Independence of drawing a labelled colored ball A bag contains $5$ red balls and $5$ blue balls. The red balls are labelled $1\cdots5$ and the blue balls are labelled similarly. Let $\text{A}$ denote the event "Ball drawn is red", and $\text{B}$ denote the event "Ball is labelled by $2$". Are $\text{A}$ and $\text{B}$ independent events?
Please clarify.
 A: $P(A) = {5\over10} = {1\over 2}$
$P(B) = {2\over10} = {1\over5}$
$P(A)\cdot P(B) ={1\over 10}$
$P(A\cap B) = {1\over 10}$
Since $P(A)\cdot P(B) = P(A\cap B)$, the mathematical definition of independence is satisfied, so events $A$ and $B$ are independent 
A: A sufficient (though not necessary) condition for independence of $A$ and $B$ is that each ball is equally likely to be drawn.  My guess is that you are to assume this to be the case (based on some principle of indifference), but you should be clear that you are in fact assuming that.  You could, in fact, have been given a completely different set of probabilities for balls to be drawn, and $A$ and $B$ could still be independent.  For instance, suppose
$$
P(R2) = 1/4, P(Rx) = 1/8 \text{ otherwise}
$$
$$
P(B2) = 1/12, P(Bx) = 1/24 \text{ otherwise}
$$
Then $P(A) = 1/4 + 4 \cdot 1/8 = 3/4$, $P(B) = P(R2)+P(B2) = 1/3$, and $P(A) \cdot P(B) = 3/4 \cdot 1/3 = 1/4 = P(R2)$, so $A$ and $B$ are independent.  As Rowan indicates in the comments, it's not in general trivial to see whether two events are independent.  In this case, it is fairly easy to tell (with some experience), but that's not generally true.
