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If two events A and B are mutually exclusive, then the probability of occurrence of either A or B is given by the sum of their probabilities, i.e.
Probability of (A or B) = Probability of A + Probability of B.
Symbolically, P(A+B) = P(A) + P(B)
This is also known as Addition Theorem.
Proof: Let us suppose that a random experiment has n possible outcomes, which are mutually exclusive, exhaustive and equally likely. If m1 of these cases are favourable to the event A, and m2 cases are favourable to the event B, then the probabilities of these events are, by the classical definition,
P(A) = m1/n,
P(B) = m2/n
However, since the events A and B are mutually exclusive (i.e. both of them cannot occur simultaneously), the m1 cases favourable to A are completely distinct from the m2 cases favourable to B. The number of cases favourable to either A or B is therefore (m1+m2).
Hence, P(A+B) = (m1+m2) / n
But, (m1+m2)/n = m1/n + m2/n = P(A) + P(B)
Thus, P(A+B) = P(A) + P(B)
Probability of (A or B) = Probability of A + Probability of B.
Symbolically, P(A+B) = P(A) + P(B)
This is also known as Addition Theorem.
Proof: Let us suppose that a random experiment has n possible outcomes, which are mutually exclusive, exhaustive and equally likely. If m1 of these cases are favourable to the event A, and m2 cases are favourable to the event B, then the probabilities of these events are, by the classical definition,
P(A) = m1/n,
P(B) = m2/n
However, since the events A and B are mutually exclusive (i.e. both of them cannot occur simultaneously), the m1 cases favourable to A are completely distinct from the m2 cases favourable to B. The number of cases favourable to either A or B is therefore (m1+m2).
Hence, P(A+B) = (m1+m2) / n
But, (m1+m2)/n = m1/n + m2/n = P(A) + P(B)
Thus, P(A+B) = P(A) + P(B)
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