If X is a binomial variate with parameters n and p, where 0 < p < 1 such that $\frac{P(X=r)}{P(X=n-r)}$ is independent of n and r, then p equal
Answer & explanation
Correct answer: option 1
We have,
$\frac{P(X=r)}{P(X=n-r)}=\frac{^nC_r \, p^r (1-p)^{n-r}}{^nC_{r-1} \, p^{n-r} (1-p)^{r}}$
$\frac{P(X=r)}{P(X=n-r)}=\left(\frac{1-p}{p}\right)^{n-2r}=\left(1-\frac{1}{p}\right)^{n-2r}$
For this to be independent of n and r, we must have $ 1- \frac{1}{p}= 1⇒ p=\frac{1}{2}$