Sum of uniform distribution transformations

Question

Let $U_1, U_2,...,U_n$ be independent uniform random variables distributed on (0,1).

Let $Z_i$ be a random variable that takes value 1 if $U_i \le \frac 14$, $0$ otherwise.

Therefore, $Z_i=1$, when $U_i \le \frac 14$, has PMF (according to my calculations) $$p_{Z_i}(z)=\frac 14, \text{when } Z_i=1$$ $$p_{Z_i}(z)=\frac 34, \text{when } Z_i=0$$ Now let $$Y_n=\sum_{i=1}^{n}Z_i$$ Find the distribution of $Y_n$.

My working: $Y_n=Z_1+Z_2+...+Z_n$

Every $Z$ is a Bernoulli random variable with mean $p=\frac 14$ (I think). So $Y_n$ must be a Binomial distribution with PMF $$p_Y(k)=P(X=k)={n \choose k}p^k(1-p)^{n-k}$$ $$=p_Y(k)=P(X=k)={n \choose k}(\frac 14)^k(\frac 34)^{n-k}$$ Am I correct in this? The thing I'm most concerned about is being wrong in my PMF for $Z_i$. Any help would be lovely.

Answer 1

You are correct. $Z_i$ are Bernoulli.

Answer 2

It is correct, but one small correction. Your solution should read: $$ p_{Z_i} (z) = 1/4 \text{, when } z = 1 $$ $$ p_{Z_i} (z) = 3/4 \text{, when } z = 0 $$