Earlier today I came up with a very complicated "proof" of the probability multiplication rule for two independent events. I used the quotation marks because I ended up obtaining a very complicated formula for the wanted probability, and only verified that it gave the right values for different parameters by using desmos. But, I don't know how to simplify the formula into the more basic form. In short, I want to show that
$$ \sum_{i=1}^R \frac{i}{N}\frac{\frac{N!}{i! (R-i)! (B-i)!(N+i-R-B)!}}{\binom{N}{R} \binom{N}{B}}= \frac{RB}{N^2} $$
Where $R\le B$, And $R+B\le N$
does anyone have any idea of how to do this?
LHS can be simplified to: $$\sum_{i=1}^R\frac{i}N\frac{\binom{R}i\binom{N-R}{B-i}}{\binom{N}B}$$
Then to be explained is the equality:$$\sum_{i=1}^R\frac{i\binom{R}i\binom{N-R}{B-i}}{\binom{N}B}=\frac{RB}N$$
The LHS of this equality can be recognized as the expectation of the number of red balls that appear if $B$ balls are taken randomly and without replacement from an urn that contains exactly $N$ balls of which exactly $R$ are red balls.
Note that the LHS is $\sum_{i=1}^Bip_i$ where $p_i$ denotes the probability that $i$ red balls are drawn.
This expectation can be found on a more elegant way using linearity of expectation and symmetry.
For $i=1,\dots,B$ let $X_i$ take value $1$ if at the $i$-th draw a red ball is chosen and let $X_i$ take value $0$ otherwise.
Then to be found is: $$\mathbb E(X_1+\dots+X_B)=B\times\mathbb EX_1=B\times\frac{R}{N}=\frac{RB}{N}$$
So this tells us that the equality is a true statement.
Maybe this does not answer your question and you are after some unraveling of the LHS that ends up in the RHS. In that case I feel a too strong reluctance to think about any answer.