Let X be a geometric random variable with parameter p, and n and m be nonnegative integers.
a. for what values of n is P(X=n) Maximum? b. what is the probability that X is even?
Solution
Let X be a geometric random variable with parameter p,
and n and m be nonnegative integers.
a. for what values of n is P(X=n) Maximum?
P(x=n) = (1-p)^n-1 p
we know that
0<=p<=1
now
=> 0<= 1- p <=1
(1-p)^n-1 < =1
maximum means equality occur only at n = 1 thus
maximum is at n = 1
P(x=1) = p
b. what is the probability that X is even?
P(x=n) = (1-p)p + (1-p)^3 p +..................
since n is even
now

P(x=n at even) = p/(1-p) ( (1-p)^2 + (1-p)^4+..........)
this is geometric distribution with ratio 1-p
now
P(x=n) = p/(1-p) *(1-p)^2/(1-(1-p)^2) = p/(1-p) *(1-p)^2(1/ 1 - p^2+2p-1) = p/(1-p) * (1-
p)^2/2p-p^2)
= p/(1-p) (1-p)^2/p(2-p)
= (1-p)/(2-p)
thus answer.