# Circuit drawings probability

• Sep 14th 2010, 09:11 AM
uberbandgeek6
Circuit drawings probability
In each of the “circuit drawings” below, every one of the numbered switches will be closed (which allows current to flow) with probability p, independently of all other switches. In each drawing, find the probability that current can flow between A and B.

http://farm5.static.flickr.com/4147/...ca41def8b0.jpg

I'm not really sure where to begin with this. For (a), the current would flow with the anything except if only (1,2), (3,4), (1), (2), (3), (4), or (none) are closed. But where do I go from here?
• Sep 14th 2010, 09:32 AM
Unknown008
(a)

For the current to pass the first switch (that is the point midway between A and B), switch 1, 2 or both should be closed.

Hence, the probability becomes P(1 close, 2 open) + P(2 close, 1 open) + P(1 and 2 close) = p(1-p) + p(1-p) + p^2

Same thing for the current to pass through the second switch.

So, the total probability is when both events occur, or: P(pass 1st) x P(pass 2nd) = [p(1-p) + p(1-p) + p^2][p(1-p) + p(1-p) + p^2]

Try out the others? (Happy)

EDIT: This simplifies to:

\$\displaystyle [p(1-p) + p(1-p) + p^2][p(1-p) + p(1-p) + p^2] = [p-p^2 + p - p^2 + p^2]^2 = (2p - p^2)^2 = 4p^2 - 4p^3 + p^4\$
• Sep 15th 2010, 08:39 AM
uberbandgeek6
For (b) then, it would be:
P(C1 n C2 n C3 n C4) + P(C1 n C2 n C3 n C4') + P(C1 n C2' n C3 n C4) + P(C1 n C2' n C3 n C4') + P(C1 n C2 n C3' n C4) + P(C1' n C2 n C3 n C4) + P(C1' n C2 n C3' n C4) = -p^4 + 2p^2
Would this be correct? (not sure if there would be a faster way to do this)
• Sep 15th 2010, 09:00 AM
Unknown008
Yes, that's what I got too (Happy)