binomial distribution ques

[Sinistral]

Let X have the pmf p(x) = (1/3) * ((2/3) ^ x ), x= 0,1,2,3...
zero elsewhere.
find the conditional pmf of X given X>=3.

[ankur]

What is the value of n- number of trials

[Sinistral]

ok. just ignore that heading "binomial distribution" and then solve.

[edited...]
by looking at the ans it seems obvious.
ans is (1/3)*[(2/3)^(x-3)] x=3,4,5...
but how to proceed towards it formally?

[wolverine]

P(X>=3)=(1/3) * ((2/3) ^ x  for x=3,4,5,....
 which  equals  (1/3) * ((2/3) ^ 3 + (1/3) * ((2/3) ^ 4 +..... i.e. an infinite  series with a=1 and r= (2/3)
                           = (2/3)^3

P(X|X>=3)= ((1/3) * ((2/3) ^ x)/ (2/3)^3=(1/3) * ((2/3) ^ (x-3)

[Sinistral]

aahaan..

thanx a lot.