# Probablity of network flow

Three vertices: X, Y, Z Three edges: edge a (Vertex X to Y), edge b (Vertex Y to Z), edge c (Vertex Y to Z [yes another edge connecting it]).

I want to know the probability of X to Z (as in flow from X to Z) if edge a has probability of .9 edge b has probability of .8 P(edge a intersects edge b) = .75 edge c has probability of .5 a and b is independent of edge c

*note you can reach vertex z from X with edge a and b OR edge a and c.

So, I was thinking P(a U b) = P(a) + P(b) - P( a intersects b); P(a Union b) or P(a)*P(c) which is adding those two 1.3 or 130% which doesnt make sense.

Anyone know how to approach this problem?

them = P(a U b) and P(a U c) = .95*.45 = .4275 which seems a bit low to gettin to Z from X

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What do you mean by "P(edge a intersects edge b) = .75" -- do you mean probability that flow occurs on these is dependent (otherwise should be $.9 \cdot .8 = .72$? – gt6989b Nov 13 '12 at 23:11
yes correct. if it was independent it would be .72. independent is for a to c – antz Nov 13 '12 at 23:12
possible duplicate of math.stackexchange.com/q/236635. – robjohn Nov 14 '12 at 10:21

Let E be flow from X to Z.

For E,as you said, 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 is the very same thing as in here Calculating the probability

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great! thank you! this makes sense now – antz Nov 13 '12 at 23:21