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Here is the question in the probability packet:
Spinner divided into 4 quadrants with 3 of the quadrants red and one blue. Bag of 6 marbles, 3 red and 3 blue.

What is the probability of spinning the spinner and getting a red OR picking a red marble?

I know the answer can't be ${3\over4} + {1\over2}$. These two events have no overlap. So, when I do P(red spinner) + P(red marble) - P(A intersection B) I would seem to get ${3\over4} + {1\over2}-0$.

How should I view this problem?

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    It isn't zero,. Notice the events are independent – 2017-01-09
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    I thought if the events are independent then the overlap would be zero. – 2017-01-09
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    Well, no. Only if one of the events is impossible. $P(A\cap B)=P(A)P(B)$ when $A$ and $B$ are independent – 2017-01-09
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    So, can you assist me is seeing the solution to this problem? – 2017-01-09
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    I thought I did. Replace $0$ with $P(A)P(B)$... You already know $P(A)$ and $P(B)$. – 2017-01-09
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    I think you are confusing the terms *independent* and *mutually exclusive*. – 2017-01-09
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    So, the answer would be the 3/4 + 1/2 - 3/8? – 2017-01-09
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    I always thought that the subtraction was for things that were counted in P(a) and ALSO counted in P(b). – 2017-01-09
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    They are double-counts. Event A is (Red, anything). Event B is (anything, Red). (Red, Red) is the overlap, that is double-counted.. – 2017-01-09
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    I'm having a difficult time seeing this. Let's say the problem was spinner with 3 red quadrants and 1 blue, OR rolling a number less than 6 on a die. – 2017-01-09
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    There would be NO overlap between the spinner and the die. Would the answer still be 3/4 + 5/6 - 15/24? Yet, there would be no overcount. – 2017-01-09

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In your example, $P(A\cap B) \ne 0.$ $P(A\cap B)$ is the probability of getting a red spin and getting a red marble. This can happen, so it's probability is not zero. In fact, since the spin and the draw are independent, $P(A\cap B) = P(A)P(B) = (3/4)(1/2)= 3/8.$ From here, you're free to use the formula $P(A\cup B) = P(A) + P(B) -P(A\cup B)$ to arrive at the answer.

Your error seems to be interpreting the words 'no overlap'. No overlap (in the sense $A\cap B=\emptyset$) means 'A and B cannot happen at the same time'. Not 'A and B have no influence on each other'.

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    OHHHH.... My book examples had always been something like chance of drawing a queen or a red card, and it was easy to see that 2 of the queens were red. So, I could see the double counting. Here, not so easy to see the double counting. – 2017-01-09