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Math Help - Circuit drawings probability

  1. #1
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    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.




    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?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    (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?

    EDIT: This simplifies to:

    [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
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  3. #3
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    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)
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Yes, that's what I got too
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