# Probability of customers arriving and waiting

Customers arrive at a bank at the mean rate of 20 per hour. the exponential probability distribution describe the time between customer arrivals. (a) What is the probability that a customer arrives within 3 mins of a previous customer? (b) What is the probability that the bank will go 6 mins without any customer?

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(a) You might think about accepting some answers –  Henry Sep 26 '12 at 16:44
(b) Part (b) might depend on how long it takes to serve a customer –  Henry Sep 26 '12 at 16:45
In (b), do you mean "without any customer arriving"? If you mean "without any customer being there", Henry's comment (b) applies. Also, presumably you mean "go without any customer for 6 particular minutes"? The probability that the bank will at some point go without any customer for 6 minutes is $1$. –  joriki Sep 26 '12 at 16:53

We need to decide between minutes and hours for our unit of time. Say it is minutes. Then the mean time between arrivals is $3$ minutes. Or else, depending on the way the exponential distribution has been introduced to you, the rate is $1/3$.

Recall that an exponential distribution with parameter $\lambda$ has mean $\frac{1}{\lambda}$. So $\frac{1}{\lambda}=3$ and therefore $\lambda=\frac{1}{3}$.

A customer has just arrived. Let $X$ be the waiting time until the next customer arrives. Then $X$ has exponential distribution with parameter $\lambda=\frac{1}{3}$. For any positive $x$, $$\Pr(X\le x)=\int_0^x \frac{1}{3}e^{-t/3}\,dt=1-e^{-x/3}.\tag{1}$$

Now we can answer the questions. Interpretation is needed, since there are some ambiguities in the questions.

(a) Interpret the question as saying: "Customer Alicia has just arrived. What is the probability that there will be a customer who arrives later than Alicia, but no more than $3$ minutes later." Then we want $\Pr(X\le 3)$. By $(1)$, this is $1-e^{-3/3}$.

(b) Interpret the question as saying that Alicia has just arrived, and we want the probability that there will be a gap of at least $6$ minutes until the next customer arrives. Then we want $\Pr(X\gt 6$. By $(1)$, this is $1-\Pr(X\le 6)$, which is $1-(1-e^{-6/3})$.

Remark: The exponential distribution is at best a crude model of the situation. For one thing, banks do close. For another, there is always a long line when one is in a hurry.

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