Expected number of same numbered balls in a box I have two boxes: A,B. The boxes A contains $n_1$ red balls which numbered from $(1, \cdots, n_1)$. The box B includes $n_2$ green balls which also numbered from $(1, \cdots, n_2)$. Throw balls from two boxes into a box C. However, some balls will drop-out with probability is $p$. **How many balls have the same number in box C?** A toy example is shown in below figure. In which, the box C will contain two balls (4,7) are the same number. Thus, the answer is 2. ![enter image description here](

Let $m:=\min(n_1,n_2)$ and for $i\in\\{1,\dots,m\\}$ let $X_i$ take value $1$ if a red and a blue ball both with number $i$ will end up in box $C$. If this does not happen then let $X_i$ take value $0$. Then $$X:=X_1+\dots+X_m$$ equals the number of numbers doubly presented in box $C$. There are $2X$ balls that have a "numbermate" in the sense that there is another ball in box $C$ having the same number.

If you are after expectation then you can use linearity of expectation and symmetry to find that: $$\mathbb EX=m\Pr(X_1=1)$$

Also the $X_i$ are independent so that $X$ will have binomial distribution with parameters $m$ and $\Pr(X_1=1)$.

Can you find $\Pr(X_1=1)$ yourself?