Respuesta :
Answer:
The probability that the sample will contain exactly 0 nonconforming units is P=0.25.
The probability that the sample will contain exactly 1 nonconforming units is P=0.51.
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Step-by-step explanation:
We have a sample of size n=4, taken out of a lot of N=12 units, where K=3 are non-conforming units.
We can write the probability mass function as:
[tex]P(x=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}[/tex]
where k is the number of non-conforming units on the sample of n=4.
We can calculate the probability of getting no non-conforming units (k=0) as:
[tex]P(x=0)=\frac{\binom{3}{0}\binom{9}{4}}{\binom{12}{4}}=\frac{1*126}{495}=\frac{126}{495} = 0.25[/tex]
We can calculate the probability of getting one non-conforming units (k=1) as:
[tex]P(x=1)=\frac{\binom{3}{1}\binom{9}{3}}{\binom{12}{4}}=\frac{3*84}{495}=\frac{252}{495} = 0.51[/tex]
