# How can I add two probabilities with the same time?

When I have two probabilities at the same time or at different time how can I add them together? I know that simply add them doesn't work because that is just for two probabilities independent or mutual exclusive. Do I need to normalize the two probabilities? Why can't I simply add them together? Is there a reason? With two probabilities I mean I've two dice with 6 side. When the probability of dice 1 and side 1 is 1:6 and it's the same for dice 2 how can I combine them together? There is this rule about mutual exclusive or independent but I also have read about the other rule of the same time. Is there a probabilities that both dice have the same side at the same time?

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What exactly do you mean by adding up two probabilities? Do you mean adding up the probabilities of two different events? – Jean-Sébastien Nov 10 '12 at 20:15
For mutually exclusive adding is correct for finding probability that (at least) one of $A$ or $B$ happens. For independent, adding is essentially always wrong. – André Nicolas Nov 10 '12 at 20:18
I don't understand a thing. I'm lost. – Phpdevpad Nov 10 '12 at 20:24
This question is really unclear. I don't understand what you mean by "time". Can you give a concrete example, stating exactly what you are trying to find? Your phrase "combine them together" is ambiguous. – Nate Eldredge Nov 10 '12 at 21:43
Is order ok for you? Order of probabilities? – Phpdevpad Nov 10 '12 at 22:00

If you want to combine the probabilities, you have to decide what event you want. If you want the chance that both dice come up $1$, the events are independent and you multiply them: $\frac 16 \cdot \frac 16=\frac 1{36}$. If you want the chance that either one comes up and they were exclusive, you would add them. In this case they are not exclusive-both dice can come up $1$. The chance that at least one of them comes up $1$ is $\frac 16 + \frac 16 - \frac 1{36}=\frac {11}{36}$ The subtraction of $\frac 1 {36}$ is because we have double counted the case where they both came up $1$-it is part of both $\frac 16$'s.
If you have a d6 and a d4, the chance of a 1 on the d6 is $\frac 16$ and the chance on the d4 is $\frac 14$. If you ask what is the chance they are both 1, it is $\frac 1{24}$ If you ask what is the chance of at least one 1, it is $\frac 16 + \frac 14 - \frac 1{24}=\frac 38$ by the same logic. You should get a basic probability book, which can explain this in much better detail. – Ross Millikan Nov 11 '12 at 1:44