#### Question

Mark the correct alternative in the following question:

Suppose a random variable *X* follows the binomial distribution with parameters *n* and *p*, where 0 < *p* < 1. If \[\frac{P\left( X = r \right)}{P\left( X = n - r \right)}\] is independent of *n* and *r*, then *p* equals

##### Options

\[\frac{1}{2}\]

\[\frac{1}{3}\]

\[\frac{1}{5} \]

\[\frac{1}{7}\]

#### Solution

\[\text{ As, X follows the binomial distribution with parameters n and p, where 0 } < p < 1\]

\[\text{ So } , P\left( X = r \right) =^{n}{}{C}_r p^r q^\left( n - r \right) , \text{ where } r = 0, 1, 2, 3, . . . ,\]

\[\text{ Now } , \]

\[\frac{P\left( X = r \right)}{P\left( X = n - r \right)} = \frac{^{n}{}{C}_r p^r q^\left( n - r \right)}{^{n}{}{C}_\left( n - r \right) p^\left( n - r \right) q^r} = p^\left( 2r - n \right) q^\left( n - 2r \right) \]

\[\text{ As } , \frac{P\left( X = r \right)}{P\left( X = n - r \right)}\text{ is independent of n and r } \]

\[\text{ So, p } = q \left[ \text{ Since, } \frac{P\left( X = r \right)}{P\left( X = n - r \right)} = p^\left( 2r - n \right) \ p^\left( n - 2r \right) = p^0 = 1 \right]\]

\[\text{ This is only possible if p } = \frac{1}{2}\]

\[ \therefore p = \frac{1}{2}\]