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Suppose that I know the probability of an event occurring by either process 1 or process 2:

process 1, p(event)=1-exp(-ax) process 2, p(event)=1-exp(-by)

With this information, is it possible to calculate the probability of the event occurring by process 1 given that process 2 also occurs?

To be more specific, my process involves a receptor that receives particles either by process 1 or process 2. It models whether or not the receptor is remotely controlled. It assumes that as x (process 1) or y (process 2) increases, there is a greater likelihood of the receptor being brought under control. I now need to know the probability that particles delivered by process 1 control the receptor when process 2 is also delivering particles (and vice versa).

So if there is a 90% chance that process 1 controls the receptor (in the absence of process 2), and an 85% chance that process 2 controls the receptor (in the absence of process 1), would it be possible to determine the probability that process 1 controls the receptor given the process 2's probability in the absence of process 1?

I should also note that I do not know the frequencies, so I could not calculate the relative abundance of particles originating from process 1 and 2.

Thank you kindly

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  • $\begingroup$ Can you be more exact in the beginning of your question? A statement like "p(event)=1-exp(-ax)" makes litte sense if we don't how this event is related to "a" and "x" $\endgroup$ – StijnDeVuyst Mar 22 '15 at 15:45
  • $\begingroup$ When there are x particles, there is a probability that process 1 "wins control". When there are y particles, there is a probability that process 2 "wins control". These probabilities are given by the functions above which apply when the two processes occur in isolation. When the two processes occur together, is there any way to calculate the probability that process 1 "wins control" given what we know about the two processes when they do not occur together. i.e., for a given value of x there is an 80% chance and for a given value of y there is a 60% chance. what is the prob of p1 winning? $\endgroup$ – rabbit1 Mar 22 '15 at 18:28
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I'll give it a shot. From what you write, I suspect the time $X$ until process 1 is under control is exponentially distributed with mean length $1/a$. Likewise, the time $Y$ until process 2 is under control is exponential with mean $1/b$. That is assuming only one of these processes is taking place.

If both of them take place concurrently, and either one of them can bring the receptor or whatever under control, the natural question to ask is of course: what is the distribution of $\min(X,Y)$, the time until there is control. As I mentioned elsewhere on this forum (can't find the link), that time is also exponential and has mean $1/(a+b)$. Moreover, if this event happens, i.e. when either 1 or 2 brings control, it will be due to process 1 with probability $a/(a+b)$ and due to process 2 with probability $b/(a+b)$.

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  • $\begingroup$ why do you say " the natural question to ask is of course: what is the distribution of min(X,Y)"? $\endgroup$ – rabbit1 Mar 22 '15 at 18:42

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