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The arrival of buses at a given bus stop follows Poisson law with rate 2. The arrival of taxis at the same bus stop is also Poisson, with rate 3. What is the probability that next time I'll go to the bus stop I'll see two taxis arriving before a bus?

We'll assume buses and taxis arrive independently.With rate lambda=2, the probability of waiting until time t for the first arrival of a bus is .With rate lambda=3, the probability of seeing exactly two taxis in time t is So, we need to integrate P(exactly 2 taxis in time t) * P(first bus arrives at time t) over all values of t --> Define k=5u and substitute:Which is 9/125 = 0.072

The probability of observing a taxi before a bus is given by 3/(3+2)=3/5 since the waiting times are independent and exponentially distributed. By the memoryless property both processes then restart and hence the probability of observing (at least) two taxis before the first bus is (3/5)^2=9/25. The probability of observing exactly two taxis before the first bus is (3/5)^2*(2/5)=18/125.

I feel, if the probability that "the first one you see is a taxi" is p,then the probability you want is p*p*(1-p).

If you let X=# taxis before a bus, then X~Ge(1-p), that is P(X=k)=p^k*(1-p), where p=P(observing a taxi). In this case we have k=2 and p=3/5. Moreover, the same reasoning can be applied if we add more (independent ) arrival processes.