Better than 98%.Explanation
The crucial datum is the chance of Murial catching no trout in her two hours. If the rate is 2 trout in 1 hour, then it is also 4 trout in 2 hours, which is the size window from which the capacity of Hidden Lake is being tested by our two lovers. Given r = 4, the probability p(0) of no trout in one hour can be seen from the Poisson Table to be 0.0183.
Those who remember earlier lessons can argue that the chance of no trout in 1 hour (see again the Poisson Table, but this time under r = 2) is 0.1353, so that her chance of getting the same result twice in a row should be (0.1353)(0.1353) = 0.01830609, or 0.0183 for short. The two arguments reach the same result. This is mentioned to reassure those who feel we have somehow entered a different world. We have not. We are just learning more about the world we already knew.
The chance of a no-trout result is thus less than 2%. This amounts to a better than 98% chance that Muriel will catch at least one fish in 2 hours. We have so reported, above. It looks, from the table, as though the marriage were almost a sure thing. On the other hand, if we even have to calculate the odds on Muriel's keeping her temper for these two luxurious and expensive hours, what about every other hour, and its irritations, between now and the wedding date? It might be unwise to bet on it.
Note that this is a Poisson problem because we cannot meaningfully ask how many fish do not bite during the two hours in question. We don't even know how many fish there are in Hidden Lake. We only know that they are added by the lake staff until the advertised average rate r = 2 is observed. The rate is really the only thing we know. In Poisson problems, the rate is the only thing that matters.
This is an "arrival" problem if you think of a trout "arriving" on Muriel's hook. There is no need to stress this business of a separate class of "arrival" problems; all Poisson problems are ultimately the same problem. But that is why this problem is in this section.
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