Combining discrete probabilities Disclaimer: I don't have a background in math. I'm a programmer searching for a solution, so I apologize in advanced for the possibly poor wording!
If Alice likes the green things with a probability of $60\%$, and Alice like armchairs with a probability of $30\%$, what is the probability that Alice likes a given green armchair?
My first approach was to use a weighted average using $\sqrt{n}$ (where $n$ is sample size) as the weight. However, the intention here is dealing with humans, which might love green or armchairs, but never a green armchair. Or maybe no green armchairs, unless it has blue stitching!
My question is: Is there a generic methodology for making these sorts of predictions that I can learn about? I suspect there are many and depending on my situation some might be better than others, but I'm not sure where to start. Any help digging in would be appreciated.
Ps: If this question doesn't fit this board, let me know, I'll look elsewhere :)
 A: Welcome to MSE! Unfortunately, just knowing these two probabilities in isolation will not pin down an exact probability.
Let $L(x) = \text{"Alice likes $x$"},G(x)=\text{ "$x$ is green"}, A(x)=\text{ "$x$ is an armchair"}$ then we can define $p_G:=P(L(x)|G(x))=60\%$ and $p_A:=P(L(x) |A(x))=30\%$. These are conditional probabilities, where we are assuming some event has happened, in this case $G(x)$ and $A(x)$, respectively.
What we want to know is $p_{GA}=P(L(x)|G(x),A(x))$. From the definition of conditional probability, we have
$$P(L(x)|G(x),A(x)) =\frac{P(L(x),G(x),A(x))}{P(G(x),A(x))}$$
As you can see from the above, knowing $p_G$ and $p_A$ doesn't tell us anything about $p_{AG}$.
It could be that among the universe of things $x\in \Omega$, the set of green armchairs that Alice likes is empty, so $p_{AG}=0$.
It could be that she only likes green armchairs, so $p_{AG}=1$, or anything in between -- all were doing is choosing how much the sets $L(x), G(x), A(x)$ overlap.
A: Insufficient information, for precisely the reason that you pointed out.  Expressing the idea differently, there is no information given on whether the two events, liking armchairs  and liking green things are independent.
More formally, first see Bayes Theorem.
If you label the two events as $E_1$ and $E_2$, then the general formula is that
$$p(E_1,E_2) = p(E_1) \times p(E_2|E_1).$$
You are given $p(E_1)$ and $p(E_2)$, but no information is given on either $p(E_2|E_1)$ (i.e. the probability of event $E_2$ occurring, given that event $E_1$ has occurred) or $p(E_1|E_2)$.
Therefore, there is insufficient information to solve the problem.
