# Probability of receiving packets

Suppose we send $n$ packets over the Internet and assume that all $n$ packets are routed along the same path, and with probability $p$, one of the links along the path fails and all $n$ packets are lost. Otherwise all packets are received..

What kind of distributions is this and how to compute the distribution and expectation of this model.

I have something like:

Let $X$ be the random variable of the number of packets loss. And $X$ can only be 0 or $n$. So $P(X=0) = (1-p)^n$ and $P(X=n) = p$. But I don't recognize this distribution and don't really know how to compute the expectation of this if this is correct distribution.

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I think $(1-p)^n$ is wrong. Why $n$? – Weltschmerz Oct 28 '11 at 19:46
because there is no packet loss when all packets go through, then $(1-p)$ is the probability for each packet going through. – geraldgreen Oct 28 '11 at 19:48
You said all the $n$ packets are lost with probability $p$. And there are only two cases: either all the $n$ packets go through, or none of them does, right? – Weltschmerz Oct 28 '11 at 19:52
The probability of losing 0 is p when n > 0, and 1 when n = 0. – smdrager Oct 28 '11 at 20:36
We can change the physical situation, and assume that with probability $p$ a given packet is lost, that there is independence (?), and that if one packet is lost, the message cannot be reconstructed, so all are lost. Then $P(X=0)=(1-p)^n$, as in the post, but $P(X=n)=1-(1-p)^n$. Again we have a modified Bernoulli, and the expectation is easy to write down. – André Nicolas Oct 28 '11 at 21:51

The $n$ packets go through, with probability $1-p$. They don't go through with probability $p$.
If there's only one trial, then this is "re-scaled" Bernoulli distribution. $X$ takes on either $0$ or $n$. $P(X=0)=1-p$, $P(X=n)=p$.
I think you've got some confusions to sort out there. You can either say that either none or all of the packets go through, or you can say that each has probability of $p$ to go through. It makes no sense to mix these two situations.
So if the situation really is as you describe it in the first paragraph, then $P(X=0)=1-p$ and $P(X=n)=p$; that's all there is to it. (Note that your distribution also couldn't have been right because it wasn't normalized to $1$.)