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