Chapter 17
Probability Models
The Geometric Distribution
1
Introduction


Suppose there is a particular toy that your neighbor wants
to get from a McDonald’s Happy Meal. Assume that the
toys are randomly distributed and the probability of
getting the toy is 15%. How many Happy Meals do you
need to buy in order to get the toy?
This situation is considered a Bernoulli trial and all
Bernoulli trials have the following three characteristics:
There are only two outcomes (success and failure). Either he
gets the toy or he doesn’t.
 The probability of success, denoted p, is the same on every
trail. In other words, the probability doesn’t change.
 The trails are independent. Getting a different toy from the
first box does not affect what will happen on the next box.


You must ALWAYS check these three assumptions when
dealing with these types of problems!
Introduction

Here are a few examples of Bernoulli trials:
1.
2.


We use a coin to see which of two football teams
gets the choice of kicking off or receiving to begin a
game.
A young couple prepares for their first child; the
possible outcomes are {boy; girl}.
Make sure that you always check your
assumptions and conditions!
There are two types of Bernoulli trials that we
will look at in this chapter: binomial and
geometric distributions. These discrete
random distribution will be distinguished by
the way you define your random variable.
Geometric Distributions
 A geometric random variable is a variable
whose domain contains the number of
trials that are needed to achieve the first
success in a repeated trial experiment.
Geometric Settings


In order to calculate probabilities using the
Geometric distribution, we must satisfy the same
three conditions for Bernoulli Trials.
The whole point of the geometric model is to
look at situations where we are dealing with
“first successes.”



Many times we wish to determine the average
number n of trials required to obtain the first success.
This is the Expected Value (or the mean) of the first
success.
Other times, we want to know the probability of
getting our first success in n number of trials.
Either way, we still are dealing with “first successes.”
Geometric Random Variable:
Mean and Standard Deviations

Let X be a geometric random variable with probability of
success p on each trial.

The mean µ, or expected value, of the random variable is
(this is the expected number of trials required to get the
first success)
 
1
p
The variance of X is 
2

Standard deviation is 
1 p
 
2
p



 or

1 p

2
p


2

 or  


q
p
2
q
p
2
or  
q
p
Rule for Calculating Geometric Probabilities
In order to calculate the probability of the first
success in n trials, we use the following:
P(X = n) = (1 - p)n-1p or P(X = n) = qn-1p
Example:
If you roll a die, what is the probability that the
first 6 will be on the fourth roll?
1

P ( X  4)   1  
6

41
3
125
1 5 1
 0 . 0965
    
 6   6   6  1296
We can also use the calculator: geometpdf (p, n)
geometpdf (1/6, 4) ≈ 0.0965
Independence




One of the most important requirements
Bernoulli Trials is that each trial must be
independent.
Most times in probability, we can assume that
events are independent (such as rolling two
dice).
However, in real-life Statistics, independence is
compromised since we choose our sample
without replacement. This means that once you
choose an individual for a study, the probability
for choosing the next individual changes.
So what do we do?
Independence



Well, as long as our population is large enough,
removing a few individual is negligible. 1 out of
a billion and one is not much different than 1 out
of a billion.
OK, but how big is large enough? There is a
condition or a criteria that we can use to
determine if a population is large enough:
The 10% condition: If independence is violated,
it is still OK to utilize Bernoulli Trials as long as
the sample is smaller than 10% of the
population.
 As
long as our sample is not too big, the
probabilities don’t change significantly.
More Probability Rules

The probability that it takes more than n trials to
see the first success is P(X > n) = (1 - p)n

Example:
 Let’s
say you pick a card with replacement from a
standard 52-card deck. What is the probability it will
take more than 10 tries before you get your first
heart?
10
10
1

P ( X  10 )   1  
4

 There
3
 
4
 0 . 0563
is approximately a 5.6% chance that it will take
more than 10 tries before we see our first heart.
Example:

Only 6% of people have type O-negative blood. If
donors line up for a blood drive, how many
people should we expect to see before we see our
first O-negative blood? What is the probability
that it will take more than 10 people before we see
the first O-negative donor? What is the
probability that a type O-negative donor is found
within the first four people in line? What is the
standard deviation?
Solution


First things first, ALWAYS….
Check your Assumptions (and conditions):
1. There are two outcomes:


2.
The probability for success is the same for each
person

3.
Success = A person has type O-negative blood
Failure = A person has different type blood
Each person has a .06 chance of success
The trails are independent

Actually, the trails are not independent!!!
 Since n is less than 10% of the population (we don’t
plan on drawing blood from more than 10% of the
population!), this condition may be compromised, so
we’re ok.
Solution

Only 6% of people have type O-negative blood. If
donors line up for a blood drive, how many
people should we expect to see before we see
our first O-negative blood? What is the
probability that it will take more than 10 people
before we see the first O-negative donor? What is
the probability that a type O-negative donor is
found within the first four people in line? What is
the standard deviation?
E(X )  X 
1
p

1
. 06
 16 . 67
Solution

Only 6% of people have type O-negative blood. If
donors line up for a blood drive, how many
people should we expect to see before we see our
first O-negative blood? What is the probability
that it will take more than 10 people before we
see the first O-negative donor? What is the
probability that a type O-negative donor is found
within the first four people in line? What is the
standard deviation?
P ( X  10 )  (1  p )  (1  . 06 )  (. 94 )  . 5386
10
10
Or you can use 1 – geometcdf(.06,10)
10
Solution

Only 6% of people have type O-negative blood. If
donors line up for a blood drive, how many
people should we expect to see before we see our
first O-negative blood? What is the probability
that it will take more than 10 people before we
see the first O-negative donor? What is the
probability that a type O-negative donor is
found within the first four people in line? What
is the standard deviation?
P ( X  4 )  P ( X  1)  P ( X  2 )  P ( X  3 )  P ( X  4 )
 . 06  (. 94 )(. 06 )  (. 94 ) (. 06 )  (. 94 ) (. 06 )  . 21925
2
Or you can use geometcdf(.06, 4)
3
Solution

Only 6% of people have type O-negative blood. If
donors line up for a blood drive, how many
people should we expect to see before we see our
first O-negative blood? What is the probability
that it will take more than 10 people before we
see the first O-negative donor? What is the
probability that a type O-negative donor is found
within the first four people in line? What is the
standard deviation?
X 
1 p
p

1  . 06
. 06

. 94
. 06
 16 . 1589
Assignment
Lesson:
Problems:
Read:
Chapter 17
Geometric and
1 – 37 (odd)
Chapter 17
Binomial Distributions
Ch. 17 WS