Let $X_1, \dots, X_c$ be independent uniform random variables on $[0,1]$. We know the number of random variables whose value will fall in the range $[0,1]$ will always be exactly $c$. Now define $Z_i = X_i + Y_i$ where $Y_i$ are independent random variables that are exponentially distribution. Let $C_Z$ be the number of random variables $Z_i$ whose value falls in the range $[0,1]$. I am interested in the distribution of $C_Z$.
If we set $c=100$ and $\lambda = 1$ then, by simulation the pdf of $C_Z$ looks like:
I fitted a normal distribution to the data and have drawn that on top of the histogram created by simulation. This is just to show the data appears to be approximately normal.
It seems that $C_Z$ has a shifted binomial distribution. Is that right and if so, why?
Update
@E-A points out that the pdf is binomial with parameters (p, c) with $p = P(X_i + Y_i \leq 1)$. But what is $P(X_i + Y_i \leq 1)$? By simulation again I plotted $P(X_i + Y_i \leq 1)$ for $0 \leq \lambda \leq 5$. Does this have a simple form?
I don't know what you mean by shifted, but assuming that $Y_i$s all have the same parameter for their distribution, the random you are looking for will be binomial with parameters (p, c) with $p = P(X_i + Y_i \leq 1)$. This is because you can think of each event $\{X_i + Y_i \leq 1 \}$ as an independent trial (say a coin toss), and you will do it $c$ times.
Note: In the question it currently says $X_i + Y$ instead of $X_i + Y_i$ but from context it seems the intention was to ask $X_i + Y_i$ so I answered it for $X_i + Y_i$; if the intended meaning is $X_i + Y$, then it will not necessarily be binomial; its distribution would be the average of binomial distributions.