Question Twoa: Textile Company Uses Three Service Companies

Question Twoa Textile Company Uses Three Service Companies Alpha Bet

Question Two A textile company uses three service companies, Alpha, Beta and Gamma, to repair machinery when it breaks down. When a piece of machinery breaks down there is a 20% chance that it is sent to Alpha and a 40% chance that it is sent to each of Beta and Gamma. A study of these companies’ service times over the last two years has revealed the following: The probability that broken machinery sent to Alpha is returned within a week is 0.5. The probability that broken machinery sent to Beta is returned within a week is 0.6 and the probability of the same return time for Gamma is 0.4. Machinery returned within a week by Alpha or Beta has an 80% probability of being satisfactorily repaired, whereas machinery returned within a week by Gamma has a 90% probability of being satisfactorily repaired. Machinery sent to Alpha or Beta, which takes longer than a week to repair has a 90% probability of being satisfactorily repaired, whereas machinery sent to Gamma and taking longer than a week is always satisfactorily repaired. If a piece of machinery has taken over a week and is not satisfactorily repaired, what is the probability that it was sent to Alpha?

Paper For Above instruction

In this problem, we analyze the conditional probability that a piece of machinery sent to Alpha was the one that was not satisfactorily repaired after taking longer than a week. To approach this, we define the relevant events and apply Bayesian probability principles.

Step 1: Define the Events

  • A: Machinery was sent to Alpha (Probability = 0.2)
  • B: Machinery was sent to Beta (Probability = 0.4)
  • G: Machinery was sent to Gamma (Probability = 0.4)

Note: The probabilities sum to 1: 0.2 + 0.4 + 0.4 = 1.

Step 2: Probabilities of Return Within a Week

  • P(Return within 1 week | A) = 0.5
  • P(Return within 1 week | B) = 0.6
  • P(Return within 1 week | G) = 0.4

Step 3: Probabilities of Return After More Than a Week

Complement probabilities, since events are mutually exclusive:

  • P(>1 week | A) = 1 - 0.5 = 0.5
  • P(>1 week | B) = 1 - 0.6 = 0.4
  • P(>1 week | G) = 1 - 0.4 = 0.6

Step 4: Probabilities of Satisfactory Repair After Return

  • P(Satisfactory | Return within 1 week, A or B) = 0.8
  • P(Satisfactory | Return within 1 week, G) = 0.9
  • P(Satisfactory | >1 week, A or B) = 0.9
  • P(Satisfactory | >1 week, G) = 1 (always satisfactorily repaired)

Step 5: Calculate the Probability of a Machinery Being Not Satisfactorily Repaired After >1 Week

  • P(Not satisfactory | >1 week, A or B) = 1 - 0.9 = 0.1
  • P(Not satisfactory | >1 week, G) = 1 - 1 = 0

Step 6: Compute the Overall Probability of a Machinery Being >1 Week and Not Satisfactorily Repaired

Using the law of total probability:

P(>1 week and not satisfactory) =

= P(A) P(>1 week | A) P(Not satisfactory | >1 week, A) +

P(B) P(>1 week | B) P(Not satisfactory | >1 week, B) +

P(G) P(>1 week | G) P(Not satisfactory | >1 week, G)

= 0.2 0.5 0.1 + 0.4 0.4 0.1 + 0.4 0.6 0

= 0.2 0.5 0.1 + 0.4 0.4 0 + 0.4 0.6 0

= 0.01 + 0 + 0 = 0.01

Step 7: Compute the Conditional Probability that the Machinery Was Sent to Alpha Given It Took Over a Week and Was Not Satisfactorily Repaired

P(Alpha | >1 week and not satisfactory) =

= [P(A) P(>1 week | A) P(Not satisfactory | >1 week, A)] / P(>1 week and not satisfactory)

= 0.2 0.5 0.1 / 0.01 = 0.01 / 0.01 = 1

Therefore, the probability that the machinery was sent to Alpha given it took over a week and was not satisfactorily repaired is 1, or 100%. This indicates that, under the model assumptions, any machinery failing to be satisfactory after more than a week is almost certainly from Alpha.

References

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