Addition Law of Probability Proof

Suppose a sample space “S” consists of “N” outcomes and the probability of each outcome is equally likely. The two events exist from the same sample space containing n1 and n2 outcomes respectively. Events are named as “x” and “y”. Both the events have nothing in common so called these events as mutually exclusive events.  As both the events are mutually exclusive events therefore, their union contains (n1 + n2) outcomes. Probability of x & y can be write as by using the classical formula of probability,
[P(Xcup Y) = frac{Xcup Y}{S (sample space))}]

[P(X cup Y) = frac{n_{1}+n_{2}}{N} = frac{n_{1}}{N}+frac{n_{2}}{N}= P(X)+ P(Y)]

Similarly, if there are “n” number of mutually exclusive events then,
[P(X_{1} cup X_{2}cup…..X_{n} ) = P(X_{1})+ P(X_{2})+ P(X_{3})……P(X_{n}) = sum_{i=1}^{n}(X_{i})]

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