Hi chiro, thanks for your help.

I'm taking my first steps in the theory of probability, and it is very important for me, at least at first, that the mathematical definitions and theorems will capture my intuition about what I already know about probability from day to day experience (and that is not much, I must admit).

So...

This is how I picture it in my mind.

Suppose that A and B are some disjoint events in the sample space:

then if A heppened, and we then want to calculate the probability that B will also happen, we have to consider, in our calculation, the fact that some outcomes (in particular - those that in event A) have already happened, thus reducing the sample space.

for example:

Suppose we have a box with 9 balls: 4 blue balls, and 5 red balls.

Two balls are randomly picked from the box.

What would be the probability to draw a blue ball and then a red ball (with NO returning them back)?

So the sample space is \(\displaystyle \Omega =\left \{ drawing\hspace{5}any\hspace{5}ball\hspace{5}from \hspace{5} the \hspace{5} box \right \}\), and then \(\displaystyle |\Omega |=9\)

and in this case, \(\displaystyle A=\left \{ drawing\hspace{5}a\hspace{5}blue\hspace{5}ball \right \}\), \(\displaystyle B=\left\{ drawing\hspace{5}a\hspace{5}red\hspace{5}ball \right \}\)

so if \(\displaystyle A\) happened (we drew a blue ball) it'll make the sample space shrink (we're not returning the blue ball back):

\(\displaystyle \mathbb{P}(A)=\frac{4}{9}\)

and then:

\(\displaystyle \mathbb{P}(B)=\frac{5}{8}\)

So

\(\displaystyle \mathbb{P}(A\cap B)=\frac{4}{9}\cdot\frac{5}{8}\)

In the above case, A and B are mutually exclusive (you can't draw blue and red both at the same time, with only one pull), therefore dependent. so far so good.

but, and this is where it gets confusing for me; what happens if there are two independent events?

can you please give me an example, based on the scenario above (with the box and the balls), of two independent events, where the occurrence of the first one does not effect the probability of the other?

I might say "well, the same as above, only this time we DO return the ball after the first pick; this way, the occurrence of the first one does not effect the probability of the other..." then we get that A and B are indeed independent, BUT(!) A and B are still mutually exclusive! and that's impossible, according to what I wrote in my first post in this thread.