Independence: Consider the system of components....

I have a problem that i'm confused about, kind of at my wits end:

Q: Consider the system of components connected as in the accompanying picture. Components 1 and 2 are connected in parallel, so that subsystem works iff either 1 or 2 works; since 3 and 4 are connected in series, that subsystem works iff both 3 and 4 work. If components work independently of one another and P(component works) = .9, calculate P(system works).

PS: The book really has it spelled iff.. not sure why?

Anyway, I am told I should assume this "System" acts like a circuit. So, in words, the diagram should mean that the system will work if 1 OR 2 works OR if BOTH 3 and 4 work. I took this to mean, translating words to probabilities:
P(1 U 2) U P(3 n 4)
= [P(1) + P(2) - P(1)P(2)] U [P(3)*P(4)]
= [.99] U [.81] = [.99 + .81 - (.99)*(.81)] = 0.9981

But this is apparently wrong. I don't understand why though. Appreciate any help you can provide.

Re: Independence

I believe "iff" stand for "if and only if".

Re: Independence

MarkSA said:

[.99] U [.81] = [.99 + .81 - (.99)*(.81)] = 0.9981
But this is apparently wrong. I don't understand why though. Appreciate any help you can provide.

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Why do you think that it is wrong? Do you know what is suppose to be correct?
I have done problem in several ways, all give that same answer.

Re: Independence

Why do you think that it is wrong? Do you know what is suppose to be correct?
I have done problem in several ways, all give that same answer.

Click to expand...

I asked my teacher about it since I wasn't totally sure how the system was supposed to work, but when I showed her my work with the answer 0.9981, she said "that makes no sense in probability. use correct notation and you will get the right answer."

Re: Independence

MarkSA said:

I asked my teacher about it since I wasn't totally sure how the system was supposed to work, but when I showed her my work with the answer 0.9981, she said "that makes no sense in probability. use correct notation and you will get the right answer."

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I would really be interested to know your teacher’s reading of this problem.
I hope that you will post it and her reasons.

It seems perfectly clear the we can change the variables to: A=”1 works”, B=”2 works”, and C=”3 and 4 work. Then assume ‘broad’ independence.
Thus we have \(\displaystyle P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C)+ P(A \cap B \cap C)=0.9981\).

thanks pka, I will ask her about it again when we have class and let you know what she says. She's not a native English speaker, so it's possible I am misunderstanding her.

MarkSA said:

thanks pka, I will ask her about it again when we have class and let you know what she says. She's not a native English speaker, so it's possible I am misunderstanding her.

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Mark, did you ever get a response from you lecturer/instructor on her reading of this problem? If you did, please post it.