Consider the example problem from Richard Durrett's Essentials Of Stochastic Processes
That is, part c.)
At quick glance the answer provided is reasonable
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However I have some objections to the solution. First, why do we count the probability of 5 cars arriving before any truck (that is $(2/3)^5)$? If we already count the probability that 4 cars arrived before two trucks, then we shouldn't need to count 5 cars in succession; we are already guaranteed that four cars have arrived before 2 trucks.
If we consider the proposition that it should be at least 4 out of 5 arrivals are cars, then this can answer the above question. However by this same logic we then should consider 6 cars before any truck $(2/3)^6$ and 7 cars $(2/3)^7$ before any truck and onward to infinity.
Is it because I need to consider a sample space of 6 vehicles?
As you suggest, he is essentially restricting the sample space to what happens in the first five arrivals. In this case, there are $2^5 = 32$ possible sequences of arrivals, since at each arrival there are two possibilities ($C$ or $T$). Of these 32 possible sequences, there are five sequences where four cars arrive before two trucks do, $(C,C,C,C,C)$, $(C,C,C,C,T)$, $(C,C,C,T,C)$, $(C,C,T,C,C)$, $(C,T,C,C,C)$ and $(T,C,C,C,C)$. The probability of the first sequence is $(2/3)^5$, and the probability of the other five is $(2/3)^4(1/3)$, which is how you get his answer. You have to include $(C,C,C,C,C)$ in this calculation, since this is one of the outcomes in the sample space where four cars arrive before two trucks do.
The reason you don't need to consider 6 cars in succession, 7 cars in succession etc. is that this is outside the above sample space. If you want to include the sixth arrival, you can do it, but you have to change the sample space to considering what happens in the first six arrivals. In that case, there are $2^6 = 120$ possible outcomes, and the outcomes where four cars arrive before two trucks do are $(C,C,C,C,C,C)$, $(C,C,C,C,C,T)$, $(C,C,C,C,T,C)$, $(C,C,C,T,C,C)$, $(C,C,T,C,C,C)$, $(C,T,C,C,C,C)$, $(T,C,C,C,C,C)$, $(C,C,C,C,T,T)$, $(C,C,C,T,C,T)$, $(C,C,T,C,C,T)$, $(C,T,C,C,C,T)$ and $(T,C,C,C,C,T)$. The probability of the first outcome is $(2/3)^6$, the probability of the next six are $(2/3)^5(1/3)$, and the probability of the last five is $(2/3)^4(1/3)$. The total probability is therefore $$ (2/3)^6 + 6(2/3)^5(1/3) + 5(2/3)^4(1/3)^2 = 0.4609. $$ You can follow the same approach if you want to consider the first seven arrivals, but that will obviously become even more complicated.
Note: I think there is a typo in the calculations you cite above. The answer should be 0.4609, not 0.4069 (try calculating (7/3)(2/3)^4).
Does this make sense?