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I've been struggling with this for a number of days now and so decided to ask this question. This may be trivial but I'm not particularly good or well-educated in statistics. So here goes nothing.

What I need to calculate is the overall probability of at least one out of a number, say N, events occurring. There is a number of things that I know about all the events:

  • all events are disjoint, i.e. there isn't a chance that any two can occur in the same sample (P(A or B) = P(A) + P(B))
  • events are not independent, specifically if one occurs all the following ones cannot occur
  • probability of any of the events occurring (computed externally, therefore not in the same sample space)

All this leads me to a conclusion that the probability of any of the events occurring should be the same as exactly one of them occurring, since that is the only possible outcome (out of a series of N events only 1 or 0 could happen).

What I've come up with so far is that for the first event out of the series the probability will be simply the probability of that particular event:

P(E_1) = P_1 // I know P_1 etc.

For the second event to happen event 1 must not have happened. So the probability of the second event is:

P(E_2) = (1-P_1)*P_2 

I have doubts about this, because I know that:

P(E_2 and E_1 complement) = P(E_2) - P(E_2 and E_1) = P(E_2) // Since P(E_2 and E_1)=0 because E_2 and E_1 cannot happen simultaneously.

Ignoring my doubts, I've generalised this to the following for N events (probability of Nth event occurring):

P(E_N) = (1-P_1)*(1-P_2)*...*(1-P_N-1)*P_N

Now, since the events are disjoint, the probability of any of them occurring (which, given the conditions, should be the same as exactly one occurring) is:

P(any event out of N) = P(E_1 or E_2 or … or E_N) = P_1 + (1-P_1)*P_2  + (1-P_1)*(1-P_2)*P_3 + … + (1-P_1)x(1-P_2)*...*(1-P_N-1)*P_N

Question 1 Is the above correct? Or am I simply complicating things too much and the combined probability of exactly one event occurring will be simply the sum of the probabilities of all the events?

Question 2 Now say I've computed the probability of one out of N events occurring correctly. If I add another, N+1th event, will any changes need to be made to account for the fact that the sample space may have changed? I worry that as N tends to infinity the probability of any of the events occurring will exceed 1 which will be awkward as hell.

Thanks a lot for the help. And apologies for not formatting the maths bits nicely, I'm new to the Maths part of Stack Exchange.

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  • $\begingroup$ My bad, mistake when writing this up (I actually do realise this). Thanks. $\endgroup$ – Aleksander Lidtke Feb 25 '14 at 17:29
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Following your and Stefanos' comments and your edit, I'm revising my answer. I think the first step is to precisely define what your events mean.

It seems like your probabilities $P_i$ are not the probabilities of event $i$ occurring in the sample space including all events. In a two event system, rather than $P_1=P(E_1)$, I think $P_1$ ought to be defined as the probability of event 1 happening in a world where all the other events can't happen. I.e., $$ P_1 = (E_1|\neg E_2), \text{ and similarly } P_2 = (E_2|\neg E_1) $$

What you want is $P(E_1 \cup E_2) = P(E_1) + P(E_2)$ because $E_1$ and $E_2$ are disjoint. By the theorem of total probability you can write $$P(E_1) = P(E_1|E_2)P(E_2) + P(E_1|\neg E_2)P(\neg E_2) = 0 + P_1(1-P(E_2)) = P_1(1-P(E_2))$$ And similarly $$P(E_2) = P_2(1-P(E_1))$$ Plugging these into the formula for one event or the other happening you get

$$\begin{align} P(E_1 \cup E_2) &= P(E_1) + P(E_2)\\ &= P_1(1-P(E_2)) + P_2(1-P(E_1)) \end{align}$$

Unfortunately you can't solve this for $P(E_1) + P(E_2)$ and the situation doesn't get any better as you add more events, although it does generalize easily. Nevertheless, your first step ought to be to define the problem correctly. For example you had

$P(E_2) = (1-P_1)*P_2$

whereas I think the following is correct based on my reasoning above

$P(E_2) = (1-P(E_1))*P_2$

and it's only by being very clear in the problem definition (which sadly means notation) that you can avoid subtle problems.


I've left some general comments from my original answer below in case they help.

events are not independent, specifically if one occurs all the following ones cannot occur

For the second event to happen event 1 must not have happened. So the probability of the second event is:

$P(E_2) = (1-P_1)*P_2$

This is not correct. $P(A \text{ and } B)=P(A)P(B)$ if and only $A,B$ are independent.

Question 1 ... the combined probability of exactly one event occurring will be simply the sum of the probabilities of all the events?

yes, exactly right, for any disjoint events $P(A\text{ or }B) = P(A) + P(B)$, as in Nicholas R. Peterson's comment.

(Note the general rule regardless of whether $A$ and $B$ are disjoint is $P(A\text{ or }B) = P(A) + P(B) - P(A\text{ and }B)$ which is easiest to see on a Venn diagram.)

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    $\begingroup$ The events come from different probability spaces. Is not as simple as adding their probabilities even if they are disjoint. I think there is something wrong with the definitions in the problem, since actually every sum of probabilities as proposed can exceed 1 $\endgroup$ – Jimmy R. Feb 25 '14 at 17:37
  • $\begingroup$ Your welcome, I just deleted mine too... $\endgroup$ – Jimmy R. Feb 25 '14 at 17:46
  • $\begingroup$ Right, I had a feeling about it, thanks a lot. I know about the general rule for union of probabilities (I've spent a lot of time reading about all this lately). What I worry (Question 2) is that if I keep adding more and more events the sum (combined probability) will tend to 1. Is this not the case? And if not, why? $\endgroup$ – Aleksander Lidtke Feb 25 '14 at 17:58
  • $\begingroup$ @Stefanos yes, that is exactly what I've been thinking. The problem is there is I can only obtain the probabilities the way I do, i.e. they'll always come form different spaces. $\endgroup$ – Aleksander Lidtke Feb 25 '14 at 18:01
  • $\begingroup$ @TooTone yes, exactly. But the probabilities are computed individually, i.e. in different spaces, so they do not "know" about the other events' probabilities, if that makes sense? Actually, probability of each individual event could reach almost 1.0 (0.9999...). SO just one more event would make the sum equal to 1.0. $\endgroup$ – Aleksander Lidtke Feb 25 '14 at 18:03

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