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- Jan 26, 2012

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I have two solutions to the this problem that I've found but they seem strange compared to each other.

__Solution I__

All possible paths that work must pass through circuit 5, so we can factor out that probability and tack it onto the end because they are all independent.

\(\displaystyle P(A \rightarrow D)=P[(1 \cap 2) \cup (3 \cap 4) \cup (1 \cap 4) \cup (3 \cap 2) \cap 5] \)

The above just shows that it can go through the path 1,2,5 or 3,4,5 or 1,4,5 or 3,2,5. It has to be at least one of them but it can be more. Now using independence and complements this simplifies to:

\(\displaystyle P(A \rightarrow D)=\left(1 - P \left[ (1 \cap 2)' \cap (3 \cap 4)' \cap (1 \cap 4)' \cap (3 \cap 2)' \right]

\right)*p_5=\boxed{\left[1 - (1-p_1p_2)(1-p_3p_4)(1-p_1p_4)(1-p_3p_2)

\right]p_5}\)

__Solution II__

Using a similar argument but instead of listing the paths I will list the stages. It has to pass through 1 or 3 initially, then 2 or 4 and finally 5.

\(\displaystyle P(A \rightarrow D)=P[(1 \cup 3) \cap (2 \cup 4) \cap 5]\).

Now using independence and the inclusion-exclusion principle I get the following answer:

\(\displaystyle P(A \rightarrow D) = \boxed{(p_1+p_3-p_1p_3)(p_2+p_4-p_2p_4)p_5}\)

So my main question is do you agree with both of these answers? Any comments?