Suppose a random variable X follows the Binomial distribution with parameters $n$ and $p$, where $0 < p < 1$. If $P(x = r)/P(x = n - r)$ is independent of $n$ and $r$, then $p$ equals to
Suppose a random variable X follows the Binomial distribution with parameters $n$ and $p$, where $0 < p < 1$. If $P(x = r)/P(x = n - r)$ is independent of $n$ and $r$, then $p$ equals to
Official Solution
….(i)
$P(X = 0) = {(1 - p)^n}$
and $P(X = n - r){ = ^n}{C_{n - r}}{(p)^{n - r}}{(q)^{n - (n - r)}}$
….(ii)
Now, $\frac{{P(x = r)}}{{P(x = n - r)}} = \frac{{\frac{{n!}}{{(n - r)!r!}}{p^r}{{(1 - p)}^{n - r}}}}{{\frac{{n!}}{{(n - r)!r!}}{p^{n - r}}{{(1 - p)}^{ + r}}}}$
[using Eqs. (i) and (ii)]
$= {\left( {\frac{{1 - p}}{p}} \right)^{n - r}} \times \frac{1}{{{{\left( {\frac{{1 - p}}{p}} \right)}^r}}}$
Above expression is independent of $n$ and $r$, if $\frac{{1 - p}}{p} = 1 \Rightarrow \frac{1}{p} = 2 \Rightarrow p = \frac{1}{2}$
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