TEXT: Nowadays, companies depend entirely on Information Technology to conduct business, regardless of the size of the company. The IT area is essential for the smooth running of the company. In this context, it is essential that companies carry out preventive maintenance of their equipment. There is a preventive maintenance checklist, as some defects are recurrent and prevention reduces the risk of your equipment having defects. Some companies have this maintenance carried out by IT professionals and others outsource this work. Some companies were analyzed for a certain period of time. The defects presented were classified into four groups: Group1 - G1 (Damaged HD; RAM memory; Defective processor) Groups 2 - G2 (Motherboard; Power supply; Virus) Groups 3 - G3 (Insufficient power; Heat; Hardware conflicts ) Groups 4 - G4 (Operating system; Hard disk; Video cards) Carrying out a careful analysis of these companies, the following scenario was reached: I. If the company does not carry out preventive maintenance on its equipment, they may present defects G1 , G2, G3 and G4 over 12 months and with the following probabilities, respectively, 5%, 5%, 7%, 7%. II. If, on the other hand, the company carries out preventive maintenance, the probability of its equipment presenting defects G1, G2, G3 and G4 decreases to 2.5%, 2.5%, 4% and 4%, respectively. III. The occurrence of defects in each group are independent events, with or without preventive maintenance. Referring to the previous text, calculate the probability that the equipment will present a defect over 12 months in 2 scenarios: 1) If preventive maintenance is not carried out; 2) If preventive maintenance is carried out.
Question
Answer:
To calculate the probability of equipment presenting a defect over 12 months in the given scenarios, we'll use the probabilities provided in the text.
Scenario 1: If preventive maintenance is not carried out
In this scenario, the probabilities of defects in each group are as follows:
- P(G1 defect) = 5% = 0.05
- P(G2 defect) = 5% = 0.05
- P(G3 defect) = 7% = 0.07
- P(G4 defect) = 7% = 0.07
Since the occurrence of defects in each group is independent, we can calculate the overall probability of a defect occurring by multiplying the probabilities of not having defects in all groups:
P(Defect without maintenance) = P(G1 defect) * P(G2 defect) * P(G3 defect) * P(G4 defect)
Scenario 2: If preventive maintenance is carried out
In this scenario, the probabilities of defects in each group are reduced:
- P(G1 defect) = 2.5% = 0.025
- P(G2 defect) = 2.5% = 0.025
- P(G3 defect) = 4% = 0.04
- P(G4 defect) = 4% = 0.04
Again, since the occurrence of defects in each group is independent, we can calculate the overall probability of a defect occurring with maintenance:
P(Defect with maintenance) = P(G1 defect) * P(G2 defect) * P(G3 defect) * P(G4 defect)
Now, you can calculate these probabilities for both scenarios.
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10 months ago
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