Question

A pharmaceutical company receives large shipments of aspirin tablets. the acceptance sampling plan is to randomly select and test 2626 ​tablets, then accept the whole batch if there is only one or none that​ doesn't meet the required specifications. if a particular shipment of thousands of aspirin tablets actually has a 44​% rate of​ defects, what is the probability that this whole shipment will be​ accepted?

Answer 1

For a really large shipment, and a relatively high "success" rate (i.e. selecting the defectives), we may assume that sampling without replacement does not change the probability, hence binomial distribution may be used, with which
$P(x)=C(n,x)p^x(1-p)^{n-x}$
where $C(n,x)=\frac{n!}{x!(n-x)!}$ and
n=26, p=0.44

In the given case, n=26, we need
P(0)+P(1)
$=C(26,0)0.44^0(1-0.44)^{26-0}+C(26,1)0.44^1(1-0.44)^{26-1} $
$=0.0000002837+0.0000057950$
=0.000006080 
with a 4% error compared with the exact answer (for a population of 5000), or 0.42% error if the population is 50,000, as follows

Exact solution by hypergeometric distribution taking into account of selection without replacement, assuming a population of only 5000 
P(x)=C(A,a)C(B,b)/(C(A+B,a+b)
a={0,1}
b={26,25}
A={2200,2199}
B={2800,2800}

P(0)+P(1)
=C(2200,0)C(2800,26)/C(5000,26)+C(2200,1)C(2800,25)/C(5000,26)
=0.0000002695+0.0000055557
=0.00058252

Similarly, for a 50000 population,
=C(2200,0)C(2800,26)/C(5000,26)+C(2200,1)C(2800,25)/C(5000,26)
=0.0000060539