Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
(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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.