Answer:
P(system) = p(subcomp1 ∩ subcomp2)
P(system) = 1.60 x 0.64
P(system) = 1.0240
Step-by-step explanation:
In the system we have 2 subsystem
Let say subsystem 1 and subsystem 2
Subsystem 1 consist of 2 components, component 1 and component 2 that subsystem 1 works if and only if either component 1 or component 2 work
I.e p(subcomp1) = p ( comp1 | comp2)
Since the P(component works) = 0.80
Therefore P(subcomp1) = p(comp1) + p(comp2)
= 0.80 + 0.80
P(subcomp1) = 1.60
Subsystem 2 consist of 2 components, component 3 and component 4 that subsystem 2 works if and only if both component 3 and component 4 work
I.e p(subcomp2) = p ( comp3 ∩ comp4)
Since the P(component works) = 0.80
Therefore P(subcomp2) = p(comp3) x p(comp4)
= 0.80 x 0.80
P(subcomp2) = 0.64
Therefore since the component are independent of one another
P(system) = p(subcomp1 ∩ subcomp2)
P(system) = 1.60 x 0.64
P(system) = 1.0240
Answer:
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
