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?

Your guess that its a conditional probability question is correct. This question is based on joint / conditional probability. So if you take A-> event where odd number occurs at first throw and let B-> be event where even number occurs at second throw. So you need a joint probability P(A,B) which is given as P(A,B)=P(B/A)*P(A). But since B is independent of A (*That's because throwing die second time doesn't depends on result you get first time). Hence we get P(A,B)=P(B)*P(A). Then as you stated we get P(A) and P(B) since getting odd number 1,3,5 are independent of each other and getting even number 2,4,5 are independent of each other. Hence you get P(A) = 0.45 and P(B) = 0.55 and hence P(A,B) = 0.2475. Please feel free to ask if you have any question regarding this description.
