
And solving we got:
[tex] P(X=0)= \frac{(102C0) (708C4)}{810C4} = 0.583[/tex]
So then for the problem given the probability that the entire bath will be accepted (none is defective among the 4) is 0.583
Step-by-step explanation:
For this case we can model the variable of interest with the hypergeometric distribution. And with the info given we can do this:
[tex]X \sim Hyp (N = 810, M=102, n=4)[/tex]
[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
And for this case we want to find the probability that none of the scales selected would be defective so we want to find this:
[tex] P(X=0)[/tex]
And using the probability mass function we got:
[tex] P(X=0) = \frac{(102C0)(810-102 C 4-0)}{810C4}[/tex]
And solving we got:
[tex] P(X=0)= \frac{(102C0) (708C4)}{810C4} = 0.583[/tex]
So then for the problem given the probability that the entire bath will be accepted (none is defective among the 4) is 0.583
