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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  • $\begingroup$ What exactly do you mean by adding up two probabilities? Do you mean adding up the probabilities of two different events? $\endgroup$ – Jean-Sébastien Nov 10 '12 at 20:15
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    $\begingroup$ 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. $\endgroup$ – André Nicolas Nov 10 '12 at 20:18
  • $\begingroup$ I don't understand a thing. I'm lost. $\endgroup$ – Gigamegs Nov 10 '12 at 20:24
  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Nov 10 '12 at 21:43
  • $\begingroup$ Is order ok for you? Order of probabilities? $\endgroup$ – Gigamegs 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.

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  • $\begingroup$ What does it mean both come up 1? Come up first? I've read this answer here math.stackexchange.com/questions/184115/…. $\endgroup$ – Gigamegs Nov 10 '12 at 20:54
  • $\begingroup$ @Skidrow: you said you were rolling two dice. The order doesn't matter, just the number each one shows. $\endgroup$ – Ross Millikan Nov 10 '12 at 20:57
  • $\begingroup$ What if my dice is different size? A cube and a tetraeder? And I've also normalize the probabilities can I add them together? Why do I need to normalize them? $\endgroup$ – Gigamegs Nov 10 '12 at 21:00
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Nov 11 '12 at 1:44
  • $\begingroup$ @Bytemain when you calculate probabilities of events it's easier to think of them as unions and intersections of finit sets. Ross-Millikan showed a sieve formula were you have to add the unions (include) and then subtract (exclude) the intersecting part to not count it twice. The math is described by the inclusion and exclusion principle. $\endgroup$ – Doomjunky Dec 20 '18 at 0:54

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