A fair coin and an unbiased die are tossed.

If a fair coin and an unbiased die are tossed, then the sample space S is given by,

$\mathrm{S}=\left\{\begin{array}{l}(\mathrm{H}, \mathrm{l}),(\mathrm{H}, 2),(\mathrm{H}, 3),(\mathrm{H}, 4),(\mathrm{H}, 5),(\mathrm{H}, 6), \\ (\mathrm{T}, 1),(\mathrm{T}, 2),(\mathrm{T}, 3),(\mathrm{T}, 4),(\mathrm{T}, 5),(\mathrm{T}, 6)\end{array}\right\}$

Let A: Head appears on the coin

$\mathrm{A}=\{(\mathrm{H}, \mathrm{l}),(\mathrm{H}, 2),(\mathrm{H}, 3),(\mathrm{H}, 4),(\mathrm{H}, 5),(\mathrm{H}, 6)\}$

$\Rightarrow \mathrm{P}(\mathrm{A})=\frac{6}{12}=\frac{1}{2}$

B: 3 on die $=\{(\mathrm{H}, 3),(\mathrm{T}, 3)\}$

$P(B)=\frac{2}{12}=\frac{1}{6}$

$\therefore \mathrm{A} \cap \mathrm{B}=\{(\mathrm{H}, 3)\}$

$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{12}$

$P(A) \cdot P(B)=\frac{1}{2} \times \frac{1}{6}=P(A \cap B)$

Therefore, A and B are independent events.