This is a fairly simple problem that can be done by considering all the elements in the sample space and picking out the favorable outcomes but I wanted to know if there was a solution using conditional probability which says for my problem :
P(multiple of 2 on one die AND multiple of 3 on other die)=$P(A).P(B|A)$ where A and B are respective events but I don't know how to proceed. Clearly $P(A)=1/2$ but how do I find the 2nd factor. I am trying to do it this way so that I can generalise the result to a N-face die, as for larger N it is not feasible to count all the favorable outcomes.Thanks!
As @Math Lover said in the comments, $\mathbb P(B \mid A) = \mathbb P(B)$ since the events are independent. To make this a full solution, consider the two possible cases for how you might fulfill the event you're looking for. For simplicity, let's color the dice (I'll use "green" and "yellow"):
Case 1: The green die is divisible by 2, and the yellow die is divisible by 3. In this case, we have $\mathbb P(A \cap B) = \mathbb P(A) \cdot \mathbb P(B \mid A) = \mathbb P(A) \cdot \mathbb P(B) = \frac 1 6.$
Case 2: The yellow die is divisible by 2, and the green die is divisible by 3. In this case, the probability is again $\frac 1 6$.
We're concerned with the union of these two cases -- but, they're not disjoint, so we need an inclusion-exclusion principle:
$$\mathbb P(C \cup D) = \mathbb P(C) + \mathbb P(D) - \mathbb P(C \cap D)$$
In other words, we've double counted the situation where cases 1 and 2 simultaneously occur (i.e. that we roll two sixes), so we need to remove that case by subtracting its probability. The final answer should be:
$$ \frac 1 6 + \frac 1 6 - \frac 1 {36} = \fbox{$\frac{11}{36}$}$$
The conditional probability part isn't the difficulty in this problem; decomposing the event into manageable pieces is the obnoxious part.