if E(X) = 1.6 and E(Y) = 2
does E (Y/X) = E(Y)/E(X)
2/1.6 ??
That doesn't bode well.
What are the possible values of Y/X ......? 1/1 = 1, 1/2, 2/1 = 2, 2/2 = 1 (already considered), 3/1 = 3, 3/2.
Pr(Y/X = 1) = Pr(Y = 1 and X = 1) + Pr(Y = 2 and X = 2) = (0.3)(0.4) + (0.4)(0.6) = 0.12 + 0.24 = 0.36.
Pr(Y/X = 1/2) = Pr(Y = 1 and X = 2) = (0.3)(0.6) = 0.18.
etc.
Then E(Y/X) = (1)(0.36) + (1/2)(0.18) + ......
This is the same calculation as CaptainB did - all I've done is simplified the arithmetic.
It uses the definition of expectation for a discrete random variable:
$\displaystyle E(U)=\sum_i u_i p(u_i)$
where the $\displaystyle u_i $'s are the possible values of $\displaystyle U$, and $\displaystyle p(u_i)$ are the probabilities of those values.
Also it uses the definition of independent RV that $\displaystyle p(a,b)=p(a)p(b).$
So in this case we have:
$\displaystyle E(Y/X)=\sum_{i \in \{1,2\}, j \in \{1,2,3\}} \frac{j}{i}\ p(x=i,y=j) =\sum_{i \in \{1,2\}, j \in \{1,2,3\}} \frac{j}{i}\ p(x=i)p(y=j) $
CB
They are random variables, not events. We are finding expectations, which are properties of RV.
Also no one has used P(Y/X) as a notation (which if they had might be mistaken for the conditional probability P(Y|X) but neither I nor MrF would do that -would we?)
Also$\displaystyle ^2$, please quote what you are replying to so we know what you are talking about (and if you don't know what you are talking about don't talk about it).
CB