Bottles of the same sizes produced in a factory are packed in boxes. Each box contains 10 bottles. If 8% of the bottles are defective, find, correct to two decimal places, the probability that box chosen at random contains at least 3 defective bottles.
Explanation
p(defective) = (frac{8}{100} = 0.08)
p(non- defective) = (1 – 0.08 = 0.92)
p(at least 3 defective) = (1 – [p(0) + p(1) + p(2)])
(p(x) = ^{n}C_{x} p^{x} q^{n – x})
(p(0) = ^{10}C_{0} (0.08)^{0} (0.92)^{10} = 0.434)
(p(1) = ^{10}C_{1} (0.08)^{1} (0.92)^{9} = 0.378)
(p(2) = ^{10}C_{2} (0.08)^{2} (0.92)^{8} = 0.148)
p(at least 3 defective) = 1 – [0.434 + 0.378 + 0.148]
= 1 – 0.960 = 0.04