In the expression StartFraction 7 m over 9 EndFraction + 6 r minus four-fifths, what are the coefficients?
2 answers:
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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
where
and
n=26, p=0.44
In the given case, n=26, we need
P(0)+P(1)
=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
where
n=26, p=0.44
In the given case, n=26, we need
P(0)+P(1)
=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