Probability of first inspector detecting all defects, P(A) = 0.97
Probability of second inspector detecting all defects, P(B) = 0.97
Probability that at least one inspector does not detect a defect, = 0.06
By De-Morgans law
a) Probability that a defective component will be detected only by the first inspector, =
= P(A) - P(A B)
= 0.97 - 0.94 = 0.03
Probability that a defective component will be detected by exactly one of the two inspector =
b) Probability that all three defective component in a batch escape detection by both inspector =
s arriving at a distributor are checked for defects by two different inspectors (each component is...
s arriving at a distributor are checked for defects by two different inspectors (each component is checked by both def thes that are present, and the second inspector does likewise. At least one impector does not detect a defect on 6% of all defective components. what is the prot that the following occur? s). The first inspector detects 97% of an tility (a) A defective component will be detected only by the first inspector A defective component will be detected...