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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
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B
12
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C
14
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D
None of these
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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 $$\displaystyle r,\frac { 1 }{ p } -1=1$$
$$\displaystyle \Rightarrow \frac { 1 }{ p } =1\Rightarrow p=\frac { 1 }{ 2 } .$$

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