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I understand the rules for finding the probability of A or B occurring. However, the rules of finding the probability of A and B happening are a bit more elusive. In the former you add, which makes sense; in the later, you multiply, which does not make as much sense. Perhaps it has to do with the fact that both events occurring simultaneously has a lower probability. Is this one valid interpretation? I would very greatly appreciate someone being able to explain the mechanics of what is going on--that is, how can I interpret abstract math symbols in this particular case.

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You can think as a branch of possibilities. For each possibility in $A$ you can consider all the possibilities in $B$, so you have to multiply. Is this the case? – Sigur Feb 4 '13 at 21:50
Eli - Don't forget to accept and/or upvote answers you find helpful! ;-) – amWhy Feb 9 '13 at 15:01
@amWhy Oh, yes. Thank you for the reminder--I've just been so busy that I forgot. – Mack Feb 9 '13 at 16:19
up vote 3 down vote accepted

First, you can add in the first case only if the events $A$ and $B$ are disjoint; if they can occur simultaneously, the probability of $A\text{ or }B$ is not the sum of the probabilities of $A$ and $B$. For example, suppose that you roll a fair die. Event $A$ is getting an even number, and event $B$ is getting a number that is not a perfect square. These events have probabilities $\frac12$ and $\frac23$, respectively, so the sum of their probabilities is $\frac76$, which is greater than $1$ and cannot be a probability of anything. The actual probability of $A\text{ or }B$ is the probability of getting something other than a $1$, so it’s $\frac56$.

The probability of getting $A\text{ and }B$ is the probability of getting $2$ or $6$, which is $\frac13$. Adding $P(A)=\frac12$ and $P(B)=\frac23$ counts this event twice, once as part of $A$ and once as part of $B$, so to get the correct value of $P(A\text{ or }B)$ you have to subtract once what was counted twice, namely, $P(A\text{ and }B)$:

$$P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)\;.$$

As for $P(A\text{ and }B)$ being $P(A)P(B)$, you’re on the right track with the idea that getting both $A$ and $B$ to occur is harder than getting either one of them to occur individually. Suppose that you perform the experiment many times; on average you expect $A$ to occur in $P(A)$ fraction of the trials, and $B$ to occur in $P(B)$ fraction of the trials. If $A$ and $B$ are independent, $B$ will occur on average in $P(B)$ fraction of the trials in which $A$ occurs, and also in $P(B)$ fraction of the trials in which $A$ does not occur. You’re interested in the former: the trials in which $A$ and $B$ both occur. Overall on average that’s $P(B)$ fraction of the $P(A)$ fraction of the trials in which $A$ occurs, or $P(A)P(B)$ fraction of all the trials. In other words, if $A$ and $B$ are independent, then $P(A\text{ and }B)=P(A)P(B)$. In the example above, $$P(A\text{ and }B)=\frac13=\frac12\cdot\frac23\;:$$ half the time on average we get an even number, and on average two-thirds of those even numbers are $2$ or $6$.

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I think you meant "or" not "of" in "A of B" at the top of your post – JoeRocc Mar 30 at 7:08
@JoeRocc: Thanks! – Brian M. Scott Mar 30 at 17:03

Probabilities are additive over disjoint unions are additive. Suppose you have events $A$ and $B$. You have the following disjoint union. $$A \cup B = (A - B) \cup (A \cap B) \cup (B - A),$$ so $$P(A\cup B) = P(A - B) + P(A\cap B) + P(B - A).$$ Since $A = (A - B) \cup (A\cap B)$, we have $P(A) = P(A - B) + P(A\cap B)$, giving $$P(A\cup B) = P(A) + P(B - A).$$ Now by symmetry of the reasoning, we have $$P(A\cup B) = P(A) + P(B - A) + P(A\cap B) - P(A\cap B) = P(A) + P(B) - P(A\cap B).$$

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Use the inclusion/exclusion principle for finding probabilities of unions and intersections in general. When the events are mutually exclusive, then you may multiply probabilities to get the probability of an intersection (i.e. "AND").

You are correct in that the probability of intersections will be less than the individual probabilities of each component event. This is because probabilities are numbers between $0$ and $1$.

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I believe the inclusion/exclusion principle would really help if I had an issue with find the probability of the events A or B, where A and B are not disjoint--that is, they can occur simultaneously. – Mack Feb 4 '13 at 21:51
That's right. In many problems, you are either told that $A$ and $B$ are disjoint, or you have to conclude it based on what you're given in a problem. – Ron Gordon Feb 4 '13 at 21:52

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