+ Section 6.3
Binomial and Geometric Random Variables
Learning Objectives
After this section, you should be able to…

DETERMINE whether the conditions for a binomial setting are met

COMPUTE and INTERPRET probabilities involving binomial random
variables

CALCULATE the mean and standard deviation of a binomial random
variable and INTERPRET these values in context

CALCULATE probabilities involving geometric random variables
Settings
When the same chance process is repeated several times, we are often interested in
whether a particular outcome does or doesn’t happen on each repetition. In some
cases, the number of repeated trials is fixed in advance and we are interested in the
number of times a particular event (called a “success”) occurs. If the trials in these
cases are independent and each success has an equal chance of occurring, we have
a binomial setting.
Definition:
A BINOMIAL SETTING arises when we perform several independent trials
of the same chance process and record the number of times that a
particular outcome occurs. The four conditions for a binomial setting are
B
• Binary? The possible outcomes of each trial can be classified as
“success” or “failure.”
I
• Independent? Trials must be independent; that is, knowing the result
of one trial must not have any effect on the result of any other trial.
N
• Number? The number of trials n of the chance process must be fixed
in advance.
S
• Success? On each trial, the probability p of success must be the
same.
+
 Binomial
Random Variable
Consider tossing a coin n times. Each toss gives either heads or tails.
Knowing the outcome of one toss does not change the probability of
an outcome on any other toss. If we define heads as a success, then
p is the probability of a head and is 0.5 on any toss.
The number of heads in n tosses is a binomial random variable X.
The probability distribution of X is called a binomial distribution.
Definition:
The count X of successes in a binomial setting is a binomial random
variable. The probability distribution of X is a binomial distribution with
parameters n and p, where n is the number of trials of the chance process
and p is the probability of a success on any one trial. The possible values of
X are the whole numbers from 0 to n.
Note: When checking the Binomial condition, be sure to check the
BINS and make sure you’re being asked to count the number of
successes in a certain number of trials!
+
 Binomial
Probabilities
+
 Binomial
having type O blood. Genetics says that children receive genes from
each of their parents independently. If these parents have 5 children,
the count X of children with type O blood is a binomial random
variable with n = 5 trials and probability p = 0.25 of a success on
each trial. In this setting, a child with type O blood is a “success” (S)
and a child with another blood type is a “failure” (F).
What’s P(X = 2)?
P(SSFFF) = (0.25)(0.25)(0.75)(0.75)(0.75) = (0.25)2(0.75)3 = 0.02637
However, there are a number of different arrangements in which 2 out of
the 5 children have type O blood:
SSFFF
SFSFF
SFFSF
SFFFS
FSSFF
FSFSF
FSFFS
FFSSF
FFSFS
FFFSS
Verify that in each arrangement, P(X = 2) = (0.25)2(0.75)3 = 0.02637
Therefore, P(X = 2) = 10(0.25)2(0.75)3 = 0.2637
Binomial and Geometric Random Variables
In a binomial setting, we can define a random variable (say, X) as the
number of successes in n independent trials. We are interested in
finding the probability distribution of X.
Example
Each child of a particular pair of parents has probability 0.25 of
Coefficient
+
 Binomial
We can generalize this for any setting in which we are
interested in k successes in n trials. That is,
P(X  k)  P(exactly k successes in n trials)
= number of arrangements  p (1 p)
k
nk
Binomial and Geometric Random Variables
Note, in the previous example, any one arrangement
of 2 S’s and 3 F’s had the same probability. This is
true because no matter what arrangement, we’d
multiply together 0.25 twice and 0.75 three times.
+ Definition:
The number of ways of arranging k successes among n observations is
given by the binomial coefficient
n 
n!

