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The number of customers arriving at an ATM is in accordance with the Poisson distribution, with a mean rate of 1 customer in every 10 mins. What is the probability that there will be atleast 2 customers in half an hour??

Solution: Acc. to Poisson distribution, $$\text{Mean}=\lambda=\frac{1}{10}$$ $$\mathbb{P}(X\geq{2})=1-\mathbb{P}(X=0)-\mathbb{P}(X=1)$$ Is this the correct way, then how to calculate for 30 mins????

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Certainly not my forte... but since no one has answered, i think you may be interested in looking at the 'erlang distribution.' According to Wikipedia: "Events that occur independently with some average rate are modeled with a Poisson process. The waiting times between k occurrences of the event are Erlang distributed." Your question appears to be considering waiting times between 2 occurrence of an event. – student101 Sep 26 '12 at 12:22
If the mean rate is 1 customer in 10 minutes then it is 3 customers in half an hour, and 3 is the parameter you have to use in the Poisson distribution. – Henry Sep 26 '12 at 12:59
up vote 1 down vote accepted

What you have is a time homogenous Poisson process with intensity $\lambda=1/10$, minutes scale. It says that the number of event in the interval $[t,t+T]$ is distributed as a Poisson RV with intensity $T\lambda$. So in your case, you find

$$ P(N\geq2)=1-P(N=0)-P(N=1)=1-\frac{e^{-\frac{30}{10}}(30/10)^0}{0!}-\frac{e^{-\frac{30}{10}}(30/10)^1}{1!}=1-4e^{-3} $$

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