Generalizing the total probability of simultaneous occurrences for independent events I want to generalize a formula and I need your help with this. This is not my homework or assignment but I need to come up with a concise formula that fits my documentation.
Background for my problem:
Considering all events to be independent of each other, 
Let the probability of $Event$ $0$ be $P_{0}$ and $Event$ $1$ be $P_{1}$ and so on ...
Then the probability that two events occur simultaneously is : $P_0P_1$ . This is nothing but the area of intersection of two circles $P_0$ and $P_1$.
Continuing the same way, 
The total probability of simultaneous occurrence in case of three events is: 
$P_0 P_1 + P_0P_2 + P_1P_2 - 2 P_{0}P_{1}P_{2}$.
Also can be visualized by drawing three intersecting circles. 
One clarification here: This gives me the total probability of any two events plus  all the three occurring at the same time right?
I cannot visualize the formula by drawing circles anymore. How can the above formula be generalized to get the probability of simultaneous occurrences when there are 4, 5, ... independent events.
I have seen that the inclusion-exclusion principle is the answer. But I am not able to get an intuition for it. The inclusion-exclusion principle gives the probability of 2,3,4 sets intersecting but isn't my question different? 
I get this doubt because, probability of four independent events occurring simultaneous is: $P_0P_1P_2P_3$. But what I need is a general formula for the total probability of two or more independent events at the same time.
Can anyone of you please throw some light?

Yes indeed I meant that the probability of "two or more events". The answer you have given is very precise and the one I was looking for. Yes it is boring to try to visualize circles when the number exceeds three. Instead, in general if use 1 - ($w_0 + w_1$) then we should land up correctly given that the events are independent. Thank you so much.
 A: $$1-\left(1+\sum_{i=1}^n\frac{P_i}{1-P_i}\right)\cdot\prod_{k=1}^n(1-P_k)$$
A: It seems that by "The total probability of simultaneous occurrence " you actually mean "the ocurrence of two or more events"
The inclusion-exclusion principle, in its most basic form, is usually apply to get the probability of "one or more" events (i.e.: an OR). It can certainly be applied here, also, but it gets a little more complicated. Your "elementary" events are now rather the pairs $A_{ij}=E_i \cap E_j$ (the intersections of two circles in your Venn diagram). Applying the inclusion-exclusion principle to the three events you'd get
$$P(A_{01} \cup  A_{12} \cup  A_{02}) =\\ P(A_{01})+P(A_{12})+P(A_{02})-
P(A_{01} \cap A_{12})-P(A_{01} \cap A_{12})-P(A_{12} \cap A_{02}) + P(A_{12} \cap A_{02} \cap A_{02})=\\
$$
which, because  $P(A_{12} \cap A_{02}) =  P(E_0\cap E_1 \cap E_2)= P(A_{12} \cap A_{02} \cap A_{12})$ equals
$$P(E_0 \cap E_1) + P(E_1 \cap E_2) +P(E_0 \cap E_2) - 2 P(E_0\cap E_1 \cap E_2)
$$
To extend this to 4 events is straighwforward, but borish and hard to visualize.
Of course, in your case, if the events are independent, it's more easy to compute the complement:
$1-(w_0 + w_1)$ where $w_0=$ probability of no event  and $w_1=$ probability of exactly one event.
$w_0 = (1-P_0)(1-P_1)(1-P_2)(1-P_3)$
$w_1 = P_0(1-P_1)(1-P_2)(1-P_3) + (1-P_0)P_1(1-P_2)(1-P_3) + (1-P_0)(1-P_1)P_2(1-P_3)+ (1-P_0)(1-P_1)(1-P_2)P_3 $
A: Inclusion-Exclusion formulas
Let $A_1, A_2,\dots, A_n$ be  events.
The probability that exactly $k$ of them occur $(k\geq 0)$ is

$$p_n(k)=\sum_{i=k}^n  (-1)^{i-k}{i\choose k}S_i,$$

while the probability that $k$ or more of them occur $(k\geq 1)$ is

$$P_n(k)=\sum_{i=k}^n (-1)^{i-k}{i-1\choose k-1}S_i.$$

Here $S_0=1$, $S_1=\sum_i P(A_i)$, $S_2=\sum_{i< j} P(A_i A_j)$, 
$S_3=\sum_{i< j<k} P(A_i A_j A_k)$, etc.
A: Just use the opposite event : one or less.
For example for 3 events (none+only $P_1$+only $P_2$+only $P_3$): $$E_3=(1-P_1)(1-P_2)(1-P_3)+P_1(1-P_2)(1-P_3)+(1-P_1)P_2(1-P_3)+(1-P_1)(1-P_2)P_3$$
Then what you want is only $1-E_n$, that is easy to compute now.
A: $P[\bigcup_{k=1}^N P_k]=\sum_{k=1}^NP_k-\sum_{j<k}^NP_jP_k+\dots+(-1)^{N+1}P_1P_2\dots P_N$
