How is this related to a Poisson distribution? In the textbook of Walpole at al. the following exercise (Ex. 5.49) appeared in the Section on the Poisson Distribution:

The probability that a person living in a certain
city owns a dog is estimated to be 0.3. Find the probability that the tenth person randomly interviewed in
that city is the ﬁfth one to own a dog.

The solution as given in the appendix is 0.0515. I arrive at this result, when I consider this as a negative binomial random variable because the question asks not only for a number of persons but also for a specific person to meet the success condition; then this number is $\binom{9}{4} p^5q^5$ for $p = 0.3, q = 0.7$ according to the formula for this distribution.
But I have no idea how to solve this exercise when I consider this as a Poisson random variable.
 A: Suppose that random variable $X$ follows a Negative Binomial distribution with parameter $r$ and $p$, in your case $X\sim NB(r=5,p=0.3)$. Then, in that case, $X$ can be viewed as a compound Poisson distribution. For further information for the compound Poisson distribution you can check here https://en.wikipedia.org/wiki/Compound_Poisson_distribution.
What is a compound Poisson distribution, is an alternative way of viewing a Negative Binomial distribution, by incorporating Poisson Distribution together with a Logarithmic Distribution.
In particular, let $Y_{n}$ independent and identically distributed random variables, each having a Logarithmic distribution, https://en.wikipedia.org/wiki/Logarithmic_distribution, with parameter $p$, also let $N$ a Poisson random variable with parameter $\lambda = -r \ ln(1-p)$. Then the random sum $\sum_{n=1}^{N}Y_{n}$ follows a $NB(r,p)$, and this sum is random because the index $N$ is random.
Hence, you can view the Negative Binomial random variable as $X= \sum_{n=1}^{N}Y_{n}$, i.e. with the use of Poisson Distribution.
Also, you can check that you can take the same result by using this alternative representation.
library(actuar)

X= c()
for(iter in 1:100000){

N = rpois(1,1.783375) #lambda = 1.1783375 = -5*ln(1-0.3)
Y = c()
for(n in 1:N){
Y[n] = rlogarithmic(1, 0.3)
}
X[iter] = sum(Y)
}

length(which(X==5))/length(X)
[1] 0.05161
```

