A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then, \(P(A \cup B)\) is

Solution :

A = {4. 5. 6}  and  B = {1, 2, 3, 4}

\(A \cap B\) = 4

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

\(\implies\) \(P(A \cup B)\) = 3/6 + 4/6 – 1/6 = 1

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