A quality control inspector is inspecting newly produced items for faults. The inspector searches an item for faults in a series of independent fixations, each of a fixed duration. Given that a flaw is actually present, let ?p ?denote the probability that the flaw is detected during any one fixation (this model is discussed in “Human Performance in Sampling Inspection,” ?Human Factors, ?1979: 99–105). a. ?Assuming that an item has a flaw, what is the probability that it is detected by the end of the second fixation (once a flaw has been detected, the sequence of fixations terminates)? b. ?Give an expression for the probability that a flaw will be detected by the end of the ?n?th fixation. c. ?If when a flaw has not been detected in three fixations, the item is passed, what is the probability that a flawed item will pass inspection? d. ?Suppose 10% of all items contain a flaw [P (randomly chosen item is flawed) = .1]. With the assumption of part (c), what is the probability that a randomly chosen item will pass inspection (it will automatically pass if it is not flawed, but could also pass if it s flawed)? e. ?Given that an item has passed inspection (no flaws in three fixations), what is the probability that it is actually flawed? Calculate for p = .5.

Solution : Step 1: It is given that the inspector searches an item for faults in a series of independent fixations, each of a fixed duration. . If flaw is actually present, let p denote the probability that flaw is detected during any one fixation. Step 2 : a) We have to find the probability that it is detected by the end of second fixation. Here we want the probability that flaw is being detected either by the first or by the second fixation. P( detected by second fixation )= P( detected first)+ P( not detected on first detect on second) = P+P(1-p) Step 3 : b) we have to give an expression for the probability that a flaw will be detected by the end of the nth fixation. Using the first formula. P( detecting by the end of nth fixation)= P+P(1-p)+p(1-p) + . . . + p(1 p) n c) we have to find the probability that flaw has not been detected in 3 fixations It will be P( not detected in 3 fixations)= (1-P) 3