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} λ^{x}/x!
(? = np = 1000 × 0.005) = 5
i. P(only one is defective) = P(1) = P(x = 1)
= (2.718^{-5} * 5^{1})/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^{-5}5^{0}/0!
= 2.718^{-5}
= 1/(2.718)^{+5}
= 0.00674
P(1) = 0.0337
P(2) = (2.718^{-5} * 5^{2})/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)]