Probability of losing packets 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 twice
A) 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.

The approximation of $10\%$ given in the exercise is not quite exact. For the exact value: Each fragment has a probability of $0.99$ to arrive. Assuming these probabilities are independent, we get a chance of $0.99^{10}$ of all fragments to arrive, which makes the probability of a packet loss $$1 - 0.99^{10} \approx 9.562\%.$$
For part A, with a probability of
$$(1 - 0.99^{10})^2 \approx 0.914\%$$
this happens twice in a row.
For part B, a single fragment is lost if it is lost in both transmissions, with occures with a probability of $0.01^2$. So it is transmitted correctly with a probability of $1 - 0.01^2$.
So all $10$ fragments are transmitted correctly with a probability of $(1 - 0.01^2)^{10}$, which yields the probability of a packet loss as
$$1 - (1 - 0.01^2)^{10} \approx 0.100\%.$$
