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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?

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    $P(A\cap B)=P(A)\cdot P(B\mid A) = \frac{1}{2}\cdot \frac{2}{5} = \frac{1}{5}$. In general $P(A\cap B)$ is **NOT** equal to $P(A)\cdot P(B)$. That is ONLY true when they are independent events. – 2017-01-13

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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, $Pr(\text{Yel1st and Blue2nd})=Pr(\text{Yel1st})\cdot Pr(\text{Blue2nd | Yel1st})=\frac{3}{6}\cdot \frac{2}{5}=\frac{1}{5}$

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    This makes complete sense to me now. In just a short paragraph you helped to clear up a great deal of mis-understanding on this topic. It seems to me that I can see and understand calculus problems FAR easier than I can probability. Thank you! – 2017-01-13
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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 $\frac{3}{6}\cdot\frac{2}{5}=\frac{1}{5}$.

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    Thank you very much. Your suggested way to look at it helps! – 2017-01-13