So, the problem is as follows: Calculate the probability that the current will flow in circuit if the chance that a light bulb will work is 0,5 and there are totally 7 bulbs. Here is the schema: 
I don't know whether the location of bulbs affects the independence of the bulbs working. But I only know how to calculate the probability if the bulb working counts as a independent happening. My solution – $P(\text {that all bulbs will work} )=0,5^{7}=0.0078125$. It is pretty low, because the chance that any bulb won't work is quite high. In fact I know that my answer is incorrect, because the next exercise asks to find the probability that the bulb "c" will work given that current is flowing in circuit. That means, that there is a possibility for a current flowing in circuit even if some bulb is not working? So, I guess there are some bulbs that must be working in order for a current to flow and some that can not work in case some other bulb is working. My guess would be that either a or b has to work but not necessary both of them the c and d must both be working and either e and f must work (am not totally sure about it), then g must work so that current flows. So it would be
$$P=(1-(((1-0,5^2)\cdot 0,5^2)\cdot 0,5^2)\cdot 0,5=0,3984375$$
Best Answer
If all bulbs work the current will flow for sure. But this is just a special case. More cases have to be considered.
Note that the current will flow if and only if $$(((a\lor b)\land c\land d)\lor e \lor f)\land g$$ where "letter" is true iff the corresponding bulb works.
As regards probabilities, $$p((a\lor b)\land c\land d)=\left(1-\left(\frac{1}{2}\right)^2\right)\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{3}{16}.$$ and therefore $$p(((a\lor b)\land c\land d)\lor e \lor f)=1-\left(1-\frac{3}{16}\right)\cdot \left(1-\frac{1}{2}\right)\cdot \left(1-\frac{1}{2}\right)=1-\frac{13}{64}=\frac{51}{64}.$$ Hence, the required probability is $$\frac{51}{64}\cdot \frac{1}{2}=\frac{51}{128}\approx 0.398.$$
P.S. In general, for independent events $a$ and $b$, $$p(a\lor b)=p(a)+p(b)-p(a\land b)=p(a)+p(b)-p(a)p(b)$$ which is the same of $1-(1-p(a))\cdot(1-p(b))$.