Consider a system consisting of three independent components. Suppose the lifetimes of these components follow an Exponential distribution with mean 300 hours, 410 hours, and 600 hours, respectively. a. Suppose that the system works only if all three components are functional (e.g., series connection). Find the probability that the system still works after 550 hours. That is, find the probability that all three components still work after 550 hours. Round your answer to 4 decimal places. "Hint": P(X1 > 550 hours AND X2 > 550 hours AND X3 > 550 hours)

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Answer:

0.3723

Step-by-step explanation:

We are given a system  consisting of three independent components.

Let the three components be A , B and C

The lifetimes of these components follow an Exponential distribution with mean 300 hours, 410 hours, and 600 hours, respectively

a) Suppose that the system works only if all three components are functional (e.g., series connection). Find the probability that the system still works after 550 hours.

Exponential Distribution : [tex]1-e^{-\lambda x}[/tex]

P(A)=P(x>550) =[tex]1-e^{-\lambda \times 550}[/tex]

[tex]\lambda=\frac{1}{\mu}[/tex]

P(A)=P(x>550) =[tex]1-e^{-\frac{1}{300} \times 550}[/tex]

P(A)=P(x>550) =[tex]0.8401[/tex]

P(B)=P(x>550) =[tex]1-e^{-\frac{1}{410} \times 550}[/tex]

P(B)=P(x>550) =[tex]0.7385[/tex]

P(C)=P(x>550) =[tex]1-e^{-\frac{1}{600} \times 550}[/tex]

P(C)=P(x>550) =[tex]0.6001[/tex]

Since P(A) , P(B) , P(C) are independents

So, the probability that all three components still work after 550 hours = P(A)*P(B)*P(C)=[tex]0.8401 \times 0.7385 \times 0.6001 = 0.3723[/tex]

Hence the probability that all three components still work after 550 hours is 0.3723

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