#### Question

If X is a binomial variate with parameters n and p, where 0 < p < 1 such that \[\frac{P\left( X = r \right)}{P\left( X = n - r \right)}\text{ is } \] independent of n and r, then p equals

##### Options

1/2

1/3

1/4

None of these

#### Solution

1/2

Given that *P*(*X*=*r*) =* **k P*(*X*=*n* -*r*), where *k* is independent of* n* and *r* .

\[^{n}{}{C}_r p^r q^{n - r} = k ^{n}{}{C}_{n - r} p^{n - r} q^r \]

\[\text{ We have } ^{n}{}{C}_r = ^{n}{}{C}_{n - r} \text{ and also q } = 1 - p\]

\[\text{ Hence, the equation changes to the following } :\]

\[ p^r (1 - p )^{n - r} = \text{ k } p^{n - r} (1 - p )^r \]

\[ \Rightarrow (1 - p )^{n - 2r} = \text{ k }p^{n - 2r} \]

\[ \Rightarrow \left( \frac{q}{p} \right)^{n - 2r} = k \]

\[ \text{ This is possible when p = q and k becomes 1 .} \]

\[\text{ Hence,} p = q = \frac{1}{2}\]