Probability

Question

a computer controlling a critical process system must have continuous power. The main power source is supplied from the local utility via overhead power lines. The plant has a backup diesel generator to supply power when the local utility has an outage. The computer has a backup battery to supply power if both the utility and diesel generator go down.

The plant has a .005 probability of suffering a local power outage on any given day. The diesel generator has a .0001 probability of failing on any given day. The battery backup has a .00001 probability of failing on any given day.

What is the probability the computer will lose power on any given day in one year?

Answer 1

If A, B and C are independent events then the probability that they will all happen is given by  

 

P(A and B and C) = P(A)*P(B)*P(C)  

 

These 3 events are (presumably) independent so you can just multiply all 3 probabilities together to get the probability that the computer will lose power  

 

P(lose power) = (.005)*(.0001)*(.00001)

                     = (5E-3)*(1E-4)*(1E-5)

                     = 5E-12  

 

where I have uses the exponential notation 5E-12 to denote 5 x 10^-12