Probability and Intersections I'm having trouble understanding the difference between conditional probability and dependent events. Even then, I'm not sure if that's what I'm having issues with.
NB This is not a homework problem. I am doing a stats course, but I've made all these variables up to help me understand.
Let's say, I have 3 "bags":
$\mathbf{A}$ = {5, 6, 7, 8}
$\mathbf{B}$ = {n, o, p, q, r}
$\mathbf{C}$ = {red, blue, green}
So, going through some basics, I think I have these right.
If:


*

*$\mathbf{A}_{5}$ = "the probability of getting a 5 from bag $\mathbf{A}$"

*$\mathbf{A}_{6}$ = "the probability of getting a 6 from bag $\mathbf{A}$"

*$\mathbf{A}_{<7}$ = "the probability of getting an element less than ** from bag $\mathbf{A}$"

*$\mathbf{A}_{5 or 6}$ = "the probability of getting a 5, or a 6 from bag $\mathbf{A}$"

*$\mathbf{B}_{consonant}$ = "the probability of getting a consonant from bag $\mathbf{B}$"

*$\mathbf{B}_{vowel}$ = "the probability of getting a vowel from bag $\mathbf{B}$"

*$\mathbf{C}_{primary}$ = "the probability of getting a primary colour (red or blue) from bag $\mathbf{C}$"


Independent events
Then P($\mathbf{A}_{5 or 6}$) = P($\mathbf{A}_{5}$ $\cup$ $\mathbf{A}_{6}$) = P($\mathbf{A}_{5}$) + P($\mathbf{A}_{6}$) = $\frac{1}{4}$+$\frac{1}{4}$ = $\frac{2}{4}$ = $\frac{1}{2}$.
Then P($\mathbf{A}_{5}$ $\cup$ $\mathbf{B}_{vowel}$) (ie, drawing once from bag $\mathbf{A}$ and once from bag $\mathbf{B}$ and getting a 5 or a vowel) = $\frac{1}{4}$+$\frac{1}{5}$ =  $\frac{5}{20}$+$\frac{4}{20}$ = $\frac{9}{20}$.
Dependent events
Drawing twice from bag $\mathbf{A}$ and getting a 5 and then a 6 = P($\mathbf{A}_{5 and 6}$) = P($\mathbf{A}_{5}$ $\cap$ $\mathbf{A}_{6}$) = $\frac{1}{4}$ . $\frac{1}{4}$  = $\frac{1}{16}$
Conditional events
Drawing twice from bag $\mathbf{A}$ and the probability of getting a 6 given I've already drawn a 5 = P($\mathbf{A}_{6}$|$\mathbf{A}_{5}$) = $\dfrac{P(\mathbf{A}_{5} \cap \mathbf{A}_{6})}{P(\mathbf{A}_{5})}$ = $\dfrac{\frac{1}{16}}{\frac{1}{4}}$ = $\frac{4}{16}$ = $\frac{1}{4}$
Problem
My problem comes in that I don't understand the semantic diference between the last two examples. How is it different to ask the probability of a 5 and then 6 as opposed to a 6 assuming I already have a 5?
I also fall apart when I have to stitch them together, eg. What is the probability that I would get a primary colour given a consonant given a 5? Is that soemthing like...
P($\mathbf{C}_{primary}$|$\mathbf{B}_{consonant}$|$\mathbf{A}_{5}$)
= P($\mathbf{C}_{primary}$|$\dfrac{\mathbf{B}_{consonant} \cap \mathbf{A}_{5}}{\mathbf{A}_{5}}$)
= P($\dfrac{\mathbf{C}_{primary} \cap \dfrac{\mathbf{B}_{consonant} \cap \mathbf{A}_{5}}{\mathbf{A}_{5}}}{\dfrac{\mathbf{B}_{consonant} \cap \mathbf{A}_{5}}{\mathbf{A}_{5}}}$)
and I'm not even sure where to start with the numerator on that.
So in summary:


*

*How is it semantically different to ask the probability of "a 5 and then a 6" as opposed to "a 6 assuming I already have a 5"?

*How does one do a $\cap$ on a fractional as in the last equation?

*Do I have this all correct?

 A: There are some inaccuracies in your understanding. The event you describe as $\mathbf{A}_{5}\cup\mathbf{B}_{vowel}$ belongs to the sample space $\Omega:\mathbf{A}\cup\mathbf{B}=\{5,6,7,8,n,o,p,q,r\}$ and is NOT the event "drawing once from bag A and once from bag B and getting a 5 or a vowel". 
$\mathbf{A}_{5}\cup\mathbf{B}_{vowel}$ describes the event of picking once from $\Omega$ and the result being a 5 or a vowel:
$P(\mathbf{A}_{5}\cup\mathbf{B}_{vowel})=\frac{1}{9}+\frac{1}{9}=\frac{2}{9}$ ->(hint: $(P(\mathbf{A}_{5}\cup\mathbf{B}_{vowel})=P(\mathbf{A}_{5}|\Omega\cup\mathbf{B}_{vowel}|\Omega)$
For sequences of (independent) experiments you have to consider the new sample space $\Omega':\mathbf{A}\times\mathbf{A}=\{(5,5),(5,6),(5,7),...\}$ containing 16 elements of all combinations. In that light the probability of the event $E_{56}$ "Drawing twice from bag A and getting a 5 and then a 6" is:
$P(\mathbf{E}_{56})=1/16$
The event "Drawing twice from bag A and the probability of getting a 6 given I've already drawn a 5" can be written as ($x$ stands for 'any' pick):
$P(\mathbf{E}_{x6}|\mathbf{E}_{5x})=\frac{P(\mathbf{E}_{x6}\cap\mathbf{E}_{5x})}{P(\mathbf{E}_{5x})}=\frac{P(\mathbf{E}_{56})}{P(\mathbf{E}_{5x})}=1/4$
The difference between the joint and the conditional probability is that the first points to the ratio of the number of desired events to the total events in $\Omega'$. It is implied that $P(\mathbf{E}_{56})=P(\mathbf{E}_{56}|\Omega')$. The conditional probability points to the ratio of the number of desired events to the number of events in the set that you condition upon $E_{5x}:\{(5,5),(5,6),(5,7),(5,8)\}$. Hope this helps for your last problem as well.
