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I was thinking about this today so I decided to ask it here. I know the rule of adding probabilities. As I was taught in grade school "OR" typically means add and "AND" typically means multiply.

Suppose there is a jar with a red, blue, green, and yellow ball. The probability of picking a red ball is 25%. The probability of picking a red ball, putting it back, AND then picking a green ball is 12.5% (25% * 25%). The probability of picking either a red OR a green ball on the first try is 50% (25% + 25%).

These are independent events. Now lets take another set of independent events.

  1. The weather man said there is an 80% chance of rain. This means it should rain 80% of the time I assume. If you looked back at the past 100 days with an 80% rain chance it should rain on ~80 of them.

  2. A girl hooked up with 5 of our 10 mutual friends. I have a 50% chance of hooking up with her too.

  3. Rolling an even number on a dice carries a 50% probability.

If it rains OR I hook up with the girl OR I roll an even number on the dice then I win. If any one of these things happen I win, just like if I pick a red or a green ball, it doesn't matter if its red or green. With that being said the probability that it rains or I hook up or I roll an even number is 180%. This is not true because it is remotely possible that none of those things happen.

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Briefly, you only add exclusive probabilities. You can't pick a red ball and a green ball at the first try. (BTW your reasoning is hopelessly optimistic in #2: if she's already passed you over 5 times you've no chance.) – Scortchi Jan 15 '13 at 14:23

Edit: Oh if you don't understand this notation feel free to ask then I will elaborate. I am assuming you are familiar with the set/event thing.

What you are learned in grade school may not always be correct

$$P(Rain) = 0,8 \\ P(Girl) = 0,5 \\ P(Dice) = 0,5$$

Also, since you assume independence

$$P(Rain \cap Girl \cap Dice) = P(Rain)*P(Girl)*P(Dice) = 0,2 \\ P(Rain \cap Girl) = P(Rain)*P(Girl) = 0,4 \\ P(Rain \cap Dice) = P(Rain)*P(Dice) = 0,4\\ P(Girl \cap Dice) = P(Girl)*P(Dice) = 0,25\\ $$

And with that we have

$$P(Rain \cup Girl \cup Dice) = P(Rain) + P(Girl) + P(Dice) - P(Rain \cap Girl) - P(Rain \cap Dice) - P(Girl \cap Dice) + P(Rain \cap Girl \cap Dice) = 0,95$$

I suggest you draw yourself a couple of Venn diagrams if this isn't clear.

I just recently posted about this topic here as well

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I would like to invite you to draw those diagrams and include them in your answer: it would make it (and the one you reference) clearer and more accessible. – whuber Jan 15 '13 at 14:24

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