Respuesta :
Answer:
Step-by-step explanation:
Given that the Products produced by a machine has a 3% defective rate.
Each product is independent of the other with a constant prob of being defective as 0.03
X - no of defects is binomial with p =0.03
a) the probability that the first defective occurs in the fifth item inspected
=Prob for first 4 non defective and 5th defective
=[tex](0.97)^4(0.03)^1\\=0.0266[/tex]
(b) the probability that the first defective occurs in the first five inspections
=P(X=1) in binomial with n=5
= 0.9915
c) the expected number of inspections before the first defective occurs
Expected defects in n trials = np
Expected number of inspection before the first defect = 1/p
= 33.33
=34
(c) What is the expected number of inspections before the first defective occurs?
