$A$ die marked $1, 2, 3$ in red and $4, 5, 6$ in green is tossed. Let $A$ be the event 'the number is even' and $B$ be the event 'the number is red'. Are $A$ and $B$ independent?

(N/A) When a die is thrown,the sample space $S$ is $S = \{1, 2, 3, 4, 5, 6\}$.
Let $A$ be the event that the number is even,so $A = \{2, 4, 6\}$.
Thus,$P(A) = \frac{3}{6} = \frac{1}{2}$.
Let $B$ be the event that the number is red,so $B = \{1, 2, 3\}$.
Thus,$P(B) = \frac{3}{6} = \frac{1}{2}$.
The intersection $A \cap B$ is the set of numbers that are both even and red,so $A \cap B = \{2\}$.
Thus,$P(A \cap B) = \frac{1}{6}$.
Now,calculate $P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
Since $P(A \cap B) = \frac{1}{6}$ and $P(A) \times P(B) = \frac{1}{4}$,we see that $P(A \cap B) \neq P(A) \times P(B)$.
Therefore,the events $A$ and $B$ are not independent.