Probability and independence I am trying to solve a particular probability question. 
I have a fair 10-sides die, whose sides are labelled 1 through 10. I am trying to find the probability of rolling a multiple of 5 or an odd number. 
I find the probability as: 
P(multiple of 5) OR  P(odd number)=P(multiple of 5) + P(odd number)-[P(multiple of 5) AND P(odd number)]=2/10+5/10-[(2/10)(5/10)]=6/10 (which is the correct answer)
Notice that I used the assumption that rolling a multiple of 5 is independent of rolling an odd number, since I essentially multiplied their probabilities to get the answer. However, rolling a multiple of 5 INVOLVES rolling an odd number in one case, namely rolling a 5. 
So did I get the correct answer by fluke? Can the multiplication rule also be used to describe non-independent events sometimes?

No, it's not a fluke, in the sense that $P(A\cap B)=P(A)P(B)$ is the definition of independence of $A$ and $B$, so if it holds, as it does here, $A$ and $B$ are necessarily independent. In another sense, it was a fluke, in that it might not have been obvious that the events multiple of 5 and odd are independent: this depends on the fact that there are exactly as many odd as non-odd multiples of 5 between 1 and 10.
So you got lucky, this time: your two events really are independent. What would happen if you'd been asked about a 9-sided die? Or a 10-sided die, but with the events multiple of 3 and even?
In general, arithmetic progressions like $(na+c)_{n\in\Bbb{N}}$ and $(nb+d)_{n\in\Bbb{n}},$ e.g. odds when $a=c=1$ or multiples of 5 when $b=5,c=0,$ are "independent" in $\Bbb{N}$ in the sense of natural density whenever $a$ and $b$ are relatively prime. To bring this fact down to the finite realm of dice with $k$ sides, we need $k$ to be a common multiple of $a$ and $b$. In fact 10 is a common multiple of 2 and 5, and that indicates a bigger reason why you lucked out with this computation.