 
k  k!(n  k)!
for k = 0, 1, 2, …, n where
n! = n(n – 1)(n – 2)•…•(3)(2)(1)

and 0! = 1.
n
 
k 
Probability
+
 Binomial
Binomial Probability
If X has the binomial distribution with n trials and probability p of success on
each trial, the possible values of X are 0, 1, 2, …, n. If k is any one of
these values,
n  k
P(X  k)   p (1 p) nk
k 
Number of
arrangements

of k successes
Probability of k
successes
Probability of
n-k failures
Binomial and Geometric Random Variables
The binomial coefficient counts the number of different ways in
which k successes can be arranged among n trials. The
binomial probability P(X = k) is this count multiplied by the
probability of any one specific arrangement of the k successes.
Inheriting Blood Type
+
 Example:
Each child of a particular pair of parents has probability 0.25 of having blood
type O. Suppose the parents have 5 children
(a) Find the probability that exactly 3 of the children have type O blood.
Let X = the number of children with type O blood. We know X has a binomial
distribution with n = 5 and p = 0.25.
5
P(X  3)   (0.25) 3 (0.75) 2  10(0.25) 3 (0.75) 2  0.08789
3
(b) Should the parents be surprised if more than 3 of their children have
type O blood?
To answer this, we need to find P(X > 3).

P(X  3)  P(X  4)  P(X  5)
5 
5
4
1
  (0.25) (0.75)   (0.25) 5 (0.75) 0
4 
5
 5(0.25) 4 (0.75)1  1(0.25) 5 (0.75) 0
 0.01465  0.00098  0.01563
Since there is only a
1.5% chance that more
than 3 children out of 5
would have Type O
blood, the parents
should be surprised!
and Standard Deviation of a Binomial
Distribution
+
 Mean
xi
0
1
2
3
4
5
pi
0.2373
0.3955
0.2637
0.0879
0.0147
0.00098
Shape: The probability distribution of X is skewed to
the right. It is more likely to have 0, 1, or 2 children
with type O blood than a larger value.
Center: The median number of children with type O
blood is 1. Based on our formula for the mean:
 X   x i pi  (0)(0.2373)  1(0.39551)  ... (5)(0.00098)
 1.25
Spread: The variance of X is  X2   (x i   X ) 2 pi  (0 1.25) 2 (0.2373)  (11.25) 2 (0.3955)  ...
 (5 1.25) 2 (0.00098)  0.9375
The standard deviation of X is  X  0.9375  0.968
Binomial and Geometric Random Variables
We describe the probability distribution of a binomial random variable just like
any other distribution – by looking at the shape, center, and spread. Consider
the probability distribution of X = number of children with type O blood in a
family with 5 children.
Notice, the mean µX = 1.25 can be found another way. Since each
child has a 0.25 chance of inheriting type O blood, we’d expect
one-fourth of the 5 children to have this blood type. That is, µX
= 5(0.25) = 1.25. This method can be used to find the mean of
any binomial random variable with parameters n and p.
Mean and Standard Deviation of a Binomial Random Variable
If a count X has the binomial distribution with number of trials n and
probability of success p, the mean and standard deviation of X are
 X  np
 X  np(1 p)
Note: These formulas work ONLY for binomial distributions.
They can’t be used for other distributions!

Binomial and Geometric Random Variables
and Standard Deviation of a Binomial
Distribution
+
 Mean
Bottled Water versus Tap Water
+
 Example:
Mr. Bullard’s 21 AP Statistics students did the Activity on page 340. If we assume the
students in his class cannot tell tap water from bottled water, then each has a 1/3
chance of correctly identifying the different type of water by guessing. Let X = the
number of students who correctly identify the cup containing the different type of water.
Find the mean and standard deviation of X.
Since X is a binomial random variable with parameters n = 21 and p = 1/3, we can
use the formulas for the mean and standard deviation of a binomial random
variable.
 X  np
 21(1/3)  7
We’d expect about one-third of his
21 students, about 7, to guess
correctly.

 X  np(1 p)
 21(1/3)(2 /3)  2.16
If the activity were repeated many
times with groups of 21 students
who were just guessing, the
number of correct identifications
would differ from 7 by an average of
2.16.
Distributions in Statistical Sampling
Suppose 10% of CDs have defective copy-protection schemes that can harm
computers. A music distributor inspects an SRS of 10 CDs from a shipment of
10,000. Let X = number of defective CDs. What is P(X = 0)? Note, this is not
quite a binomial setting. Why?
The actual probability is
9000 8999 8998
8991


