# Communications engineering, call arrival rate, poisson

I am taking a course in communication engineering which involves some probability theory that I'm not quite familiar with. Please see the attached image for my problem. I don't get why that equation is valid. The two events are dependent (call arrival and call failure) so I initally thought an application of Baye's rule would solve it. Can someone please explain this to me?

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stats.stackexchange.com may be a better place for this question. –  JonnyBoats Apr 7 '12 at 13:45
Communication theory has always been considered acceptable here. This is definitely a boundary question, but I think one that is still on topic. If you would like to continue discussion of it @JonnyBoats ask on meta. Right now as a moderator I am not going to take action as I support this as on topic, at its heart all communication theory is applied statistics. –  Kortuk Apr 7 '12 at 16:38
ahh great, cheers Kortuk! Posted this on maths.stackexchange too but I hoped that by posting it here I will perhaps get more engineering oriented perspectives. –  xupv5 Apr 8 '12 at 12:23

The key is actually that the results of the individual trials are independent of each other. If you get a successful "arrival" on the first trial, it doesn't affect what you get on the 2nd or later trials. This gives the binomial distribution, which is described by the probability mass function given in your question. It's easily derived:

You're making n trials. The result of each trial will be either an "arrival" or a "failure". The probability of failure in each trial is given by p.

The total number of (ordered) outcomes is $2^n$. For example, for 3 trials you could have aaa, aaf, afa, aff, faa, faf, ffa, or fff. The probability of each of these ordered results happening is $p^k(1-p)^{(n-k)}$, because if p is the probability of an arrival, then (1-p) must be the probability of failure.

So then you apply combinatorics. You've decided (or been told) you don't care about the order of the results, just how many arrivals. So out of the total $2^n$ ordered results, the number that meet your requirement is given by $n \choose k$, which is $\frac{n!}{k!(n-k)!}$.

Taken together, you have $n\choose{}k$ results that meet your requirements, each with probability $p^k(1-p)^{(n-k)}$, so the total probability is

${n\choose{}k} p^k(1-p)^{n-k}$

which is the result you were looking for.

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Just had a look at binomial distributions. Cheers for this explanation. –  xupv5 Apr 8 '12 at 12:36