When a coin is thrown, head or tail will occur.
∴ Probability of getting head P(A) = \(\frac{1}{2}\).
when a dice is tossed 1, 2, 3, 4, 5, 6 one of them will appear.
∴ Probability of getting 3 = P(B) = \(\frac{1}{6}\).
When a coin and a dice are tossed, total number of cases are
H1, H2, H3, H4, H5, H6
T1, T2, T3, T4, T5, T6
Head and 3 will occur only in 1 way.
∴ Probability of getting head and 3 = \(\frac{1}{12}\).
i.e; P(A ∩ B) = \(\frac{1}{12}\).
∴ P(A) × P(B) = \(\frac{1}{2}\) × \(\frac{1}{6}\) = \(\frac{1}{12}\)
∴ P(A ∩ B) = P(A) × P(B)
⇒ Events A and B are independent