Question

# Suppose $$X$$ follows a binomial distribution with parameters $$n$$ and $$p$$, where $$0<p<1$$. If $$\displaystyle \frac { P\left( X=r \right) }{ P\left( X=n-r \right) }$$ is independent of $$n$$ and $$r$$, then $$p$$ is equal to

A
13
B
12
C
14
D
None of these

Solution

## The correct option is D $$\displaystyle \frac { 1 }{ 2 }$$$$\displaystyle \frac { P\left( X=r \right) }{ P\left( X=n-r \right) } =\frac { _{ }^{ n }{ { C }_{ r } }{ p }^{ r }{ \left( 1-p \right) }^{ n-r } }{ _{ }^{ n }{ { C }_{ n-r }{ p }^{ n-r } }{ \left( 1-p \right) }^{ r } } =\frac { \left( 1-p \right) ^{ n-2r } }{ { p }^{ n-2r } } ={ \left( \frac { 1 }{ p } -1 \right) }^{ n-2r }$$For $$\displaystyle { \left( \frac { 1 }{ p } -1 \right) }^{ n-2r }$$ to be independent of $$n$$ and $$\displaystyle r,\frac { 1 }{ p } -1=1$$$$\displaystyle \Rightarrow \frac { 1 }{ p } =1\Rightarrow p=\frac { 1 }{ 2 } .$$Maths

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