Download Chapter 4

Chapter
4
Discrete Probability
Distributions
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All rights reserved.
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Chapter Outline
• 4.1 Probability Distributions
• 4.2 Binomial Distributions
• 4.3 More Discrete Probability Distributions
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Section 4.1
Probability Distributions
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Section 4.1 Objectives
• Distinguish between discrete random variables and
continuous random variables
• Construct a discrete probability distribution and its
graph
• Determine if a distribution is a probability
distribution
• Find the mean, variance, and standard deviation of a
discrete probability distribution
• Find the expected value of a discrete probability
distribution
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Random Variables
Random Variable
• Represents a numerical value associated with each
outcome of a probability distribution.
• Denoted by x
• Examples
 x = Number of sales calls a salesperson makes in
one day.
 x = Hours spent on sales calls in one day.
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Random Variables
Discrete Random Variable
• Has a finite or countable number of possible
outcomes that can be listed.
• Example
 x = Number of sales calls a salesperson makes in
one day.
x
0
1
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2
3
4
5
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Random Variables
Continuous Random Variable
• Has an uncountable number of possible outcomes,
represented by an interval on the number line.
• Example
 x = Hours spent on sales calls in one day.
x
0
1
2
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3
…
24
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Example: Random Variables
Decide whether the random variable x is discrete or
continuous.
1. xx = The number of Fortune 500 companies that
lost money in the previous year.
Solution:
Discrete random variable (The number of companies
that lost money in the previous year can be counted.)
{0, 1, 2, 3, …, 500}
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Example: Random Variables
Decide whether the random variable x is discrete or
continuous.
2. xx = The volume of gasoline in a 21-gallon
tank.
Solution:
Continuous random variable (The amount of
gasoline in the tank can be any volume between 0
gallons and 21 gallons.)
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Example
• Discrete or Continuous?
• X represents the number of shares of stock in the
Dow Jones Industrial Average that have price
increases on a given day
• X represents the volume of bottled water in a 32
ounce container
• The length of time it takes to take a test
• The number of home runs hit during a baseball game
Hint: Discrete Random Variables are usually counted data, while
Continuous Random Variables are usually measured data
Discrete Probability Distributions
Discrete probability distribution
• Lists each possible value the random variable can
assume, together with its probability.
• Must satisfy the following conditions:
In Words
In Symbols
1. The probability of each value of the
discrete random variable is between
0 and 1, inclusive.
0 ≤ P (x) ≤ 1
2. The sum of all the probabilities is 1.
ΣP (x) = 1
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Constructing a Discrete Probability
Distribution
Let x be a discrete random variable with possible
outcomes x1, x2, … , xn.
1. Make a frequency distribution for the possible
outcomes.
2. Find the sum of the frequencies.
3. Find the probability of each possible outcome by
dividing its frequency by the sum of the frequencies.
4. Check that each probability is between 0 and 1,
inclusive, and that the sum of all probabilities is 1.
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Example: Constructing a Discrete
Probability Distribution
An industrial psychologist administered a personality
inventory test for passive-aggressive traits to 150
employees. Individuals were given a score from 1 to 5,
where 1 was extremely passive and 5 extremely
aggressive. A score of 3 indicated
Score, x Frequency, f
neither trait. Construct a
1
24
probability distribution for the
2
33
random variable x. Then graph the
3
42
distribution using a histogram.
4
30
5
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21
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Solution: Constructing a Discrete
Probability Distribution
• Divide the frequency of each score by the total
number of individuals in the study to find the
probability for each value of the random variable.
P(1) 
24
 0.16
150
30
P(4) 
 0.20
150
P(2) 
33
 0.22
150
21
P(5) 
 0.14
150
• Discrete probability distribution:
P(3) 
42
 0.28
150
We read this as: If X
occurs, the probability of
X is…
x
1
2
3
4
5
P(x)
0.16
0.22
0.28
0.20
0.14
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Solution: Constructing a Discrete
Probability Distribution
x
1
2
3
4
5
P(x)
0.16
0.22
0.28
0.20
0.14
This is a valid discrete probability distribution since
1. Each probability is between 0 and 1, inclusive,
0 ≤ P(x) ≤ 1.
2. The sum of the probabilities equals 1,
ΣP(x) = 0.16 + 0.22 + 0.28 + 0.20 + 0.14 = 1.
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Solution: Constructing a Discrete
Probability Distribution
• Histogram
Passive-Aggressive Traits
Probability, P(x)
0.3
0.25
0.2
0.15
0.1
0.05
0
1
2
3
4
5
Score, x
Because the width of each bar is one, the area of
each bar is equal to the probability of a particular
outcome.
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Guidelines
• What is the probability of having a 1 or a 2 on the
survey?
• Just add the probabilities: .16 + .22 = .38
• Verifying probability distributions:
 Each probability should be between 0 and 1
 The sum of the probabilities = 1
Example
• Verify that this is a probability distribution
• What is the probability that a randomly selected person
involved in a traffic accident is in the 16-34 age group
• ∑ P(ages) = 1 -check
• .30 + .27 = .57
Chart Title
• 57% probability
That a randomly
selected person is
in the 16-34 age group
0.35
0.3
Axis Title
0.25
0.2
0.15
0.1
0.05
0
16-24
25-34
35-44
Axis Title
45-54
55-64
Mean
Mean of a discrete probability distribution
• μ = ΣxP(x)
• Each value of x is multiplied by its corresponding
probability and the products are added.
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Example: Finding the Mean
The probability distribution for the personality
inventory test for passive-aggressive traits is given. Find
the mean score.
