{"id":4826,"date":"2021-08-29T12:06:59","date_gmt":"2021-08-29T12:06:59","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4826"},"modified":"2021-11-27T23:09:39","modified_gmt":"2021-11-27T17:39:39","slug":"addition-theorem-of-probability","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/addition-theorem-of-probability\/","title":{"rendered":"Addition Theorem of Probability – Statement and Proof"},"content":{"rendered":"
Here you will learn addition theorem of probability for two and three events with statement and proof.<\/p>\n
Let’s begin –<\/p>\n
For two events :<\/strong><\/p>\n If A and B are two events associated with a random experiment, then<\/p>\n P(\\(A \\cup B\\)) = P(A) + P(B) – P(\\(A \\cap B\\))<\/p>\n<\/blockquote>\n Proof : <\/strong>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.<\/p>\n Let \\(m_1\\), \\(m_2\\) and m be the number of elementary events favourable to A, B and \\(A \\cap B\\) respectively then,<\/p>\n P(A) = \\(m_1\\over n\\), P(B) = \\(m_2\\over n\\) and P(\\(A \\cup B\\)) = \\(m\\over n\\)<\/p>\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<\/p>\n \\(m_1\\) – m + \\(m_2\\) – m + m = \\(m_1\\) + \\(m_2\\) – m<\/p>\n So, P(\\(A \\cup B\\)) = \\(m_1 + m_2 – m\\over n\\) = \\(m_1\\over n\\) + \\(m_2\\over n\\) – \\(m\\over n\\)<\/p>\n \\(\\implies\\) P(\\(A \\cup B\\)) = P(A) + P(B) – P(\\(A \\cap B\\))<\/p>\n Corollary : <\/strong>If A and B are mutually exclusive events, then<\/p>\n (\\(A \\cup B\\)) = 0<\/p>\n \\(\\therefore\\) P(\\(A \\cup B\\)) = P(A) + P(B)<\/p>\n This is the addition theorem for mutually exclusive events.<\/p>\n For three events :<\/strong><\/p>\n If A, B, C are three events associated with a random experiment, then<\/p>\n 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\\))<\/p>\n<\/blockquote>\n Corollary : <\/strong>If A, B, C are mutually exclusive events, then<\/p>\n P(\\(A \\cup B\\)) = P(\\(B \\cup C\\)) = P(\\(A \\cup C\\)) = P(\\(A \\cap B\\cap C\\)) = 0<\/p>\n \\(\\therefore\\) P(\\(A \\cup B\\cup C\\)) = P(A) + P(B) + P(C)<\/p>\n This is the addition theorem for three mutually exclusive events.<\/p>\n\n\n\n
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