How to calculate intersection and union of probabilities? lets say I have a switch A with 3 legs, each leg has 0.8 chance to be connected (and then electricity will flow), we need only 1 leg connected for A to transfer the electricity (so sorry I didn't explain it that well I'm having hard time to translate this problem)
So I calculated the chance of A to transfer electricity by doing $1-(0.2)^3$  which is $124\over125$ which I think is true.
The problem is I wanted to say that A will transfer elecricity only if 1 of his legs will be connected so its like saying $0.8 + 0.8 + 0.8$ which is obviously wrong(over $1$) so I used the weird formula that says to do like this: $0.8 + 0.8 + 0.8 - (0.8)^2 -(0.8)^2 - (0.8)^2 + (0.8)^3 = {124\over125}$ too.
My only problem is that I didnt understand why I had to use that formula and why I could multiple probabilities for the intersection but couldn't just sum them for the union.
Thanks in advance
 A: This is the inclusion-exclusion principle. We have that
$$
\mathbb{P}(A_1\cup A_2\cup A_3)=\mathbb{P}(A_1)+\mathbb{P}(A_2)+\mathbb{P}(A_3)-\mathbb{P}(A_1\cap A_2)-\mathbb{P}(A_1\cap A_3)-\mathbb{P}(A_2\cap A_3)+\mathbb{P}(A_1\cap A_2\cap A_3)
$$
for any three events $A_1$, $A_2$ and $A_3$. The following picture illustrates the idea behind this formula:

A: The Product Rule applies to events which are independent.  Only then is the probability of the intersection equal to the product of the probabilities of the events.
$\mathsf P(A\cap B) ~=~ \mathsf P(A)~\mathsf P(B)~$ only when events $A$ and $B$ are independent.
(When dealing with more than two events we require mutual independence.)
Otherwise conditional probability must be used: $\mathsf P(A\cap B)~=~\mathsf P(A)~\mathsf P(B\mid A)\\\qquad\qquad~=~\mathsf P(A\mid B)~\mathsf P(B)$

The Addition Rule applies only when the events are mutually exclusive (also known as disjoint).   Only then is the probability of the union equal to the sum of probabilities of the event.
$\mathsf P(A\cup B)~=~\mathsf P(A)+\mathsf P(B)$
Otherwise if the events are not disjoint (ie they have common outcomes) then we would be over measuring and must exclude the measure of the intersection.
$\mathsf P(A\cup B)~=~\mathsf P(A)+\mathsf P(B) - \mathsf P(A\cap B)$
When dealing with more than two events, the principle of inclusion and exclusion is required
$\begin{align}\mathsf P(A\cup B\cup C)~=~&\mathsf P(A)+\mathsf P(B)+\mathsf P(C) - \mathsf P(A\cap B)-\mathsf P(A\cap C)-\mathsf P(B\cap C)+\mathsf P(A\cap B\cap C)\end{align}$
... and so on.
$\Box$
A: Hint One of the three legs running isnt dependent on the running of other two legs. So the events are independent events. So we just multiply their probabilities instead of adding . 
