Let there be three cash registers within a small market. Each of these registers serves an average of k customers per hour, independently of the other registers. If all three cash registers work for an hour, estimate the total number of customers served.
My solution:
Let $X_n$ denote the number of customers served at the $n$-th cash register within an hour. Then $X_n \sim P(k)$, where $P(k)$ is the Poisson distribution with a parameter $k$. The total number of customers served within one hour would then be a random variable $X$, such that $X=\sum_{i=1}^3 X_i$ and therefore $X \sim P(\sum _{i=1}^3 k)$, i.e. $X \sim P(3k)$. So, we're looking for $P\{X=1\}$, which is $3*k*e^{-3k}$.
I want to ask if my solution is correct, since my textbook doesn't have an answer sheet.
The problem is somewhat idiosyncratically phrased – one usually considers the expected value of a random variable, rather than estimating its value. As Mick A pointed out, the expected value could serve as an estimate of the value, whereas the probability of the variable taking the value $1$ is not a useful estimate of its value.
You determined the correct distribution for $X$, but note that this was a) only possible because the individual distributions had the propitious property of adding up nicely, and b) not necessary, since by the linearity of expectation you could have just added up the expected values of the individual distributions.