Physics
Physics, 18.08.2021 01:20, abbeygrace13

All ships travel at the same speed through a wide canal. Each ship takes days to traverse the length of the canal. Eastbound ships (i. e., ships traveling east) arrive as a Poisson process with an arrival rate of ships per day. Westbound ships (i. e., ships traveling west) arrive as an independent Poisson process with an arrival rate of ships per day. A pointer at some location in the canal is always pointing in the direction of travel of the most recent ship to pass it. In each part below, your answers will be algebraic expressions in terms of and/or . Enter 'LE' for and 'LW' for , and use 'exp()' for exponentials. Do not use 'fac()' or '!' for factorials; instead, calculate out the numerical value of any factorials. Follow standard notation.

For parts (1) and (2), suppose that the pointer is currently pointing west.

1) What is the probability that the next ship to pass will be westbound?
2) Determine the PDF, , of the remaining time, , until the pointer changes direction.
For the remaining parts of this problem, we make no assumptions about the current direction of the pointer.

3) What is the probability that an eastbound ship does not pass by any westbound ships during its eastward journey through the canal?
4) Starting at an arbitrary time, we monitor the cross-section of the canal at some fixed location along its length. Let be the amount of time that will have elapsed (since we began monitoring) by the time we observe our seventh eastbound ship. Determine the PDF of .
5) What is the probability that the next ship to arrive causes a change in the direction of the pointer?

6) If we begin monitoring a fixed cross-section of the canal at an arbitrary time, determine the probability mass function for , the total number of ships we observe up to and including the seventh eastbound ship we see. The answer will be of the form , for suitable algebraic expressions in place of and .

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All ships travel at the same speed through a wide canal. Each ship takes days to traverse the length...

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