 ...
 0.3485
10000 9999 9998
9991
10
P(X  0)   (0.10)0 (0.90)10  0.3487
0 
P(no defectives ) 
Using the binomial distribution,
In practice, the
binomial distribution gives a good approximation as long as we don’t
sample more than 10% of the population.
Sampling
Without Replacement Condition

When taking an SRS of size n from a population of size N, we can use a
binomial distribution to model the count of successes in the sample as
long as
1
n
10
N
Binomial and Geometric Random Variables
The binomial distributions are important in statistics when we want to
make inferences about the proportion p of successes in a population.
+
 Binomial
Approximation for Binomial Distributions
As n gets larger, something interesting happens to the shape of a
binomial distribution. The figures below show histograms of
binomial distributions for different values of n and p. What do
you notice as n gets larger?
+
 Normal
+
Normal Approximation for Binomial Distributions
Suppose that X has the binomial distribution with n trials
and success probability p. When n is large, the
distribution of X is approximately Normal with mean and
standard deviation
X  np
 X  np(1 p)
As a rule of thumb, we will use the Normal approximation when n is so
large that np ≥ 10 and n(1 – p) ≥ 10.
That is, the expected number of successes and failures are both at
least 10.


Example: Attitudes Toward Shopping
+

Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a
nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying
new clothes, but shopping is often frustrating and time-consuming.” Suppose that
exactly 60% of all adult US residents would say “Agree” if asked the same question. Let
X = the number in the sample who agree. Estimate the probability that 1520 or more
of the sample agree.
1) Verify that X is approximately a binomial random variable.
B: Success = agree, Failure = don’t agree
I: Because the population of U.S. adults is greater than 25,000, it is
reasonable to assume the sampling without replacement condition
is met.
N: n = 2500 trials of the chance process
S: The probability of selecting an adult who agrees is p = 0.60
2) Check the conditions for using a Normal approximation.
Since np = 2500(0.60) = 1500 and n(1 – p) = 2500(0.40) = 1000 are both at least 10, we may
use the Normal approximation.
+
The number X of switches that fail
inspection has approximately the binomial
distribution with n = 10 and p = 0.1.

What is the probability that no more than one
switch fails?
P(X ≤ 1) = P(X = 0) + P(X = 1)
 10 
=   (0.1)0(0.9)10
0 
= 0.7361
+
 10 
1
9
  (0.1) (0.9)
1 
The mailing list of an agency that markets
scuba-diving trips to the Florida Keys contains
70% males and 30% females. The agency calls
30 people chosen at random.
a)
What is the probability that 20 of the 30
are men?
b)
What is the probability that the first
woman is reached on the fourth call? (That
is, the first 4 calls give MMMF.)
Let X = the number of men called. X is B(30, 0.7)

What is the probability that 20 of the 30 are men?
a)

P(X = 20) =
 30 
 
 20 
(0.7)20(0.3)10 = 0.1416
What is the probability that the first woman is reached on the
fourth call? (That is, the first 4 calls give MMMF.)
b)
P(MMMF) = (0.7)3(0.3)1 = 0.1029
Binomial probabilities on the
calculator

P(X = k) = binompdf (n, p, k)
pdf  probability distribution function 
 Assigns a probability to each value of a discrete
random variable, X.

P(X < k) = binomcdf (n, p, k)
cdf  cumulative distribution function 
 for R.V. X, the cdf calculates the sum of the
probabilities for 0, 1, 2 … up to k.
binompdf( n,p,k)
 Allows to find the probability of an individual
outcome.
- specify n and p
( calculator: 2nd Dist)
Ex: Suppose a player's picture is in 20% of the
cereal boxes. You want to know the probability
of finding that player is exactly twice among 5
boxes. Then,
n= 5, p = 0.2, k= 2
binompdf( 5, 0.2, 2) = 0.2048
( means 20% chance)
binomcdf ( n, p, k)
 Allows to find the total probability of getting k or
fewer successes among the n trials
Ex: Suppose a player's picture is in 20% of the
cereal boxes. We have 10 boxes of cereal and
now we wonder about the probability of finding - “up to 4 pictures” of the Player.
So, that’s the probability of 0, 1, 2, 3 or 4
successes. ( at most 4 pictures)
binomcdf( 10, 0.2, 4) = 0.9672065025
What is the probability of getting at least 4
pictures of the player in 10 boxes?

