Poisson Distribution Lambda, Probability, and Looking for Exactly k Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with a rate of $ \lambda $ = 10 per hour. Suppose that with probability 0.5 an arriving vehicle will have no equipment violations.
What is the probability that exactly 5 have no violations?
I wanted to know if this process was correct?
And I'm a bit confused as to how the probability comes into play here.
$ p_X(k) = e^{-\lambda} \cdot \frac{\lambda^k}{k!} $
Then, $~ p_X(5) = e^{-10} \cdot \frac{10^5}{5!} = 0.0378 = 3.78\%$
 A: If $X$ has Poisson distribution with parameter $\lambda$, and $Y$ has binomial distribution with the number of trials equal to the random variable $X$, and $p$ any fixed probability $\ne 0$, then the number of "successes" has Poisson distribution with parameter $\lambda p$. 
This has been proved repeatedly on MSE, at least twice by me. Here is a link to a proof.    
In our case $\lambda=10$ and $p=0.5$, so the required probability is $e^{-5}\frac{5^5}{5!}$. (It is not the number you obtained.)
A: In my previous comment I said it was correct, but I was a bit too quick and didn't read the question properly. My apologies for that.
What you need to do is to consider the hierarchical model
$$
p(Y|X=k)
$$
where
$$
\begin{align}
Y|X=k &\sim Bin(k, 0.5)\\
X &\sim Po(10)
\end{align}
$$
Then to find $p(Y=5)$ you need to find the marginal, i.e.
$$
p(Y=y)=\sum_{k=0}^\infty p(Y=y|X=k)p(X=k)=\sum_{k=y}^\infty p(Y=y|X=k)p(X=k).
$$
The last equality follows from the fact that if $k<y$ then the probability is 0. For example, you cannot have 5 vehicles with violations if there are only 4 vehicles to begin with.
