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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
\[\text{ As, X follows the binomial distribution with parameters n and p, where 0 } < p < 1\]
\[\text{ So } , P\left( X = r \right) =^{n}{}{C}_r p^r q^\left( n - r \right) , \text{ where } r = 0, 1, 2, 3, . . . ,\]
\[\text{ Now } , \]
\[\frac{P\left( X = r \right)}{P\left( X = n - r \right)} = \frac{^{n}{}{C}_r p^r q^\left( n - r \right)}{^{n}{}{C}_\left( n - r \right) p^\left( n - r \right) q^r} = p^\left( 2r - n \right) q^\left( n - 2r \right) \]
\[\text{ As } , \frac{P\left( X = r \right)}{P\left( X = n - r \right)}\text{ is independent of n and r } \]
\[\text{ So, p } = q \left[ \text{ Since, } \frac{P\left( X = r \right)}{P\left( X = n - r \right)} = p^\left( 2r - n \right) \ p^\left( n - 2r \right) = p^0 = 1 \right]\]
\[\text{ This is only possible if p } = \frac{1}{2}\]
\[ \therefore p = \frac{1}{2}\]
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