Respuesta :
Answer:
If the lot has 4 defective covers out of 25 total covers, the probability of accepting the lot is P=0.98.
Step-by-step explanation:
We have a population of N=25 skylight covers, were K=4 are defective.
We sample n=5 covers, and we will accept the lot if k=2 or fewer are defective.
We will use the hypergeometric distribution to model this probabilities.
First, to be accepted, the sample can have 2, 1 or 0 defective covers, so the probability of being accepted is:
[tex]P(accepted)=P(k\leq2)=P(k=0)+P(k=1)+P(k=2)[/tex]
The probability that there are k defective covers in the sample is:
[tex]P(k)=\dfrac{\dbinom{k}{k}\dbinom{N-k}{n-k}}{\dbinom{N}{n}}[/tex]
Then, we can calculate the individual probabilities as:
[tex]P(k=0)=\dfrac{\dbinom{4}{0}\cdot \dbinom{25-4}{5-0}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{0}\cdot \dbinom{21}{5}}{\dbinom{25}{5}}\\\\\\P(k=0)=\dfrac{1\cdot 20349}{53130}=0.38[/tex]
[tex]P(k=1)=\dfrac{\dbinom{4}{1}\cdot \dbinom{25-4}{5-1}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{1}\cdot \dbinom{21}{4}}{\dbinom{25}{5}}\\\\\\P(k=1)=\dfrac{4\cdot 5985}{53130}=0.45[/tex]
[tex]P(k=2)=\dfrac{\dbinom{4}{2}\cdot \dbinom{25-4}{5-2}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{2}\cdot \dbinom{21}{3}}{\dbinom{25}{5}}\\\\\\P(k=2)=\dfrac{6\cdot 1330}{53130}=0.15[/tex]
If we add this probabilities, we have:
[tex]P(accepted)=P(k\leq2)=P(k=0)+P(k=1)+P(k=2)\\\\P(accepted)=0.38+0.45+0.15=0.98[/tex]
If the lot has 4 defective covers out of 25 total covers, the probability of accepting the lot is P=0.98.
