Suppose that X is a geometric random variable with parameter (probability of success) p. Show that Pr(X>a+b∣X>a)=Pr(X>b).

madeeha1d8

madeeha1d8

Answered question

2022-09-19

Geometric Distribution Probability Problem
Suppose that X is a geometric random variable with parameter (probability of success) p.
Show that Pr ( X > a + b X > a ) = Pr ( X > b ).

Answer & Explanation

ruinsraidy4

ruinsraidy4

Beginner2022-09-20Added 17 answers

Step 1
There are two ways to get Pr ( X > n ), the easy way and the hard way. Since there is something to be learned from both, we do it both ways.
The hard way: The probability that X = n + 1 is the probability of n failures in a row, then success. This is ( 1 p ) n p
Similarly, the probability that X = n + 2 is the probability of n + 1 failures in a row, then success. This has probability ( 1 p ) n + 1 p.
Continue, and add up. There is a ( 1 p ) n p "in" each term. Take it out as a common factor. So we get ( 1 p ) n p ( 1 + ( 1 p ) + ( 1 p ) 2 + ) .
The geometric series inside the brackets has sum 1 1 ( 1 p ) = 1 p
Step 2
The easy way: We have X > n precisely if we get n failures in a row, probability ( 1 p ) n .
koraby2bc

koraby2bc

Beginner2022-09-21Added 2 answers

Step 1
Here's how to handle an infinite geometric series when the index starts at n instead of 0:
k = n a x k = a x n + a x n + 1 + a x n + 2 + a x n + 3 + = ( a x n ) + ( a x n ) x + ( a x n ) x 2 + ( a x n ) x 3 + = b + b x + b x 2 + b x 3 +
Now it starts at index 0. And of course b is a x n .
It looks as if you can do the rest.
Second method: You mentioned that the probability of "success" is p. That means the probability of success on each trial is p.
If X is defined as the number of trials needed to get one success, then the event X > n is the same as the event of failure on all of the first n trials, so that probability of that is ( 1 p ) n .
If X is defined as the number of trials needed to get one failure, then the event X > n is the same as the event of success on all of the first n trials, so the probability of that is p n .

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school geometry

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?