Here you will learn addition theorem of probability for two and three events with statement and proof.
Let’s begin –
Addition Theorem of Probability
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.