“at least 4” means “ not 3 or fewer”. That is the
complement of 0,1,2 or 3 successes
1-binomcdf( 10, 0.2, 3) = 0.1208738816
Using the calculator for calculating
binomial probabilities

An engineer chooses an SRS of 10 switches from
a shipment of 10,000 switches. Suppose that
(unknown to the engineer) 10% of the switches in
the shipment are bad. What is the probability
that no more than 1 of the 10 switches in the
sample fail inspection?
The calculator can calculate the
probabilities of each value X
using the command
binompdf(n, p, X)
Using the calculator

What’s the probability that no more than 1
fails ( or at most 1 fails) ?
= P(X = 0) + P(X = 1)
= binompdf(10, 0.1, 0) + binompdf(10, 0.1, 1)
= 0.3487 + 0.3874
= 0.7361

The suffix pdf stands for probability
distribution function.





Corrine makes 75% of her free throws over
the course of a season. In a key game,
she shoots 12 free throws and only makes
7 of them. The fans think she failed
because she is nervous. Is it unusual for
Corinne to perform this poorly? (Assume
that each shot is independent of one
another)
What is the probability of making a basket on
at most 7 free throws?
On calculator: binomcdf(n, p, x)
B(12, 0.75)
P(X ≤ 7) = binomcdf(12, 0.75, 7) = 0.1576436
Cumulative Binomial Probability




In applications we frequently want to find the
probability that a random variable takes a
range of values. The cumulative binomial
probability is useful in these cases.
On calculator: binomcdf(n, p, x)
B(12, 0.75)
P(X ≤ 7) = binomcdf(12, 0.75, 7) = 0.1576
Cumulative Binomial Probability
It is very important that you understand the
difference between binomcdf and binompdf
Among employed women, 25% have never been
married. Select 10 employed women at random.
1.
What is the probability that exactly 2 of
the 10 women in your sample have never
been married?
binompdf(10, 0.25, 2) = 0.2816
2.
What is the probability that 2 or fewer
have never been married?
binomcdf(10, 0.25, 2) = 0.5256
Gretchen is a 60% free throw shooter. In a
season, she shoots an average of 75 free throws.
What is the probability that Gretchen
1)
makes exactly 50 out of 75 free throws?
P(X = 50) = binompdf(75, 0.6, 50) = 0.0479
2)
makes more than 50 free throws?
P(X > 50) = 1 - binomcdf(75, 0.6, 50) = 0.0963
3)
makes no more than 40 free throws?
P(X ≤ 40) = binomcdf(75, 0.6, 40) = 0.1446
Caution…..



On the AP Exam you are not allowed to just
write “calculator talk.”
You must write out the formula for each
problem, but feel free to use your calculator
for the calculations
For cumulative, use “+ … +” to show the first
two and last part of the formula.
Geometric Distribution
Getting Started
Suppose Sam is bashful and has trouble getting girls
to say “Yes” when he asks them for a date. Sadly, in
fact, only 10% of the girls he asks actually agree to
go out with him. Suppose that p = 0.10 is the
probability that any randomly selected girl will agree
to go out with Sam when asked.
1.
What is the probability that the first four girls he
asks say “No” and the fifth says “Yes?”
2.
How many girls can he expect to ask before the
first one says “Yes?”
The Geometric Distribution



There are situations in which the goal is to
obtain a fixed number of successes.
If the goal is to obtain one success, a random
variable X can be defined that counts the
number of trials needed to obtain that first
success.
Geometric distribution – a distribution of a
random variable that counts the number of
trials needed to obtain a success.

