There are 3 machines in a factory that perform all production. Let this independently be working with probabilities 0.8, 0.9, and 0.95. a. Find the probability distribution for this if defined as the number of machines running at any one time. b. According to the maintenance agreement made by the factory, the repair price of a machine is fixed at 500 TL. When maintenance service is given to repair the damaged machines at any time, each one of them is maintained for 50 TL. In this case, how much is this maintenance agreement expected to cost to the factory?

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There are 3 machines in a factory that perform all production. Let this independently be working with probabilities 0.8, 0.9, and 0.95.
a. Find the probability distribution for this if defined as the number of machines running at any one time.
b. According to the maintenance agreement made by the factory, the repair price of a machine is fixed at 500 TL. When maintenance service is given to repair the damaged machines at any time, each one of them is maintained for 50 TL. In this case, how much is this maintenance agreement expected to cost to the factory?

Expert Solution
Step 1

Given,

There are 3 machines in a factory and their respective working probabilities are 0.8, 0.9 and 0.95

 

 

 

 

To find,

a) Find the probability distribution for this if defined as the number of machines running at any one time.

b) How much is this maintenance agreement expected to cost to the factory

 

Step 2

a) Find the probability distribution for this if defined as the number of machines running at any one time.

Answer:

Let x = number of machines working 

Let machines be A1,A2, A3

For x = 0 no machinne is working  

P [X = 0] = (1 - 0.8) × (1 - 0.9) × (1 - 0.95) = [ 0.2 × 0. 1 ×  0.05] = 0.001

For x = 1 one machine is working

p [x = 1] = 0.8 × (1 -  0.9) × (1-0.95) + (1 - 0.8) × 0.9 × (1 - 0.95) + (1 - 0.8) × (1 - 0.9) ×0.95

              = 0.032

For x = 2 two machine are working

P [x = 2] = 0.8 × 0.9 × (1-0.95) + 0.8 × (1 - 0.9) ×0.95 + (1 - 0.8) × 0.9 × 0.95

              = 0.283

For x = 3 Three machine are working

P [x = 3] = 0.8 × 0.9 × 0.95

              = 0.684

 

 

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