Both $\mathrm{A}$ and $\mathrm{B}$ are $d$-dimensional vector with all the elements being $1$. Then we randomly and uniformly pick one element from $\mathrm{A}$ with replacement, and let it be $0$. We repeat this procedure for $m$ times. Then we also independently do the same thing for vector $\mathrm{B}$. Assuming $2m\leq d$. Then, what is the expectation for how many zero elements in $\mathrm{A}+\mathrm{B}$? Thanks very much for your help. We know that we at most have $2m$ zero elements.
2026-03-30 00:19:56.1774829996
How many 'zero' elements in $\mathrm{A}+\mathrm{B}$ after randomly letting one entry in $\mathrm{A, B}$ be $0$ for $m$ times,respectively?
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The probability a particular element of $A$ is $1$ after $d$ picks is $\left(\dfrac{m-1}{m}\right)^d$.
The probability a particular element of $A$ is $0$ after $d$ picks is $1-\left(\dfrac{m-1}{m}\right)^d$.
The probability a particular element of $A+B$ is $0$ after $d$ picks is $\left(1-\left(\dfrac{m-1}{m}\right)^d\right)^2$.
The expected number of zero elements of $A+B$ after $d$ picks is $m\left(1-\left(\dfrac{m-1}{m}\right)^d\right)^2$.