## Solution :

Let \(A_1\), \(A_2\), \(A_3\) be the events of match winning in first, second and third matches respectively and whose probabilities are

P(\(A_1\)) = P(\(A_2\)) = P(\(A_2\)) = \(1\over 2\)

\(\therefore\) Required Probability

= P(\(A_1\))P(\(A_2’\))P(\(A_3\)) + P(\(A_1’\))P(\(A_2\))P(\(A_3\))

= \(({1\over 2})^3\) + \(({1\over 2})^3\) = \(1\over 8\) + \(1\over 8\) = \(1\over 4\)

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