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 works) = 0.80, calculate P(system works). (Round your answer to four decimal

0.9856

Step-by-step explanation:

Let A₁, A₂, A₃, and A₄ denote the four components and P(A₁), P(A₂), P(A₃) and P(A₄) denote the probability that the components work and P(A₁) = P(A₂) = P(A₃) = P(A₄) = 0.8. For the system to work, either A₁ or A₂ works and A₃ and A₄ works. This is denoted by (A₁ U  A₂) U (A₃ ∩ A₄). The probability that the system works is calculated as ]follows. Let S₁ = (A₁ U  A₂) represent system 1 and S₂ = (A₃ ∩ A₄) represent system 2. P[(A₁ U  A₂) U (A₃ ∩ A₄)] = P(S₁ U S₂), the probability that the system works.

The probability of any two events is P(A U B)= P(A) + P(B) - P(A ∩ B). Since the events are independent, P(A ∩ B) = P(A)P(B).

P(A₁ U A₂) = P(A₁) + P(A₂) - P(A₁ ∩ A₂) = 0.8 + 0.8 -0.8² = 1.6 - 0.64 = 0.96. P(A₃ ∩ A₄) = P(A₃)P(A₄) = 0.8 × 0.8 = 0.64.

P(S₁ U S₂)= P(S₁) + P(S₂) - P(S₁ ∩ S₂) = 0.96 + 0.64 - 0.96 × 0.64 = 1.6 + 0.6144 = 0.9856

yes. we do proportions in probabilities.

a

step-by-step explanation: