. Advertisement .

..3..

. Advertisement .

..4..

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) = 0.73 for * i* = 1, 2 and = 0.65 for

*= 3, 4, calculate*

*i**P*(system works). (Round your answer to four decimal places.)

......... ADVERTISEMENT .........

..8..

Revelant knowledge

Independent Events: The probability of either in the events, A and B, which are distinct from each other is not dependent on the event that occurs both. Probability *P(AB)* is calculated by multiplying the probabilities that are independent of B and A.

Answer:_{1}work or c_{2}work or both c_{3}and c_{4}work)_{1}∪ c_{2}∪ ( c_{3}∩c_{4})) = P(c_{1}) + P(c_{2}) + P(c_{3}∩c_{4}) – P(c_{1}∩c_{2}) – P(c_{1}∩c_{3}∩c_{4}) – P(c_{2}∩c_{3}∩c_{4}) + P(c_{1}∩c_{2}∩c_{3}∩c_{4})_{1}) + P(c_{2}) + P(c_{3}).P(c_{4}) – P(c_{1}).P(c_{2}) – P(c_{1}).P(c_{3}).P(c_{4}) – P(c_{2}).P(c_{3}).P(c_{4}) + P(c_{1}).P(c_{2}).P(c_{3}).P(c_{4})^{2}– 0.73^{2}– 0.73 x 0.65^{2}– 0.73 x 0.65^{2}+ 0,73^{2}x 0,65^{2}0.9579