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?
Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
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Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
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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?
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
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] =
For x = 1 one machine is working
p [x = 1] =
= 0.032
For x = 2 two machine are working
P [x = 2] =
= 0.283
For x = 3 Three machine are working
P [x = 3] =
= 0.684
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