# Addition Theorem of Probability – Statement and Proof

Here you will learn addition theorem of probability for two and three events with statement and proof.

Let’s begin –

For two events :

If A and B are two events associated with a random experiment, then

P($$A \cup B$$) = P(A) + P(B) – P($$A \cap B$$)

Proof : Let S be the sample space associated with the given random experiment. Suppose the random experiments results in n mutually exclusive ways. Then, S contains n elementary events.

Let $$m_1$$, $$m_2$$ and m be the number of elementary events favourable to A, B and  $$A \cap B$$ respectively then,

P(A) = $$m_1\over n$$, P(B) = $$m_2\over n$$ and P($$A \cup B$$) = $$m\over n$$

The number of elementary events favourable to A only is $$m_1$$ – m. Similarly, the number of events favourable to B only is $$m_2$$ – m. Since m events are favourable to both A and B. Therefore, the number of elementary events favourable to A or B or both i.e, $$A \cup B$$ is

$$m_1$$ – m + $$m_2$$ – m + m = $$m_1$$  + $$m_2$$ – m

So, P($$A \cup B$$) = $$m_1 + m_2 – m\over n$$ = $$m_1\over n$$ + $$m_2\over n$$ – $$m\over n$$

$$\implies$$  P($$A \cup B$$) = P(A) + P(B) – P($$A \cap B$$)

Corollary : If A and B are mutually exclusive events, then

($$A \cup B$$) = 0

$$\therefore$$  P($$A \cup B$$) = P(A) + P(B)

This is the addition theorem for mutually exclusive events.

For three events :

If A, B, C are three events associated with a random experiment, then

P($$A \cup B\cup C$$) = P(A) + P(B) + P(C) – P($$A \cap B$$) – P($$B \cap C$$) – P($$A \cap C$$) + P($$A \cap B\cap C$$)

Corollary : If A, B, C are mutually exclusive events, then

P($$A \cup B$$) = P($$B \cup C$$) = P($$A \cup C$$) =  P($$A \cap B\cap C$$) = 0

$$\therefore$$  P($$A \cup B\cup C$$) = P(A) + P(B) + P(C)

This is the addition theorem for three mutually exclusive events.