# How do i calculate the probability of the relay in the circuits?

I am trying to solve my following probability question but I can't see how to make any progress. Any help will be highly appreciated

Question: The probability of the closing of the $i$-th relay in the circuits shown is given by $p_i$ for $i = 1,2,3,4,5$. If all relays function independently, what is the probability that a current flows between $A$ and $B$ for the respective circuits? • Have you tried systematically listing all the different combinations of closed switches for current to flow? – David Quinn Sep 4 '15 at 17:54
• The way i tried is Let E1 be the event in which the current flows through (1,2,3) and E2 be the event in which the current flows through (4,5). So the current can flow through either E1 or E2. So P(E1 U E2) = P(E1)+ P(E2) - P(E1 n E2). But not sure if that is correct – 0ptimus Sep 4 '15 at 18:27
• But you have to allow for the possibility that, for example, in part a), current will flow if 4 and 5 are closed as well as any of the other switches being closed or open – David Quinn Sep 4 '15 at 18:31
• I didn't thought about that. Thanks i will try – 0ptimus Sep 4 '15 at 18:51

Using $1-$ open, $1'-$closed: $$a) \ \ P(1,2,3)+P(4,5)-P(1,2,3,4,5).\\ b) \ \ P(1,2,5)+P(3,4,5)-P(1,2,3,4,5).\\ c) \ \ P(1,4)-P(1,3,4,5)+P(2,5)-P(2,3,4,5)+P(1,3,5)-P(1,3,4,5)+\\ P(2,3,4)-P(2,3,4,5).$$

There may be a more elegant way of doing this, but one way is simply to list the different combinations of gates which can be closed to allow current to flow. For example, for part b), we have

1. Three gates: 1,2,not3,not4,5; not1,not2,3,4,5

2. Four gates: 1,2,3,not4,5; 1,2,not3,4,5; 1,not2,3,4,5; not1,2,3,4,5

3. Five gates: 1,2,3,4,5

The other parts of the question can be done in the same way

I hadn't seen the "bumped to homepage" before.

For case a) I'd suggest you look for the probability that no current flows. So \begin{aligned} \mathbb{P}(I = 0)\ =\ &(1-p_1) \cdot (1-p_2) \cdot (1-p_3) + (1-p_4) \cdot (1-p_5)\\ &-\ (1-p_1) \cdot (1-p_2) \cdot (1-p_3)\cdot(1-p_4) \cdot (1-p_5) \end{aligned}

which follow from $\mathbb{P}(\{1,2,3\}\cup\{4,5\})=\mathbb{P}(\{1,2,3\})+\mathbb{P}(\{4,5\})-\mathbb{P}(\{1,2,3\}\cap\{4,5\})$.

At the end, calculate $\mathbb{P}(I = 0) =1-\mathbb{P}(I \neq 0)$.