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asked ago in Probability by Veteran (11.7k points)  
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A random variable $x$ takes values $0,1,2,3,4,5...............$ with probability proportional to 

$\left ( x+1 \right )\left ( \frac{1}{5} \right )^x$

Hence, I can write $P(x|x\geq i:i=0,1,2,3,4........) \propto \left ( x+1 \right )\left ( \frac{1}{5} \right )^x$

OR, In other words

  • $P(x|x\geq i:i=0,1,2,3,4........) = M \left ( x+1 \right )\left ( \frac{1}{5} \right )^x$

Hence, $P(x= 0) = M \left ( 0+1 \right )\left ( \frac{1}{5} \right )^0 = M.1$

$P(x= 1) = M \left ( 1+1 \right )\left ( \frac{1}{5} \right )^1 = M.\frac{2}{5}$

$P(x= 2) = M \left ( 2+1 \right )\left ( \frac{1}{5} \right )^2 = M.\frac{3}{25}$

This thing ends upto infinity ....

Now, Applying the property, that $P(x|x\geq i:i=0,1,2,3,4........) = 1$

  • $M.1.\frac{1}{5}^{0} + M.2.\frac{1}{5}^{1} + M.3.\frac{1}{5}^{2}............$ = $1$
  • $M\left ( 1 + 2.x + 3.x^{2}............... \right ) = 1$

Now, If I apply the last formula from the given figure,

 

I would end up getting $M\left ( 1 + 2.x + 3.x^{2} + 4.x^3 +5.x^4............... \right )$

OR 

This is an arithmetico - geometric progression also. So, that technique also helps here.

$A = 1 + 2x + 3x^{2} + 4x^{3}.............$   

Now, Multiply by $x$,

$Ax = 1x + 2x^{2} + 3x^{3} + 4x^{4}.............$

  • $A(1-x) = 1 + x + x^{2} + x^{3}.............$
  • $A(1-x) = \frac{1}{1-x}$

Apply, any of the above methods.

Hence, Solving this , I get 

  • $M.\left ( 1-x \right )^{-2} = 1$
  • Put, $x = \frac{1}{5} => M= \frac{16}{25}$

Finally, that proportionate probability becomes,

$P(x|x\geq i:i=0,1,2,3,4........) = \left ( \frac{16}{25} \right ) \left ( x+1 \right )\left ( \frac{1}{5} \right )^x$


The Probability that $x\leq 5$ is

$P(x\leq 5) = P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5) = 0.9997$


 

answered ago by Veteran (40.1k points)  
edited ago by
this was dope !
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