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#1




ASM lesson 14 example 14B
could someone help explain the logic in ASM lesson 14 example 14B?
'express trains arrive in a poisson process at a rate of 4 per hour. Local trains arrive in a poisson rate of 12 per hour. Calculate the probability that at least 4 locals will arrive before 3 express' I understand the probability that a local arrives before express is 4/(4+12) and that the probability of 4 local before 3 express is (6 C 4) (.75^4)(.25^2) but why does he lock it at 6 trains? ...he says the p of 4 or more of the next 6 trains are local is.. (6 C 4) (.75^4)(.25^2) + (6 C 5) (.75^5)(.25^1) + (6 C 6) (.75^6) why do we always expect 6? Couldn't there be 5 local and 2 express...6 local and 2 express...7 local and 2 express? 
#2




When the sixth train shows up, you know whether you won or not. If you have seen 4 (or more) locals, you won. If you have seen 3 or fewer, then you have seen 3 or more expresses, so you lost.
There are 3 ways to win: exactly 4 at time 6, exactly 5 at time 6, and exactly 6 at time 6. These are disjoint so the probabilities add. Presto.
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#4




As CE said, after 6 trains you know, so it is by far the easiest way to calculate what you want.
You want to be sure that 5 local and 2 express is covered? To get that, you either need the first 6 to be 4 local and 2 express, then a local (included in the 6C4 term) or 5 local and 1 express, then an express (included in the 6C5 term). Can you think of any order of arrivals that is not included in one of those terms? (Or that is included in more than one, which would create a doublecounting problem)(or any order than is included in one of those terms that should not be included at all) 
#5




I read a similar example where it asks for 2 of type A before 3 of type B. The solution is similar in that it basically requires 2 of A out of 4, 3 of A out of 4, 4 of A out of 4.
The difference is in the wording. The original example asks for 'Pr at least 4 local before 3 express' where as this example asks 'Pr 2 of A before 3 of B' but both cases seem to be asking an 'At least' type question. 
#6




Yes, they are the same. One wording version focuses on counting A´s and B’s until you get enough A’s or get too many B´s. The other focuses on counting them until you get too many B’s, then deciding if you had enough A’s. Same conclusion either way, even though the second is more counting until you decide whether you got enough.
And same solution to either wording. Instead of a infinite sum, with complications of cases being in more than one term, do a binomial calculation with (the A target plus the B target number  1) trials. 
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