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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?

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 $mathbb{P}(I = 0) =1-mathbb{P}(I neq 0)$.

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