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 $P\left(\frac{X=r}{X+Y=r+s}\right)=$

• A. $\frac{\left({}^m{C}_r\right)\left({}^n{C}_s\right)}{{}^{m+n}{C}_{r+s}}$
• B. $\frac{\left({}^m{C}_s\right)\left({}^n{C}_r\right)}{{}^{m+n}{C}_{r+s}}$
• C. $\frac{\left({}^m{C}_s\right)}{{}^{m+n}{C}_r}$
• D. none of these

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 $P\left(\frac{X=r}{X+Y=r+s}\right)=\frac{P[(X=r)\cap (X+Y=r+s)]}{P(X+Y=r+s)}$
$=\frac{P[(X=r)\cap (Y=s)]}{P(X+Y=r+s)}=\frac{P(X=r)P(Y=s)}{P(X+Y=r+s)}$
$P(X+Y=r+s)=\sum _{k=0}^{r+s}P[(X=k)\cap (Y=r+s-k)]$
$=\sum _{k=0}^{r+s}({}^n{C}_k.p^k.q^{n-k})({}^m{C}_{r+s-k}.p^{r+s-k}.q^{m-r-s+k})$
$=p^{r+s}.q^{m+n-r-s}.\sum _{k=0}^{r+s}\left({}^n{C}_k\right)\left({}^m{C}_{r+s-k}\right)$
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 ${}^{m+n}{C}_{r+s}$.
Therefore $P(X+Y=r+s){=}^{m+n}{C}_{r+s}.p^{r+s}.q^{m+n-r-s}$ so that
$P\left(\frac{X=r}{X+Y=r+s}\right)=\frac{({}^m{C}_r.p^r.q^{n-r})({}^n{C}_s.p^s.q^{m-s})}{{}^{m+n}{C}_{r+s}.p^{r+s}.q^{m+n-r-s}}=\frac{\left({}^m{C}_r\right)\left({}^n{C}_s\right)}{{}^{m+n}{C}_{r+s}}$