Assume that the flaws on a magnetic tape follow a Poisson distribution with a mean of $0.2$ flaws per metre.
Let $X$ denote the distance between $2$ successive flaws.
What is the probability that the first time the distance between $2$ flaws exceeds $8$ metres is at the $5$th flaw?
I don't seem to be able to make any headway into this question... Any help would be extremely appreciated...
ASSUMING THE 8M DISTANCE HAPPENS BETWEEN CONSECUTIVE FAILULRES: If you know the distribution for the waiting time until the next failure after a failure occurs (and note that these waiting times are independent), then let $X_j$ be the waiting time after the $j$th failure until the next failure. Note all the $X_j$ have the same distribution (i.e., the distribution for waiting time until a failure). You want $X_1,X_2,X_3$ to all be less than $8$ and $X_4$ to be greater than 8. Multiply the probabilities for those 4 things happening and you've got your answer.
UPDATE: For a Poisson process with rate $c$, the distribution of the number of events over an interval of length $r$ is standard Poisson with parameter $rc$. That settles the first question, you want exactly one failure in an interval of length 8m, (and for this to happen for 3 different intervals of length 8m), and then you want exactly zero failures in an interval of length 8m.
If you want the probability that the distance between the 1st and 5th failures is at least 8, and no other failures are 8m apart, that makes things different. The location of the first failure isn't important, you just want the next 3 failures to all occur within 8m after the first one. So you want exactly 3 failures in an interval of length 8m. Then the position of the 5th failure isn't important, because it will be at least 8m after the first.