Question

X follows a binomial distribution with parameters n and p and Y follows a binomial distribution with parameters m and p. Then if X and Y are independent and r, s $\ge$ 0, then the given expression also equals
$P(X=r|X+Y=r+s)=$

• A. $\frac{\left({}^m{C}_r\right)\left({}^n{C}_s\right)}{{}^{m+n}{C}_{r+s}}$
• B. $\frac{{}^n{C}_r{.}^{m-1}{C}_s}{{}^{m+n}{C}_{r+s}}$
• C. $\frac{{}^n{C}_r{.}^m{C}_s}{{}^{m+n-2}{C}_{r+s}}$
• D. $\frac{{}^n{C}_r{.}^{m+1}{C}_s}{{}^{m+n}{C}_{r+s}}$

Answer 1

The correct option is A $\frac{\left({}^m{C}_r\right)\left({}^n{C}_s\right)}{{}^{m+n}{C}_{r+s}}$

We have, $P(X=r/X+Y=r+s)$
$=\frac{P[(X=r)\cap (X+Y=r+s)]}{P(X+Y=r+s)}$
$=\frac{P[(X=r)\cap (Y=s)]}{P(X+Y=r+s)}=\frac{P(X=r)P(Y=s)}{P(X+Y=r+s)}$
And, $P(X+Y=r+s)={\sum }_{k=0}^{r+s}P[(X=k)\cap (Y=r+s-k)]$
$={\sum }_{k=0}^{r+s}\left({}^n{C}_k{p}^k{q}^{n-k}\right)\left({}^m{C}_{r+s-k}{p}^{r+s-k}{q}^{m-r-s+k}\right)$
$=p^{r+s}{q}^{m-r-s}{\sum }_{k=0}^{r+s}\left({}^n{C}_k\right)\left({}^m{C}_{r+s-k}\right)$
Now the last sum is the expression for the number of ways of choosing
$r+s$ persons out of $n$ men and $m$ women, which is ${}^m{C}_{r+s}$.
Therefore,
$P(X+Y=r+s){=}^{m+n}{C}_{r+s}{p}^{r+s}{q}^{m+n-r-s}$
so thast $P(X=r/X+Y=r+s)$
$=\frac{\left({}^m{C}_r{p}^r{q}^{n-r}\right)\left({}^n{C}_s{p}^s{q}^{m-s}\right)}{{}^{m+n}{C}_{r+s}{p}^{r+s}{q}^{m+n-r-s}}=\frac{\left({}^m{C}_r\right)\left({}^n{C}_s\right)}{{}^{m+n}{C}_{r+s}}$