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.
The (a) part could be simply solved by binomial. Although, Poisson is still an option, it just gets a little messy.
Binomial will simply be: Binomial(20,1/12). Try to check by solving. The answer will be (11/12)^32
Best Answer
Your reasoning is correct.
The joint conditional distribution for the arrival times in the interval, under the condition that there are exactly two, is that for the order statistics for two identical and independent uniformly distributed samples over the interval.