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I read on my textbook that in a certain scenario, the sum of the probabilities is supposed to equal to one.

However, I read an example of an event on this site, and it says in some cases, independent events can add to greater than one. For example, if you add the probability of the chance of getting a red card from a pack of cards OR getting a tails on a coin flip, these events could be greater than one, and the sample space could therefore be greater than one.

Is this example, could the probabilities add to greater than one, and is it possible for the sample space to be greater than one? Or is this impossible?

chris
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    Well, the probabilities of overlapping events could certainly add to something $>1$. Say you throw a fair die. Let $A$ be the event "you throw $>1$ " and $B$ be the event "you throw $<6$." Then $P(A)+P(B)>1$. – lulu Feb 18 '19 at 22:03
  • Probabilities of disjoint events add to at most 1. Probabilities of independent events can add to more than one. Independent and disjoint are different kinds of things. Never mix them up. – kimchi lover Feb 18 '19 at 22:06
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    Think about what it means if I say "The Probability it we rain in Papeete today is 60% and probability that I flip a head is 50% so if I add them up that is 110%". It doesn't actually affect or have anything to do with either event happening. That is not the probability of both occurring (30%) or of one or the other occurring (80%), it's just a rather pointless observation about two separate events. At least as far as I can tell. – fleablood Feb 18 '19 at 22:10
  • For the 80%, you're talking about the P(AUB), right? – chris Feb 19 '19 at 05:50

2 Answers2

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Of course the sum of probabilities of events can be greater than $1$... however the probability of the union of events can never be greater than $1$. Reworded, $Pr(A)+Pr(B)$ is allowed to be greater than one but $Pr(A\cup B)$ is never greater than $1$.

As for the follow-up question "is it possible for the sample space to be greater than one" I assume you mean the probability of the sample space. No. No probability is ever allowed to be greater than one, ever.

Note: $Pr(A\cup B) = Pr(A)+Pr(B)\color{red}{-Pr(A\cap B)}$. The probability of a union of events is equal to the sum of the probabilities of the events only in the situation that the intersection of those events is impossible (i.e. $Pr(A\cap B)=0$)

JMoravitz
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  • Sorry, but how can Pr(A) + Pr(B) be greater than one? Could you give me an example? If Pr(A) and Pr(B) are inside the sample space, how would it be possible? – chris Feb 19 '19 at 04:02
  • A sum of probabilities does not necessarily represent a probability itself. Let our sample space represent results of drawing a card from a standard playing card deck. Let A and B both be events with probability equal to 0.75. For instance, A the event that you draw a non-spade card from a playing card deck and B the event you draw a non-heart. Then Pr(A) + Pr(B) = 0.75+0.75>1. Note that this is not the same as the probability of at least one of those things happening. – JMoravitz Feb 19 '19 at 04:08
  • Note also how "non-spade" and "non-heart" has overlap, namely they both include the possibility of being clubs or diamonds. – JMoravitz Feb 19 '19 at 04:10
  • Sorry for asking a whole new question, but how would you find the probability of at least one of those two things happening? – chris Feb 19 '19 at 04:15
  • By either examining the possibilities directly or by doing so indirectly as described in my original post as $Pr(A)+Pr(B)-Pr(A\cap B)$. In the card example note that every card is either non-heart or non-spade (some are both) so for that problem the probability of falling into at least one of those categories is $1$. Indirectly you notice the overlap is all of the clubs and diamonds which accounts for half of the cards so the probability is $0.75+0.75-0.5=1$ – JMoravitz Feb 19 '19 at 04:21
  • Got it, thanks! – chris Feb 19 '19 at 04:48
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Suppose we have three balls in a bag, one red (R), one blue (B), and one yellow (Y), well mixed, and you draw one from the bag.

The probability of drawing R is $1/3$. The probabilities of drawing B and drawing Y are the same. The probability of drawing R or B is $2/3$, as is the probabilities of drawing R or Y, and B or Y. The probability of drawing R, B, or Y is $1$. If you add all of these probabilities, you get $$\frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{2}{3} + \frac{2}{3} + \frac{2}{3} + 1 = 4.$$

It's easy to get probabilities that sum to more than $1$. The question is, if they're summing to more than $1$, then why are you summing them? It doesn't represent anything meaningful in terms of what you're trying to model. In the above example, I can't think of any use I would have for the sum of the probability that I would draw R $(1/3$) and the probability that I would draw R, B, or Y ($1$) to make $4/3$. I don't see how this number makes anything about the above situation clearer.

There is a situation where adding two probabilities does tell you something: when you add mutually exclusive events. If it is impossible for two events to happen at the same time (or at least, the probability of them happening together is $0$), then adding the probabilities of these events will tell you the probability of one or the other happening. For example, because it's impossible to draw B and Y at the same time, the probability of drawing B or Y is $1/3 + 1/3$, the probability of drawing B plus the probability of drawing Y. Whereas, the probability of drawing R or (R, B, or Y) is not $4/3$, because the event of drawing R and the event of drawing R, B, or Y are definitely not mutually exclusive, because they can both be simultaneously satisfied by drawing R.

So, yes, it's possible to sum probabilities to more than $1$, but such numbers are not relevant, and are not even probabilities. It's impossible sum the probabilities of mutually exclusive events to be more than $1$, since the result is a relevant probability: the probability that one of the events will occur.

Theo Bendit
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