Maybe Poisson Distribution? The number of visitors I'm trying to solve the problem, can you help me?

To study the behavior of its costumers, a small shop registers the number of visitors that it receives per hour during a day. The results show that, on average, it has 1 visitors every hour. If we are told that in a given hour the shop has received 1 visitor at least, what is the probability that it will receive exactly 3 visitors within that hour? 
HINT: Enter your answer with at least 3 decimal digits.
Let $N$ be the random variable giving the number of visitors in a given hour.

$$P(N=k)=e^{-\lambda}\frac{\lambda^k}{k!}$$

Here we know the expectation is $1$, so $\lambda=1$, since $E(N)=\lambda$.

The event $N\ge1$ the complement of $N&lt;1$, which is $N=0$. So:

$$P(N\ge1)=1-P(N=0)=1-e^{-1}$$

Now, you want to know the probability that $N=3$ given that $N\ge1$, that is $P(N=3|N\ge1)$:

$$P(N=3|N\ge1)=\frac{P(N=3\wedge N\ge1)}{P(N\ge1)}$$

But of course, $N=3\wedge N\ge1$ is the same event as $N=3$, so

$$P(N=3|N\ge1)=\frac{P(N=3)}{P(N\ge1)}=\frac{\frac{e^{-1}}{3!}}{1-e^{-1}}=\frac{1}{6(e-1)}$$
Probability $P(3visitors)=e^{-\lambda}\frac{\lambda^k}{k!}=\frac{1}{6e}=0.06131$
where $\lambda=$no. of visitors per hour
$k=$ required no. of visitors.