(N/A) The sample space $S$ for tossing a fair coin and an unbiased die is given by:
$S = \{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)\}$
The total number of outcomes is $n(S) = 12$.
Let $A$ be the event 'head appears on the coin':
$A = \{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)\}$
$P(A) = \frac{n(A)}{n(S)} = \frac{6}{12} = \frac{1}{2}$.
Let $B$ be the event '$3$ on the die':
$B = \{(H, 3), (T, 3)\}$
$P(B) = \frac{n(B)}{n(S)} = \frac{2}{12} = \frac{1}{6}$.
The intersection $A \cap B$ is the event 'head appears on the coin and $3$ appears on the die':
$A \cap B = \{(H, 3)\}$
$P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{12}$.
Now,check for independence:
$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$.
Since $P(A \cap B) = P(A) \times P(B)$,the events $A$ and $B$ are independent.