Respuesta :
Answer:
The probability that a box containing 3 defectives will be shipped is 0.6696.
The probability that a box containing only 1 defective will be sent back for screening is 0.12
Step-by-step explanation:
Consider the provided information.
Boxes of 25 items are readied for shipment, and a sample of 3 items is tested for defectives.
Part (A) What is the probability that a box containing 3 defectives will be shipped?
Let x represents the number of defective item among the 3 item.
According to hyper geometric distribution:
[tex]h(x; N, n, k) = \frac{\binom{k}{x} \binom{N-k}{n-x}}{\binom{N}{n}}[/tex]
Where n is randomly selected without replacement from a population of N items. k items can be classified as successes, and N - k items can be classified as failures.
Therefore, N = 25, n = 3 and k=3 in above formula.
The probability that box deliver 3 defective item.
P(No defective item among selected 3)=P(x=0)
[tex]P(x=0) = \frac{\binom{3}{0} \binom{25-3}{3-0}}{\binom{25}{3}}[/tex]
[tex]P(x=0) =0.6696[/tex]
Therefore, the probability that a box containing 3 defectives will be shipped is 0.6696.
Part (B) What is the probability that a box containing only 1 defective will be sent back for screening?
Substitute, N = 25, n = 3 and k=1 in above formula.
The probability that a box containing only 1 defective will be sent back for screening
[tex]P(x=1) = \frac{\binom{1}{1} \binom{25-1}{3-1}}{\binom{25}{3}}[/tex]
[tex]P(x=1) =0.12[/tex]
Hence, the probability that a box containing only 1 defective will be sent back for screening is 0.12
