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?

In general, whoever tells you that "OR" means adding and "AND" means multiplying is not telling you the whole story.
OR is not always just adding:  $P(A\cup B) = P(A)+P(B)-P(A\cap B)$ and this is equal to $P(A)+P(B)$ if and only if $P(A\cap B)=0$, for example when $A$ and $B$ are disjoint.
AND is not always just multiplying:  $P(A\cap B) = P(A)\cdot P(B\mid A)$ and this is equal to $P(A)\cdot P(B)$ if and only if $P(B\mid A)=P(B)$, i.e. when $A$ and $B$ are independent events.
In this case, $Pr(\text{Yel1st and Blue2nd})=Pr(\text{Yel1st})\cdot Pr(\text{Blue2nd | Yel1st})=\frac{3}{6}\cdot \frac{2}{5}=\frac{1}{5}$
