Math, asked by maheshkuhar, 6 months ago

An appliance manufacturer maintain a repair center for its customers. Based on past experience, 30% of all appliances sent in for repair have mechanical problems and 70% have an electrical problem. If the problem is mechanical 90% of the appliances can be repaired and returned to the customer in the remaining 10% appliances are replaced with new unit. if the problem is electrical, 60% can be repaired and 40% are replaced. If an appliance at repair center is selected at random and it needs to be replaced with a new unit, what is the probability that the appliance have an electrical problem? Round your answer to 4 decimal places.​

Answers

Answered by amitnrw
0

Given : An appliance manufacturer maintain a repair center for its customers. Based on past experience, 30% of all appliances sent in for repair have mechanical problems and 70% have an electrical problem. If the problem is mechanical 90% of the appliances can be repaired and returned to the customer in the remaining 10% appliances are replaced with new unit. if the problem is electrical, 60% can be repaired and 40% are replaced.

To Find : If an appliance at repair center is selected at random and it needs to be replaced with a new unit, what is the probability that the appliance have an electrical problem?

Solution:

Mechanical problem = 30 %  = 0.3

Electrical Problem = 70% = 0.7

Mechanical problem Repaired = 0.3 (0.9) = 0.27

Mechanical problem Replaced with new unit  = 0.3 (0.1) = 0.03

Electrical problem Repaired = 0.7 (0.6) = 0.42

Electrical problem Replaced with new unit  = 0.7 (0.4) = 0.28

Mechanical problem Replaced with new unit  =  0.03

Electrical problem Replaced with new unit    = 0.28

Total Replaced with new unit   = 0.03 + 0.28 = 0.31

If an appliance at repair center is selected at random and it needs to be replaced with a new unit,  

probability that the appliance have an electrical problem = 0.28/0.31

= 28/31

= 0.9032

probability that the appliance have an electrical problem = 0.9032

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