Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# 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. - Mathematics

#### 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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