Asked by Chibuzo Ekenna Ekejiuba
Consider the communication network shown in the figure below and suppose that each link can fail with probability p. Assume that failures of different links are independent.
There is a point A, and two line segments originate from point A to the right. The top line segment is labeled 'Link 1' and the bottom line segment is labeled 'Link 3'. Link 1 is then connected on its right to another line segment labeled 'Link 2', and Link 3 is connected on its right to another line segment labeled 'Link 4'. Link 2 and Link 4 are joined on their right ends to a single point, so that Links 1,2,3 and 4 together make a diamond-shape. From the single point at the ends of Links 2 and 4 is another line segment labeled 'Link 5', and its right end is labeled 'B'.
Assume that p=1/3. Find the probability that there exists a path from A to B along which no link has failed. (Give a numerical answer.)
Given that exactly one link in the network has failed, find the probability that there exists a path from A to B along which no link has failed. (Give a numerical answer.)
There is a point A, and two line segments originate from point A to the right. The top line segment is labeled 'Link 1' and the bottom line segment is labeled 'Link 3'. Link 1 is then connected on its right to another line segment labeled 'Link 2', and Link 3 is connected on its right to another line segment labeled 'Link 4'. Link 2 and Link 4 are joined on their right ends to a single point, so that Links 1,2,3 and 4 together make a diamond-shape. From the single point at the ends of Links 2 and 4 is another line segment labeled 'Link 5', and its right end is labeled 'B'.
Assume that p=1/3. Find the probability that there exists a path from A to B along which no link has failed. (Give a numerical answer.)
Given that exactly one link in the network has failed, find the probability that there exists a path from A to B along which no link has failed. (Give a numerical answer.)
Answers
Answered by
PsyDAG
I applaud your attempt to describe the network, but it is so complex, I cannot make "heads or tails" of it. Nice try.
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