1. Bag Has 7Red and 3Green : You Reach in and Grab Two Tiles !
(Same Idea as: Pick 1st One, Keep it, then Pick 2nd One)
SLIDE 1
RED GREEN
My first draw has 70% chance of being Red (7 out of 10 Tile in the Bag are Red), and 30% chance of being Green. So from my whole “Unit Rectangle” (100% Probability), I have a 7/10 Red Rectangle and 3/10 Green Rectange…

2. …And, that 7/10 Red Rectangle and 3/10 Green Rectangle in SLIDE 1
correspond to the first two “Branches” of my Tree, representing my
first pick of a tile…
P(R) 7/10
1st Pick
P(G) 3/10
…Which could be Red (with a 7/10 Chance) or Green (3/10 Chance)

3. SLIDE 3
RED GREEN
If my first pick is Red… Then, on my second pick, there’s only 6 out of 9 Tiles left that are Red. So, within my “Red Rectangle” (First Pick), I have shaded 6/9 Red (This is shown by 6 out of 9 Horizontal “Red Bars”). Similarly, on my second pick there are 3/9 Green Tiles (Shown by 3 out of 9 Horizontal “Green Bars”)

4. And those Horizontal “Red Bars” and “Green Bars” of SLIDE 3 correspond
to the two branches of the tree which stem from the First Red Branch.
P(R|R) 6/9
2nd Pick
P(R) 7/10
P(G|R) 3/9
1st Pick
P(G) 3/10
Specifically, using the “A | B” notation to mean “A given that B has occurred”, we have P(R on 2nd | R on 1st ) = 6/9 and P(G on 2nd | R on 1st ) = 3/9. Note that these two new branches for the 2nd pick represent conditional probabilities.

5. SLIDE 5
RED GREEN
If my first pick is Green… Then, on my second pick, there’s only 2 out of 9 Tiles left that are Green. So, within my “Green Rectangle” (First Pick), I have shaded 2/9 Green (This is shown by 2 out of 9 Horizontal “Green Bars”). Similarly, on my second pick there are 7/9 Red Tiles (Shown by 7 out of 9 Horizontal “Red Bars”)

6. SLIDE 6
And those Horizontal “Green Bars” and “Red Bars” from SLIDE 5 correspond
to the two branches of the tree which stem from the First Green Branch.
P(R) 7/10
1st Pick
P(G|G) 2/9
P(G) 3/10
2nd Pick
P(R|G) 7/9
Specifically, using the “A | B” notation to mean “A given that B has occurred”, we have P(G on 2nd | G on 1st ) = 2/9 and P(R on 2nd | G on 1st ) = 7/9. Note that these two new branches for the 2nd pick represent conditional probabilities.

7. Now looking at the Whole Tree, we see “ “ can represent P(R & R),
SLIDE 7
Now looking at the Whole Tree, we see “ “ can represent P(R & R),
the probability that both Tile are Red, and this is equal to 42/90
P(R 1st & R 2nd ) = 42/90
P(R|R) 6/9
2nd Pick
P(R) 7/10
P(G|R) 3/9
1st Pick
P(G 1st & G 2nd ) = 6/90
P(G|G) 2/9
P(G) 3/10
2nd Pick
P(R|G) 7/9
Similarly, P( G & G) = 6/90, and P(Tiles are Different Colors) =
(21/90) + (21/90), which is equal to 42/90. One thing we could
wonder is: For P(R & R) = 42/90, where do we “see” this in our
Area Model?

8. 42 out of 90 Small Rectangles are in the “R 1st and R 2nd Zone”
SLIDE 8
Well, we can see 42/90 = P(R & R) by noting there are 90 Small Rectangle Shapes in my “Unit Rectangle” below. Looking where the 1st Pick is Red followed by 2nd Pick Red, we definitely can count Small Rectangles out of Total Small Rectangles for P(R & R)
RED GREEN 1 2 3 … … 40 41 42
42 out of 90 Small Rectangles are in the “R 1st and R 2nd Zone”