GATE CSE
First time here? Checkout the FAQ!
x
+5 votes
141 views

asked in Probability by Veteran (14.7k points)  
retagged by | 141 views

1 Answer

+7 votes
Best answer

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 by Veteran (47.4k points)  
edited by
this was dope !


Top Users Jun 2017
  1. Bikram

    2512 Points

  2. Hemant Parihar

    1480 Points

  3. junaid ahmad

    1432 Points

  4. Arnab Bhadra

    1334 Points

  5. Niraj Singh 2

    1311 Points

  6. rahul sharma 5

    1060 Points

  7. Rupendra Choudhary

    1042 Points

  8. Debashish Deka

    888 Points

  9. Arjun

    856 Points

  10. srestha

    836 Points

Monthly Topper: Rs. 500 gift card
Top Users 2017 Jun 19 - 25
  1. Niraj Singh 2

    1306 Points

  2. Bikram

    768 Points

  3. junaid ahmad

    502 Points

  4. akankshadewangan24

    252 Points

  5. joshi_nitish

    250 Points


23,325 questions
29,999 answers
67,196 comments
28,337 users