Probability with an "OR" Marbles:  3 Yellow, 1 Purple, 2 Blue
Probability of choosing a yellow first or a blue second.
I understand that the OR situation will result in my adding the 2 probabilities and then subtracting the $P(A\text{ and }B)$.
For me, I can see that P(yellow first or blue second) = $1/2 + 1/3 -P(A\text{ and }B)$.
If I physically draw out the sample space I can see that $P(A\text{ and }B) = 1/5$.  My question is how do I see this without drawing the sample space and physically counting the choices --  far too time consuming for a test.  I would have thought that $P(A\text{ and }B)$ would be = $(1/2)(1/3)=1/6$ since that is $P(A)P(B)$.
Can you help me see this?

The simplest way, I think, is to follow what happens: first pick, 3 good balls in 6, second pick, 2 good balls among the remaining 5, and $\frac{3}{6}\cdot\frac{2}{5}=\frac{1}{5}$.
