Answer:
[tex]Pr = 0.7282[/tex]
Step-by-step explanation:
Given
[tex]n = 2100[/tex] --- sample size
[tex]r = 160[/tex] -- defective
[tex]x = 4[/tex] --- selected
Required
The probability that the batch will be accepted
When the first is selected, the probability that it is okay is:
[tex]Pr(1) = \frac{2100-160}{2100} = \frac{1940}{2100}[/tex]
The subtraction is done to get the number of items that are not defective
Now, there are 2099 left
For the second selection, the probability is:
[tex]Pr(2) =\frac{2099-160}{2099} = \frac{1939}{2099}[/tex]
Now, there are 2098 left
For the third selection, the probability is:
[tex]Pr(3) =\frac{2098-160}{2098} = \frac{1938}{2098}[/tex]
Now, there are 2097 left
For the fourth selection, the probability is:
[tex]Pr(4) =\frac{2097-160}{2097} = \frac{1937}{2097}[/tex]
So, the probability that the batch will be selected is:
[tex]Pr = Pr(1) * Pr(2) * Pr(3) * Pr(4)[/tex]
[tex]Pr = \frac{1940}{2100} * \frac{1939}{2099} * \frac{1938}{2098} * \frac{1937}{2097}[/tex]
[tex]Pr = 0.7282[/tex]