# If X is a binomial variate with parameters n and p, where 0 < p < 1 such that P ( X = r ) P ( X = n − r ) is independent of n and r, then p equals - Mathematics

#### 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 .

$^{n}{}{C}_r p^r q^{n - r} = k ^{n}{}{C}_{n - r} p^{n - r} q^r$

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

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

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

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

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

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

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

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If X is a binomial variate with parameters n and p, where 0 < p < 1 such that P ( X = r ) P ( X = n − r ) is independent of n and r, then p equals Concept: Bernoulli Trials and Binomial Distribution.