Sta107 Online Tutorial

Geometric and Negative Binomial Random Variables

The geometric and negative binomial random variables are based on a sequence of independent and identical Bernoulli trials.

Definition 3.2-1:

The geometric random variable is the number of independent and identical Bernoulli trials it takes to obtain the first success. Given an integer k>1, a negative binomial random variable is the number of independent and identical Bernoulli trials it takes to obtain the kth success.

Theorem 3.2-1 :

The probability mass function of the geometric random variable X, with probability p of a success is

Example 1:

Consider Exercise 3.2-2 on page 115. Each time customers visit a resturant they are given a game card. Suppose the probability of winning a prize with the game card is 0.2. Let X represent the number of visits to a resturant before winning a prize with the game card.What is the probability that a customer will win a prize for the first time on the 6th visit?

Solution:

The probability of a success is p=0.2, and the probability of a failure is 1-p=0.8. The first success it to occur on the x = 6th trial. So p(x)= (0.2)(0.8)^(5)=0.06554

What is the probability that it will take the customer 6 or more visits to win a prize for the first time?

Solution:

Note: The sum of a geometric series is

P(X>=6) = S(0.2)(0.8)^(x-1) for x =6,7,....

= (0.2)(0.8)^6 [1/(1-0.8)]

=0.262144

Example 2:

Consider Exercise 3.2-1 on page 115. In a game of billiards, a player shoots until a miss occurs. A player misses on any given shot with a probability of 0.15. Let the random variable X denote the number of shots taken to obtain the first miss.

A: What is the probability that the player will have thier first miss on the 3rd shot?

(Round to 4 decimal places)

B: What is the probability that the player will take at more than 5 shots?

(Round to 4 decimal places)

Solution

Theorem 3.2-2:

The mean and variance of a geometric random variable, with probability p of a success are

Example 3:

Refer to Example 1 again.

What is E(X)?

What is VAR(X)?

Solution

Explore the Geometric Distribution by using the following applet! Click on Open Applet. Enter the value of p for p.

Theorem 3.2-3

The probability mass function of a negative binomial random variable Y, with probability p of a success and k>1 is

The kth success occurs on the yth trial, then k-1 successes and y-k failures occured during the first y-1 trials.

Example 4:

The probability of winning a prize in a raffle is 0.34. Let Y denote the number of raffles a person needs to buy before they win 2 prizes. The probability that it takes exactly 12 raffles to win 2 prizes is:

P(Y=12) =(12-1)C(2-1) * (0.34)^2 * (1-.34 )^(12-2)

= (11C1)(0.1156)(0.015683368)

=0.0199

Solution

Theorem 3.2-4:

The mean and variance of a negative binomial random variable, with probability of success p and k>1, are

Example 5:

Refer to Example 4 once again. Let Y denote the number or raffles a person needs to buy before they win 5 prizes.

What is E(Y)?

(Round to 2 decimal places)

What is VAR(Y)?

(Round to 2 decimal places)

Solution

Explore the Negative Binomial Distribution by using the following applet! Click on Open Applet. Enter the value of p for p and the value of k for r.

Useful Web Resources

Hammack's Statistics Class: Geometric Random Variables

Geometric Distribution- Wikipedia

Negative Binomial Distribution- Wikipedia

Solutions

Solution to Example 1

A: p=0.15 and 1-p=0.85 and the first miss is to occur on the x=3rd trial. So p(x)=(0.15)(0.85)^2=0.1084

B: We want to find the probability that the player takes more than 5 shots.

So P(X>5)= P(Y>=6) = S(0.15)(0.85)^(x-1) for x = 6,7,8,.....

=(0.15)(0.85)^(5) [1/(1-0.85)]

= 0.4437

Solution to Example 2

A:E(Y)= 1/p=1/0.2= 5

B: VAR(Y)=(1-p)/p^2= (1-0.2)/0.2^2=20

Solution to Example 4

P(Y=20) =(20-1)C(4-1) * (0.34)^4* (1-.34 )^(20-4)

= (19C3)(0.01336336)(0.0012963)

=0.016786=0.0168

Solution to Example 5

E(Y)=k/p= 5/0.34=14.705=14.71

VAR(Y)=k(1-p)/p^2 = (5)(1-0.34)/(0.34)^2=0.38148= 0.38

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