If we have (n) independent events with probability P1, P2, .... Pn, what is the probability of the union of events (1, 2, .... n) ? is their a closed form expression for the same ?
We all know the famous rule when we have two independent events:
P(1 union 2) = P1 + P2 - P1 . P2 ,
However the question is :
P(1 union 2 union .... union n) = ?
See formulas (6.4) and (6.6) at next:
https://math.dartmouth.edu/archive/m19w03/public_html/Section6-2.pdf
P(1 union 2) = P1 + P2 - P1 . P2 ,
replace now P2 by P2 + P3 - P2 . P3 ,
you get
P(1 union 2 union 3) = P1 + (P2+P3-P2.P3) - P1 . (P2+P3-P2P3) ,
=P1+P2+P3-P1P2-P1P2-P2P3+P1P2P3
and so on (induction proof) you get the Formula attached by Sundar Vs.
good luck
If the family (A_i)_i is independent, so is (A_i^c)_i. We can thus compute the probabilty of of complement of the union. We should find $1-\prod_{i=1}^n(1-p_i)$.
Dear Akram
If you are asking about the independent events, then P(1 union 2 union... n)=P(1) + P(2) + P(3) + ..... + P(n)
Hi Ravi, the events are independent by not mutually exclusive, your answer is correct for mutually exclusive events.
You can calculate this as 1 - P(complement) where the complement is the intersection of de complements of these events, en since these are independent, you can write this as one minus the product of the probabilities of the complements.
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