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, $~ p_X(5) = e^{-10} \cdot \frac{10^5}{5!} = 0.0378 = 3.78\%$

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.)
