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?
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.