# How do I find the probability for this circuit to run current?

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? 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?

• I think it's right, but I wouldn't swear to it. You've got the right idea. – saulspatz Mar 3 '19 at 23:53

## 1 Answer

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 $$p=p_3\cdot p^{(1)} + (1-p_3)p^{(2)}$$

• 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? – Jaigus Mar 4 '19 at 0:19
• @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. – Haris Gušić Mar 4 '19 at 0:24
• 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))? – Jaigus Mar 4 '19 at 0:30
• @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. $p=p_3 \cdot p(\text{flows }|p_3) + \bar p_3 \cdot p(\text{flows }|\bar p_3)$. – Haris Gušić Mar 4 '19 at 0:35