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A company has an electric network with three vertices i, j and k. One line goes from i to j, which we will denote as "a" and denote "A" if the event works. There is a line b from j to k and another line c from j to k (there are no other lines). Denote B if b works and denote C if the event c works. Denote E as the event that electricity can flow from i to k. Imagine a powerplant in i and an electrical company in k. A and B are not independent, instead:

$$P(A) = 0.9, P(B) = 0.8, P(A \cap B) = 0.75$$

we guess $P(C) = 0.5$ and that any event related to a and b is independent of C. Calculate the probability of E.

My attempt at a solution:

I know that $A \cup B, A \cap B, A, B,A - B, B-A$ are all independent of C. I also know in order for A and B to be independent then $(P\cap B) = P(A)P(B)$, which is not the case here and already stated in the question. Knowing these then is $$P(E) = P((A \cap B) \cup C) = P(A \cap B) +P(C) - P(A\cap B \cap C) = [(.9)(.8) + (.5) - [(.9)(.8)(.5)]]= .86$$

Is this correct?

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Hint: There are $8$ possibilities for the three links to be working/not working. So, set up a truth table and see if you can figure out an expression for the event $E$ in terms of $A$, $B$, and $C$. As helopticor says, your expression $E = (A\cap B) \cup C$ is not correct. – Dilip Sarwate Nov 13 '12 at 21:14
@DilipSarwate can you further elaborate. What do you mean helopticor and my conclusion about $P(E)$ is not correct? – Q.matin Nov 13 '12 at 21:37
@Q.matin The event you want (from the description) is $E=A\cap\left(B\cup C\right)$. – Daryl Nov 13 '12 at 22:42
@Daryl how will I be able to calculate $(B\cup C)$? – Q.matin Nov 13 '12 at 22:47
@Q.matin Your expression $E = (A\cap B)\cup C$ is not correct; as Daryl pointed out, it should have read $E = A\cap (B \cup C)$ since the link between $i$ and $j$ must work (event $A$) and at least one of the two links between $j$ and $k$ must also work (event $B \cup C$). A user named helopticor posted an answer (actually more of a comment) saying that your expression was incorrect, but has apparently deleted the answer. – Dilip Sarwate Nov 14 '12 at 3:30
up vote 3 down vote accepted

For E you need A and at least one of B or C so

$P(E) = P (A\cap(B \cup C))= $

by De Moivre rule $=P ((A\cap B )\cup (A\cap C))= P(A \cap B) + P(A \cap C) -P((A\cap B) \cap (A \cap C))= $

because of independence of C from all other events $=P(A \cap B) + P(A)P(C) -P(A\cap B\cap C) = P(A \cap B) + P(A)P(C) -P(A\cap B)P(C)=$

inputing values $.75 + .9*.5-.75*.5=.825$

which indeed is greater then $ P(A \cap B) $ as expected since we have C, a backup line.

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Thank you! I am confused because I have gotten three diferent answers for this problem and I do not know who is right? – Q.matin Nov 13 '12 at 23:17
Thank you, Golob! – Q.matin Nov 13 '12 at 23:23
As Daryl said, the event you want to happen is to get flow from i to j AND flow from j to k. For first one you need A to happen. For the second one you need either of B or C to happen. That leads us back to $P(E) =P(A \cap ( B \cup C))$, and math is given above. – Golob Nov 14 '12 at 7:39

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