Concept explainers
A quality control inspector is examining 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
- 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 nth 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 is 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.
a.
Find the probability of detecting the flaw at the end of the second fixation.
Answer to Problem 85E
The probability of detecting the flaw at the end of the second fixation is
Explanation of Solution
Given info:
The series of independent items were detected for faults by the quality control inspector. The items were inspected for the faults at each fixation for the each fixed time. Here, p denotes the probability of detecting the flaw at any one fixation.
Calculation:
Define the events as given below:
The probabilities of the events are given below:
Independent events
If two events are said to be independent then the occurrence of one does not affect the occurrence of the other.
Multiplication rule for two independent events:
If the events A and B are independent then,
The probability of detecting the flaw at the end of the second fixation is obtained as shown below:
The probability of the event is obtained as given below:
Thus, the probability of the event is
b.
Find the probability of detecting the flaw at the end of the
Answer to Problem 85E
The probability of detecting the flaw at the end of the
Explanation of Solution
Calculation:
Define the event as given below:
The probability of detecting the flaw at the end of the
Based on geometric series:
The expression
The probability of the event is obtained as given below:
Thus the probability of the event is
c.
Obtain the probability of not detecting the flaw in three fixations.
Answer to Problem 85E
The probability of not detecting the flaws in three fixations is
Explanation of Solution
Calculation:
The probability of not detecting the flaws in three fixations is obtained as shown below:
Thus the probability of not detecting the flaws in three fixations is
d.
Obtain the probability of selecting the item that passes the inspection.
Answer to Problem 85E
The probability of selecting the item that pass the inspection is
Explanation of Solution
Calculation:
Here, the probability of items flawed is 0.1, the probability of the item not flawed is 0.9, the probability of not detecting the flaws in 3 fixation is
The probability of selecting the item that pass the inspection is obtained as shown below:
Thus the probability of not detecting the flaws in three fixations is
e.
Obtain the probability that the items flawed given that the item has passed the inspection.
Answer to Problem 85E
The probability that the items flawed given that the item has passed the inspection is 0.0137.
Explanation of Solution
Calculation:
Here, the probability of detecting the flaws at any one fixation is 0.5.
From part (d), the probability of selecting the item that pass the inspection is
The probability that that the items flawed given that the item has passed the inspection is obtained as given below:
Substitute p as 0.5,
Thus, the probability that the items flawed given that the item has passed the inspection is 0.0137.
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