Let me give you an example. Then you take what you learn from my example, and apply it to your actual problem, ok?
Let's say there are ten houses with families who may have pets on a street. Four have them have only a dog, three have only a cat, and two have both a cat and a dog.
Consider these representations of the facts I gave. Are they true?
The probability that there is a dog living at a randomly selected house along this street is six out of ten.
The probability that there is a cat living at a randomly selected house along this street is five out of ten (or one-half).
The probability that there are both cats and dogs living at a randomly selected house along this street is two out of ten (or one fifth).
Yes, these are all true.
Ok, suppose I asked you these questions.
What is the probability that a randomly selected house along this street contains a cat or a dog?
... well, we know six houses contain dogs. Out of those six we know four contain only dogs and two more also contain cats. But a total of five houses contain cats, so there must be three houses which contain cats but do not contain dogs. So the total number of houses which contain a cat or a dog (or both) is six + three = nine, and thus the probability is nine out of ten, or 0.9.
Notice how the formula becomes 0.6 + (0.5 - 0.2). The number containing dogs + the number not containing dogs which contain cats. This latter number is computed by taking the number which contains cats and discounting it by the number which also contain dogs, because those houses were already included in the count when we totaled up houses with dogs in them.
Venn diagrams show this situation visibly, but are very time-consuming to describe here. I'd be happy to teach you the graphical Venn diagram approach if you schedule a brief session with me this weekend. (I can teach it to you over Skype)
