# What is the probability that the current will flow in the circuit, if P(1/2, that the bulb will work) and there are 7 bulbs.

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 $$P=(1-(((1-0,5^2)\cdot 0,5^2)\cdot 0,5^2)\cdot 0,5=0,3984375$$

• Your probability is still low. Note that if f and g work then current will flow. So the probability has to be $\geq 1/2^2=0.25$! – Robert Z Oct 12 '19 at 13:37
• Finally, it seems that now we agree. Was my answer of any help? – Robert Z Oct 12 '19 at 13:59

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 $$1-(1-p(a))\cdot(1-p(b))$$.

• Yes, I understood it after posting and so I updated the result, is it now correct? – user Oct 12 '19 at 13:04
• No, $a\lor b$ is not $1/2+1/2=1$, it is $1-(1/2)^2$ i.e. the complement that they are both broken. – Robert Z Oct 12 '19 at 13:07
• @user I edited my answer. Is it clear now? Any further comment? – Robert Z Oct 12 '19 at 13:35
• Ok, my questions: Then the probability use of sets and operations with them is largely different from those operations within logic and set theory? Because How can you multiply (conjuagate) sets that don't have any point in common? $(1-p(a))\cdot (1-p(b)).$ Secondly as far as I can understand your given formula differs from your calculations because you subtract from 1 a multiplication of $(a \lor b) \land c \land b$ and $e \lor f$, but in the formula you should separately subtract the expressions from one and then again subtract the result from 1. – user Oct 12 '19 at 14:13
• If you expand the product $1-(1-p(a))\cdot (1-p(b))$ you will find $p(a)+p(b)-p(a)p(b)$. As regards your second comment, note that $(1-1/2)=1/2$ and $$p(a\lor b\lor c)=1-p(\lnot a\land \lnot b\land \lnot c)=1-(1-p(a))(1-p(b))(1-p(b))$$. – Robert Z Oct 12 '19 at 14:17