Geometric Settings
In a binomial setting, the number of trials n is fixed and the binomial random variable
X counts the number of successes. In other situations, the goal is to repeat a
chance behavior until a success occurs. These situations are called geometric
settings.
Definition:
A geometric setting arises when we perform independent trials of the same
chance process and record the number of trials until a particular outcome
occurs. The four conditions for a geometric setting are
B
• Binary? The possible outcomes of each trial can be classified as
“success” or “failure.”
I
• Independent? Trials must be independent; that is, knowing the result
of one trial must not have any effect on the result of any other trial.
T
• Trials? The goal is to count the number of trials until the first success
occurs.
S
• Success? On each trial, the probability p of success must be the
same.
All four of these conditions must be met
in order to use a geometric distribution.
1.
Each observation falls into one of two
categories: “success” or “failure”
2.
The n observations are independent
3.
The probability of success, call it p, is the
same for each observation.
4.
The variable of interest is the number of
trials required to obtain the first success.
The Geometric Distribution

Let X = number of trials required for
the first success
X

is a geometric random variable
How is this distribution different than a
binomial distribution?
Be aware of the key differences between
binomial and geometric distributions.
 Binomial:
Finds the probability that k success will
occur for in n number of attempts.
 Geometric:
Finds the probability that a success will
occur for the first time on the nth try.
 equation: P(x=n) = ( 1 - p )n-1 p
Is this a Geometric Setting?
A game consists of rolling a single die. The
event of interest is rolling a 3; this event is
called a success. The random variable is
defined as X = the number of trials until a 3
occurs.
1.
Yes.
Success = 3, Failure = any other number.
 p = 1/6 and is the same for all trials.
The observations are independent.
We roll until a 3 is obtained.

Geometric Setting?
2.
Suppose you repeatedly draw cards without
replacement from a deck of 52 cards until
you draw an ace. There are two categories
of interest: ace = success; not ace = failure.
No. It fails the independence requirement. If
the first card is not an ace then the 2nd card
will have a greater chance of being an ace.
Using example 1, let’s calculate some probabilities
A game consists of rolling a single die. The event of
interest is rolling a 3; this event is called a success.
The random variable is defined as X = the number of
trials until a 3 occurs.
X  1 : P  X  1  P  success on first roll  1
6
X  2 : P  X  2   P  success on second roll 
 P  failure on 1st roll and success on 2nd roll
 P  failure on 1st roll  P  success on 2nd roll
 6   16
 5
Example 1 continued…
X  3 : P  X  3  P  failure on 1st roll   P  failure on 2nd roll
 P  succes on 3rd roll 
5  1


 6  6  6
 5
 6  16
X  n : P X  n   5
n 1
The Birthday Game
+
 Example:
Read the activity on page 398. The random variable of interest in this game is Y = the
number of guesses it takes to correctly identify the birth day of one of your teacher’s
friends. What is the probability the first student guesses correctly? The second? Third?
What is the probability the kth student guesses corrrectly?
Verify that Y is a geometric random variable.
B: Success = correct guess, Failure = incorrect guess
I: The result of one student’s guess has no effect on the result of any other guess.
T: We’re counting the number of guesses up to and including the first correct guess.
S: On each trial, the probability of a correct guess is 1/7.
Calculate P(Y = 1), P(Y = 2), P(Y = 3), and P(Y = k)
P(Y  1)  1/7
P(Y  2)  (6 /7)(1/7)  0.1224
P(Y  3)  (6 /7)(6 /7)(1/7)  0.1050
Notice the pattern?
Geometric Probability
If Y has the geometric distribution with probability p of
success on each trial, the possible values of Y are
1, 2, 3, … . If k is any one of these values,
P(Y  k)  (1  p) k 1 p
Understanding Probability



In rolling a die, it is possible that you will have
to roll the die many times before you roll a 3.
In fact, it is theoretically possible to roll the
die forever without rolling a 3 (although the
probability gets closer and closer to zero the
longer you roll the die without getting a 3)
P(X = 50) = (5/6)49(1/6) = 0.0000
Probability distribution table for the
geometric random variable never ends….


