$P(X \geq 2Y) = ?$ where $P(n) = 2^{-n} , n =1,2,3,...$. Let $X,Y$ are independent identically distributed random variables.Then $P(X \geq 2Y) = ?$ where $P(n) = 2^{-n} , n =1,2,3,...$.
So it is the discrete case, So in order to calculate this - 
$P(X \geq 2Y) = \sum_{y=1}^{\infty} \sum_{x = 2y}^{\infty} 2^{-x-y} = \sum_{y=1}^{\infty}2^{-y} (\frac{2^{-2y}}{1 - 2^{-2}}) = \sum_{y=1}^{\infty}  \frac{4}{3}. 2^{-3y} =\frac{4}{3}. \frac{2^{-3}}{1 - 2^{-3}} = \frac{4}{21}.$
Is the above correct?  

Above Answer is not Correct. It should be
$P(X \geq 2Y) = \sum_{y=1}^{\infty} \sum_{x = 2y}^{\infty} 2^{-x-y}
 = \sum_{y=1}^{\infty}2^{-y} (\frac{2^{-2y}}{1 - 2^{-1}}) 
 = \sum_{y=1}^{\infty}  \frac{2}{1}. 2^{-3y} =\frac{2}{1}. \frac{2^{-3}}{1 - 2^{-3}}
 = \frac{2}{7} $
