Question

Consider the system of components connected as in the accompanying picture. Components 1 and 2 are connected in parallel, so that subsystem works if and only if either 1 or 2 works; since 3 and 4 are connected in series, that subsystem works if and only if both 3 and 4 work. If components work independently of one another and P(component i works) = 0.77 for i = 1, 2 and = 0.61 for i = 3, 4, calculate P(system works). (Round your answer to four decimal places.)

Answer 1

Answer:

P [ A U B ] = 0.9668

Step-by-step explanation:

Given:

- The probability of component i works:

                               P ( i ) = 0.77  ,  i = 1 , 2     ...... parallel

                               P ( i ) = 0.61  ,   i = 3 , 4    ...... series

Find:

Calculate P(system works)

Solution:

- Calling i = 1, 2 as subsystem A and i = 3 ,4 as subsystem B.

- So,

             P [ A ] = Either 1 or 2 components work = P [ A_1 U A_2 }

             P [ B ] = Both 3 & 4 components work = P [ B_1 ] * P [ B_2 ]

So the probability of system working is the union of A and B for the system to work, remember that each component i working is independent of each other. We have:

             P [ A U B ] = P [ A ] + P [ B ] - P [ A ] * P [ B ]

             P [ A U B ] = P [ A_1 U A_2 ] + P [ B_1 ] * P [ B_2 ] - P [ A ] * P [ B ]

             P [ A U B ] = P [ A_1 ] + P [ A_2 ] - P [ A_1 ] * P [ A_2 ] + P [ B_1 ] * P [ B_2 ] - P [ A ] * P [ B ]

             P [ A U B ] = (0.77 + 0.77 - 0.77^2) + (0.61^2) - P [ A ] * P [ B ]

             P [ A U B ] = (0.9471) + (0.3271) - P [ A ] * P [ B ]

             P [ A U B ] = (0.9471) + (0.3271) - (0.9471) * (0.3271)

             P [ A U B ] = 0.9668