I am currently enrolled in an Intro to Networking course and I have been studying for an upcoming exam by doing practice problems in the course textbook. I came across this question that stumped me. It deals with probability, which I have always had a hard time understanding and never really studied in depth.
Suppose an IP packet is fragmented into 10 fragments, each with a 1%
(independent) probability of loss. To a reasonable approximation, this
means there is a 10% chance of losing the whole packet due to loss of
a fragment. What is the probability of net loss of the whole packet if
the packet is transmitted twiceA) Assuming all fragments received must have been part of the same
transmission?B) Assuming any given fragment may have been part of either
transmission?
I assume for part A there is a 20% chance of losing the whole packet, because since we send two transmissions the probability doubles. What I don't understand is how the probability is affected in part B. How does the probability change if the two packets are mixed within each transmission? Can someone explain this to someone who is not a math major? Thanks in advance.
Best Answer
With all respect, never ask a mathematician to solve an engineering problem; the lack of precision inherent in the real world gives them heart palpitations:-)
As the question states, " to a reasonable approximation ... 10% chance of losing the whole packet" (mathematicians can't read either:-)). There is also an inherent assumption that the loss of fragments is independent; in a real network this is unlikely to be the case.
So the chance of losing both is $.1\times.1=.01$. So the chance of getting at least 1 is $1-0.01=0.99$.
For part B, the chance of losing any single fragment is $0.01\times0.01=0.0001$. Now, if any of the 10 fragments is lost twice the whole packet is lost. So we must get 10 good fragments, this happens with a probability of $(1-0.0001)^{10}\approx0.999$ which gives a loss probability of $0.001$.