Given: A fair coin and an unbiased die are tossed.

Let A be the event head appears on the coin. So the sample space of the event will be:

$\Rightarrow A={(\mathrm{H}, 1),(\mathrm{H}, 2),(\mathrm{H}, 3),(\mathrm{H}, 4),(\mathrm{H}, 5),(\mathrm{H}, 6)}$

$\Rightarrow P(A)=6 / 12=1 / 2$

Now, Let $B$ be the event 3 on the die. So the sample space of event will be:

$\Rightarrow \mathrm{B}={(\mathrm{H}, 3),(\mathrm{T}, 3)}$

The probability of the event will be $\Rightarrow P(B)=2 / 12=1 / 6$

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

Thus evaluating the value of parameter required to proof that the events are independent.

$\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=1 / 12 \ldots \ldots$ (1)

And
$P(A) . P(B)=1 / 2 \times 1 / 6=1 / 12 \ldots \ldots . .(2)$

From (1) and (2) $P(A \cap B)=P$ (A). $P(B)$

Therefore, A and B are independent events.