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The probability of the closing of the ith relay in the circuit below is given by $p_i$, $i$ = 1,2,3,4,5. If all the relays function independently, what is the probability that a current flows between $A$ and $B$ for the circuit below?

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So far, I have broken it down as: the 4 events needed for the current to flow from $A$ to $B$ is: $P(p_1 p_4)$, $P(p_2 p_5)$, $P(p_1p_3p_5)$, and $P(p_2p_3p_4)$. Therefore the probability we're looking for is:

[ $P(p_1 p_4) \bigcup P(p_2 p_5)$] $\bigcup$ [$P(p_1p_3p_5) \bigcup P(p_2p_3p_4)$]

=[$P(p_1 p_4) + P(p_2 p_5)- P(p_1p_4p_2p_5)$] $\bigcup$ $P(p_3)[P(p_1p_5)+P(p_2p_4)-P(p_1p_5p_2p_4)]$

=[$P(p_1 p_4) + P(p_2 p_5)- P(p_1p_4p_2p_5)$] $+$ $P(p_3)[P(p_1p_5)+P(p_2p_4)-P(p_1p_5p_2p_4)]$ - ( [$P(p_1 p_4) + P(p_2 p_5)- P(p_1p_4p_2p_5)$]*$P(p_3)[P(p_1p_5)+P(p_2p_4)-P(p_1p_5p_2p_4)]$ )

Is this correct?

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  • $\begingroup$ I think it's right, but I wouldn't swear to it. You've got the right idea. $\endgroup$ – saulspatz Mar 3 '19 at 23:53
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There is an easier way.

  1. Suppose that "3" is closed. $$p^{(1)}=(p_1\cup p_2) \cap (p_4 \cup p_5) = (p_1+p_2-p_1p_2)(p_4+p_5-p_4p_5)$$

  2. Suppose that "3" is open. $$p^{(2)}=(p_1 \cap p_4) \cup (p_2 \cap p_5) = p_1p_4+p_2p_5-p_1p_4p_2p_5$$

Finally, you get $$p=p_3\cdot p^{(1)} + (1-p_3)p^{(2)}$$

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  • $\begingroup$ I like your logic; but if gate 3 is closed, then shouldn't the logic for $p^(1)$ be: 1 and 5 OR 2 and 4? $\endgroup$ – Jaigus Mar 4 '19 at 0:19
  • $\begingroup$ @Jaigus If "3" is closed, then we have (1 in parallel with 2) in series with (4 in parallel with 5). For a parallel combination to conduct we need that at least one conducts, hence OR, and for a series combination to conduct we need that both conduct, hence AND. $\endgroup$ – Haris Gušić Mar 4 '19 at 0:24
  • $\begingroup$ I don't think this problem is considering parallel and series in mind; I believe its simply considering the walking the path. And, for the final equation at the bottom, should we subtract (p3 * p1)*(1-p3(p2))? $\endgroup$ – Jaigus Mar 4 '19 at 0:30
  • $\begingroup$ @Jaigus Well, it is a circuit problem, so all the rules for circuits apply. The last equation is simply the formula for total probability, i.e. $p=p_3 \cdot p(\text{flows }|p_3) + \bar p_3 \cdot p(\text{flows }|\bar p_3)$. $\endgroup$ – Haris Gušić Mar 4 '19 at 0:35

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