Question

If x is a geometric variate taking values 1,2,...,infinity, find p(x is odd)

Answer 1

Answer:

answer is ofcourse 23 for this question

Answer 2

The answer will be 1/q+1

Given,

P(X is odd) = P(X = 1) + P(X=3) + P( X = 5) + .... and so on.

To find:

P(X)

Solution:

P(X is odd) = P(X = 1) + P(X=3) + P( X = 5) + .... and so on.

The pmf for the given geometric equation is $q^{k-1}p$.

Therefore, for P(X is odd)  = $q^{0}p$ + $q^{1}p$ + $q^{3} p$ + ..... and so on.

This will be equal to p/$1-q^{2}$

=> 1/q+1.