8.2 The Geometric Distribution

1. What is the geometric setting?2. How do you calculate the

probability of getting the first success on the nth trial?

3. How do you calculate the means and variance of a geometric distribution?

4. How do you calculate the probability that it takes more than n trials to see the first success for a geometric random variable?

The Geometric Distribution

•Suppose an experiment consists of a sequence of trials with the following conditions:

1. The trials are independent.2. Each trial can result in one of two possible

outcomes, success and failure.3. The probability of success is the same for all

trials.

A geometric random variable is defined asx = number of trials until the first success is observed (including the success trial)

The probability distribution of x is called the geometric probability distribution.

The Geometric Distribution

• If x is a geometric random variable with probability of success = for each trial, then p(x) = (1 – )x-1 x = 1, 2, 3, …

Example• Over a very long period of time, it has been noted that on Friday’s 25% of the customers at the drive-in window at the bank make deposits.

• What is the probability that it takes 4 customers at the drive-in window before the first one makes a deposit.

Example - solution• This problem is a geometric distribution problem with = 0.25.

• Let x = number of customers at the drive-in window before a customer makes a deposit.

The desired probability is4 1p(4) (.75) (.25) 0.0117

Geometric Probability

pp-1 deviation standard

p1 mean

)1()( trial nth on success of yProbabilit

1

nppnP

A sharpshooter normally hits the target 70% of

the time.• Find the probability that her first hit is on the second shot.

• Find the mean and the standard deviation of this geometric distribution.

A sharpshooter normally hits the target 70% of the

time.• Find the probability that her first hit is on the second shot.

P(2)=p(1-p) n-1 = .7(.3)2-1

= 0.21• Find the mean

= 1/p = 1/.7 1.43• Find the standard deviation78.07.

7.1pp-1