This should remind you of an infinite
geometric sequence from Algebra II
In order for it to be a true pdf, the sum of
the second row must add up to 1.
Verifying it is a valid pdf

Recall the infinite sum formula:
a1
S
1r

In this case, a1 = p and r = 1 – p
p
p
S
 1
1  (1  p) p
Calculating Geometric Probabilities
P(X = k) = (1 – p)k – 1p
 P(X =k)  geometpdf (p, k)


“Probability that the first success occurs
on the kth trial”

P(X < k)  geometcdf (p, k)
The Expected Value

The mean or expected value of the geometric
random variable is the expected number of
trials needed for the first success.
μ = 1/p
The formula for Standard Deviation of a
Geometric series in not on the AP Statistics
course syllabus. In case you are wondering,
the formulas is:
The variance of X is σ2 = (1 - p)/p2
The State Fair

Glenn likes the game at the state fair where
you toss a coin into a saucer. You win if the
coin comes to a rest in the saucer without
sliding off. Glenn has played the game many
times and has determined that on average he
wins 1 out of every 12 times he plays. He
believes that his chances of winning are the
same for each toss. He has no reason to think
that his tosses are not independent. Let X be
the number of tosses until a win. Find his
expected value.
The State Fair

This is a geometric distribution

p = 1/12

μ = 1/(1/12) = 12
This means that he can expect to
play 12 games before he wins
Sometimes we are interested in the
probability that is takes more than a
certain number of trials to achieve success.

The probability that it takes more than n
trials to reach the first success is
P(X > n) = (1 - p)n
Roll a die until a 3 occurs.

What is the probability that it takes more
than 6 rolls to observe a 3?
P(X > 6) = (1 – 1/6)6 = 0.3349
Game of chance

Three friends each toss a coin. The odd man
wins; that is, if one coin comes up different
from the other two, that person wins the
round. If the coins all match, then no one
wins and they toss again. We’re interested in
the number of times the players will have to
toss the coins until someone wins.
Game of chance.

What is the probability that no one will win on
a given coin?
Out of 8 outcomes, HHH and TTTdo not
produce winners. So,
P(no winner) = 2/8 or 0.25
Define a success as “someone wins a given coin
toss.” What is the probability of success?
P(winner) = 1 – P(no winner)
1 – 0.25 = 0.75
Game of chance.



Define the random variable of interest:
X = number of tosses until someone wins
Is X binomial or geometric?
geometric
What is the probability that is takes no more
than 2 tosses for someone to win?
P(X ≤ 2) = 1 – P(X > 2) = 1 – (0.25)2 = 0.9375
Game of chance.

What is the probability that is takes more
than 4 tosses for someone to win?
P(X > 4) = (0.25)4 = 0.0039

What is the expected number of tosses
needed for someone to win?
X = 1 = 1
= 1.333
p
0.75
+ Section 6.3
Binomial and Geometric Random Variables
Summary
In this section, we learned that…

A binomial setting consists of n independent trials of the same chance
process, each resulting in a success or a failure, with probability of success
p on each trial. The count X of successes is a binomial random variable.
Its probability distribution is a binomial distribution.

The binomial coefficient counts the number of ways k successes can be
arranged among n trials.

If X has the binomial distribution with parameters n and p, the possible
values of X are the whole numbers 0, 1, 2, . . . , n. The binomial probability
of observing k successes in n trials is
n  k
P(X  k)   p (1 p) nk
k 
+ Section 6.3
Binomial and Geometric Random Variables
Summary
In this section, we learned that…

The mean and standard deviation of a binomial random variable X are
 X  np
 X  np(1 p)

The Normal approximation to the binomial distribution says that if X is a
count having the binomial distribution with parameters n and p, then when n
is large, X is 
approximately Normally distributed. We will use this
approximation when np ≥ 10 and n(1 - p) ≥ 10.
+ Section 6.3
Binomial and Geometric Random Variables
Summary
In this section, we learned that…

A geometric setting consists of repeated trials of the same chance process
in which each trial results in a success or a failure; trials are independent;
each trial has the same probability p of success; and the goal is to count the
number of trials until the first success occurs. If Y = the number of trials
required to obtain the first success, then Y is a geometric random variable.
Its probability distribution is called a geometric distribution.

If Y has the geometric distribution with probability of success p, the possible
values of Y are the positive integers 1, 2, 3, . . . . The geometric probability
that Y takes any value is
P(Y  k)  (1  p) k 1 p

The mean (expected value) of a geometric random variable Y is 1/p.
+
Looking Ahead…
In the next Chapter…
We’ll learn how to describe sampling distributions that
result when data are produced by random sampling.
We’ll learn about
 Sampling Distributions
 Sample Proportions
 Sample Means