Suppose that we want to show that P(X leq n)=(1-p)^n. <br> Where X is a geometric random variable with probability of success p.

latinoisraelm1

latinoisraelm1

Open question

2022-08-16

Geometric probability expressed as a geometric series
Suppose that we want to show that P ( X n ) = ( 1 p ) n
Where X is a geometric random variable with probability of success p.
P ( X > n ) can be written as a geometric series as follows:
Σ X = n + 1 ( 1 p ) n 1 = ( 1 p ) n
But how do I proof this? What is a and r in this particular geometric series?

Answer & Explanation

Alexia Mata

Alexia Mata

Beginner2022-08-17Added 15 answers

Step 1
It helps to start from the right place.
When X is geometric with success rate p, then X = n is the event of n 1 failures before the first success, so:
P ( X = n ) = ( 1 p ) n 1 p
Step 2
So: P ( X > n ) = k = n + 1 P ( X = k ) = k = n + 1 ( 1 p ) k 1 p = p ( 1 p ) n   r = 0 ( 1 p ) r    

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