Conditional probability and the intersection

Assuming you have an unfair die where the probabiity of rolling a certain number is defined as follows: $$\begin{array}\\ \text{Die Face} & 1 &2&3&4&5&6\\ \hline\\ \text{Probability} & .05 &.2 & .1 &.2 &.3 & .15 \end{array}$$ Find the probability that we roll an even number followed by an odd number.

I would think this is a conditional probability question, however, in the solutions it simply said that rolling an odd number first and rolling an even number second are two events $A$ and $B$ respectively, and then by independence said: $P(A)=.05+.1+.3=.45$ and $P(B)=.2+.2+.15=.55$ and $P(A\cap B)=P(A)P(B)=(.45)(.55)=.248$. When I worked this out first I thought this would be done by conditional probability. Why would my reasoning be wrong here? And why can we solve this problem like this?

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Maybe you can provide the exact formulation of the problem? So far, I would agree it is ambiguous –  Ilya Jan 5 '13 at 14:47
@Ilya This is pretty close to what it is exactly, I'll change it a little. The solution i posted is exactly what the solutions say too... It gave the distribution and asked precisely to find the probabilities that we roll an even number following an odd number. –  TheHopefulActuary Jan 5 '13 at 15:10
@Kyle, by the way note that product is commutative, so the probability of rolling odd followed by even equals that of rolling even followed by odd. –  alancalvitti Jan 5 '13 at 15:25
It is a conditional probability problem, but a specially easy one. By independence, $\Pr(B|A)=\Pr(B)$. –  André Nicolas Jan 5 '13 at 17:05
@AndréNicolas: but $P(B)$ seems to me different from the answer $.248 = P(A)P(B)$ –  Ilya Jan 6 '13 at 11:02

The content of your answer is fine, but it's incorrect to say that "it's not a problem in conditional probability" - since Bayes theorem is derived from the product rule. It just happens that $P(B|A)=P(B)$ in this case b/c of independence, as Ahshay writes below. –  alancalvitti Jan 5 '13 at 15:20