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A manufacturer assures his customers that the probability of having defective item is as 0.005. A sample of 1000 items was inspected. Determine the probabilities of having the given possible outcomes
i. Simply one is defective
ii. Mostly 2 defective
iii. More than 3 defective
P(x) = e-λ λx/x!
(? = np = 1000 × 0.005) = 5
i. P(only one is defective) = P(1) = P(x = 1)
= (2.718-5 * 51)/1! Note that 2.718-5= 1/(2.718)5
= 5/(2.718)5
= 5/(148.33)
ii. P(at most 2 defective) = P(x ≤ 2)
= P(0) +P(1)+P(2)
P(x=0) = e-550/0!
= 2.718-5
= 1/(2.718)+5
= 0.00674
P(1) = 0.0337
P(2) = (2.718-5 * 52)/2!
= 0.08427
P(x<2) = 0.00674 + 0.0337 + 0.08427
= 0.012471
iii. P(more than 3 defective) = P(x > 3)
= 1 - [P(0)+P(1)+P(2)+P(3)]
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