On the product of Bernoulli random variables Let $X$ and $Y$ be two independent random variables which take only the values $0$ and $1$ with identical positive probabilities. Can $X$ and $Y$ be chosen such that $P(XY=0)=1$?
My attempt:
Let $P(X=0)=P(Y=0)=p$.
Then,$P(XY=0)=2p-p^2$.
Therefore $P(XY=0)=1 \Rightarrow p=1$.
Therefore not possible as $1-p=0$.

Since $X,Y$ have the same Bernoulli distribution, we can write $P(X=0)=P(Y=0)=p_1$ and $P(X=1)=P(Y=1)=p_2$ where $p_1,p_2\in[0,1]$.
Now, the probability $P(XY=0)=1-P(X\ne0 , Y\ne 0)=1-P(X=1,Y=1)$.
And since $X,Y$ are independent, $P(X=1,Y=1)=P(X=1)P(Y=1)=p_2^2$.
So, yes, $X,Y$ can be chosen such that $P(XY=0)=1$, by choosing $p_2=0$.
