Query about formula for independent events According to my lecture notes (beginner probability & stats student here!), I am told that if events X and Y are independent, then P (X ∩ Y) = P(X)P(Y). I am further told that, in general however, P(X ∩ Y) is usually not equal to P(X)P(Y).
I am given a very simple example problem: a bag contains 8 green marbles and 3 blue marbles. Two marbles are drawn from the bag one after another, with the second marble being drawn only after the first marble is replaced. Find the probability of drawing a green marble in the first draw and then a blue marble in the second draw (one after another, with replacement). Answer is simple: just take 8/11 multiplied by 3/11 to get the answer. I am told that I can do this because the events are independent hence I can use P(X ∩ Y) = P(X)P(Y). Here, X is the event that you draw a green marble in the first draw and Y is the event that you draw a blue marble in the second draw. 
My problem arises from something I observed from another problem I am given. In this example, the marble is NOT replaced. If the question remains the same i.e. find probability that I will draw a green marble on the first draw and then a blue marble in the second draw, then here I am told the answer is 8/11 x 3/10 since the total number of marbles decreases by 1 after the first draw. I am taught to visualise this by drawing a tree diagram. The thing is, in this case, aren't the two draws NOT independent, precisely because my first draw decreases the total number of marbles in the bag thus affecting the probability of my second draw? And yet, the equation P(X ∩ Y) = P(X)P(Y) still holds - and I'm told this equation is only for independent events and doesn't happen if that's not the case. Here, X is the event that you draw a green marble in the first draw and Y is the event that you draw a blue marble in the second draw. 
Please help me reconcile this apparent contradiction, thanks :)
 A: You are right in noting that $\mathbf{X}$ and $\mathbf{Y}$ are not independent in the second example. However, the identity $P(\mathbf{X} \cap \mathbf{Y}) = P(\mathbf{X})P(\mathbf{Y})$ has not been used in this example. Instead, the answer is evaluated using the expression $P(\mathbf{X} \cap \mathbf{Y}) = P(\mathbf{X})P(\mathbf{Y|X})$, which is the definition of conditional probability. Note that $3/10$ is not equal to $P(\mathbf{Y})$ in the second example, instead it is equal to $P(\mathbf{Y}|\mathbf{X})$. You can calculate $P(\mathbf{Y})$ using the total probability theorm; however, this I will leave to you as an exercise :).    
A: You are probably confusing the difference between $P(X \cap Y)=P(X)P(Y|X)$ where $P(Y|X)$ is probablity of $Y$ happening given that $X$ happened and independence equation.. This equation, whether or not $X,Y$ are independent, always holds.
In the second case, in calculating $P(X\cap Y)$ You did not use $P(X\cap Y)=P(X)P(Y)$ but rather $P(X \cap Y)=P(X)P(Y|X)$. When you found probability of picking blue marble, you already assumed that you picked a green marble on first draw. That corresponds to $P(Y|X)$, not $P(Y)$.
Let's calculate $P(Y)$. Drawing a blue marble in second draw can happen in two different ways: you either have $GB$ or $BB$ (green first, blue second and two blues in a row). So
$$
P(Y)=P(GB)+P(BB)=P(\text{second draw is $B$} | \text{first draw is $G$}) \cdot P(G) + P(\text{second draw is $B$} | \text{first draw is $B$}) \cdot P(B)=\frac{3}{10}\cdot\frac{8}{11}+\frac{2}{10}\cdot \frac{3}{11}=\frac{3}{11}
$$
Then you can see, $P(X\cap Y)=\frac{8}{11}\cdot \frac{3}{10} \neq \frac{8}{11}\cdot \frac{3}{11}=P(X)P(Y)$. So indeed the two events are not independent.
The moral is, always identify what events you are referring to. In this case you confused the events $Y$ and $Y|X$, leading to a false contradiction.

In fact, you can see that if $X,Y$ are independent, $P(X)P(Y)=P(X \cap Y)=P(X)P(Y|X)$ so if $P(X)\neq 0$ then $P(Y)=P(Y|X)$. So in literal terms, $X,Y$ are indepent if and only if $P(Y|X)=P(Y)$, that is, probability of $Y$ does not depend on whether or not $X$ happened (ie independent in intuitive sense).
