Disjoint Union and Total Probability Theorem question There is some clarification I need with respect to a question on the following formula. 


  
*For any sequence of events $B_1,B_2,\dotsc,B_n$,
  $$\mathbb P\left(\bigcup_{i=1}^nB_i\right) =\sum_{i=1}^n\mathbb P(B_i)- \sum_{i<j} \mathbb P(B_i\cap B_j)+\sum_{i<j<k}\mathbb P(B_i\cap B_j\cap B_k)+\dotsb+(-1)^{n+1}\mathbb P\left(\bigcap_{i=1}^n B_i\right).$$
  

Image.
Would someone mind clarifying the last part with $(-1)^{n+1}P(\bigcup_{i = 1 }^n B_i)$, I cant seem to find where this is coming from and I have tried looking on wikipedia for both union and disjoint union but did not have success in finding this specific definition. However, I know that it the first 3 cases are true such as the union of 3 events. 
I would appreciate any explanation, thank you. 
 A: It is the Inclusion Exclusion Principle.  The equation can be broken up to say:
$$\begin{array}{c:l}
\Bbb P(\bigcup\limits_{i=1}^n B_i) & \text{The probability of the union of all $n$ events}
\\[2ex] =  & \text{equals}
\\[2ex] \sum\limits_{i=1}^n \Bbb P(B_i) & \text{the sum of the probability of each of the events}
\\[1ex] - & \text{minus (to exclude)}
\\[2ex] \sum\limits_{1\leq i < j\leq n}\Bbb P(B_i\cap B_j) & \text{ the sum of the probabilities of all intersections of pairings of events}
\\[1ex] + & \text{plus (to include)}
\\[2ex] \sum\limits_{1\leq i < j < k\leq n}\Bbb P(B_i\cap B_j\cap B_k) & \text{the sum of the probabilities of all intersections of triplets of events}
\\[1ex] +\cdots + & \text{summing et cetera in that pattern of alternating exclusion/inclusion}
\\[2ex] (-1)^{n+1}\Bbb P(\bigcap\limits_{i=1}^n B_i) & {\text{with the final term of the probability of the intersection of all $n$ events}\\ \text{ added if $n$ is odd or substracted is $n$ is even.} }
\end{array}$$
The reason for the process is making sure we don't over measure the probabilities of overlapping events (that is: don't over count outcomes). 