Solution:
x
P(x)
xP(x)
What does this
mean? It means
that for this group
of 50 people, they
average a little
towards the passive
side (remember 1 is
passive, 5 is
aggressive)
1
2
3
4
0.16
0.22
0.28
0.20
1(0.16) = 0.16
2(0.22) = 0.44
3(0.28) = 0.84
4(0.20) = 0.80
5
0.14
5(0.14) = 0.70
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μ = ΣxP(x) = 2.94
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Variance and Standard Deviation
Variance of a discrete probability distribution
• σ2 = Σ(x – μ)2P(x)
Standard deviation of a discrete probability
distribution
•    2  ( x   ) 2 P ( x )
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Example: Finding the Variance and
Standard Deviation
The probability distribution for the personality
inventory test for passive-aggressive traits is given. Find
the variance and standard deviation. ( μ = 2.94)
x
P(x)
1
2
3
4
0.16
0.22
0.28
0.20
5
0.14
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Solution: Finding the Variance and
Standard Deviation
Recall μ = 2.94
x
P(x)
x–μ
(x – μ)2
(x – μ)2P(x)
1
0.16
1 – 2.94 = –1.94
(–1.94)2 ≈ 3.764
3.764(0.16) ≈ 0.602
2
0.22
2 – 2.94 = –0.94
(–0.94)2 ≈ 0.884
0.884(0.22) ≈ 0.194
3
0.28
3 – 2.94 = 0.06
(0.06)2 ≈ 0.004
0.004(0.28) ≈ 0.001
4
0.20
4 – 2.94 = 1.06
(1.06)2 ≈ 1.124
1.124(0.20) ≈ 0.225
5
0.14
5 – 2.94 = 2.06
(2.06)2 ≈ 4.244
4.244(0.14) ≈ 0.594
Variance: σ2 = Σ(x – μ)2P(x) = 1.616
Standard Deviation:     1.616  1.3
2
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Blaise Pascal
• (June 19, 1623, at Clermont, France – August 19, 1662) was a French
mathematician, physicist, and religious philosopher. He was a child prodigy
who was educated by his father, a civil servant. Pascal's earliest work was
in the natural and applied sciences where he made important contributions
to the construction of mechanical calculators, the study of fluids, and
clarified the concepts of pressure and vacuum by generalizing the work of
Evangelista Torricelli. Pascal also wrote in defense of the scientific method.
• Pascal was a mathematician of the first order. He helped create two major
new areas of research. He wrote a significant treatise on the subject of
projective geometry at the age of sixteen, and later corresponded with
Pierre de Fermat on probability theory, strongly influencing the
development of modern economics and social science.
• Pascal is also generally known as the father of Expected Value
Expected Value
• A mathematical way to use probability to determine
what to expect in various situations over the long run
• For example used to determine premiums on
insurance companies, payouts on gambling games,
risk/benefit analysis in business propositions
• The standard way to find expected value is to
multiply each possible outcome by its probability and
then add these products
• We use E as the symbol for Expected Value
• Expected Value is also used in Gaming Theory –such
as the Prisoner’s Dilemma
Prisoner’s Dilemma
Prisoner A Stays Silent
Prisoner A Betrays
Prisoner B Stays Silent
Each serves six months
(Here both are silent)
Prisoner A goes free
Prisoner B serves ten years
(here A talks, B is silent)
Prisoner B Betrays
Prisoner A serves ten years
Prisoner B goes free
(Here A is silent, but B talks)
Each serves five years
(here, both talk)
• The dilemma arises when the person analyzing this assumes that both
prisoners only care about minimizing their own jail terms.
• Each prisoner has two options: to cooperate with his accomplice and stay
quiet, or to break the code and squeal on his accomplice in return for a
lighter sentence.
• The outcome of each choice depends on the choice of the
accomplice, but each prisoner must choose without knowing what
his accomplice has chosen.
Larson/Farber 5th ed
26
Expected Value
Expected value of a discrete random variable
• Equal to the mean of the random variable.
• E(x) = μ = ΣxP(x)
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Example
• Find the expected value for the outcome of one roll of the
dice
• E = 1 x 1/6 + 2 x 1/6 + 3 x 1/6 + 4 x 1/6 + 5 x 1/6 + 6 x
1/6 =each number 1 through 6 is the side of a die, and
each side has a probability of 1/6 of being the side which
lands up
• = (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 = 3.5
• The expected value -the most likely number to show if I
roll the dice- is 3.5
• While this value cannot occur in 1 roll of the dice, it
should emerge as the average over a long period of tests
• This idea comes up a lot – “over a long period of time”
Example
• Find the expected value for the number of girls for a family
with 3 children –in other words, how many girls would you
expect to see in a family with 3 children?
• No Girls: Boy/Boy/Boy  1 way
• 1 Girl: GBB/BGB/BBG  3 ways
• 2 Girls: GGB/BGG/GBG  3 ways
• 3 Girls: GGG  1 way
• 8 ways total to have 3 children
• E = 0 x 1/8 + 1 x 3/8 + 2 x 3/8 +3 x 1/8 = 12/8 = 3/2 = 1.5
• Again, this makes sense. If you had 3 children, then you would
expect half to be boys, and half to be girls
Example
• An automobile insurance company has determined
the probabilities for various claim amounts for drivers
age 16-21, as shown below:
Amount of Claim
Probability
$0
0.70
$2,000
0.15
$4,000
0.08
$6,000
0.05
$8,000
0.01
$10,000
0.01
• Calculate the expected value and describe what this means in
practical terms
• How much should the company charge for premiums?
Example
• E = $0 x .70 + $2,000 x .15 + $4,000 x
.08 + $6,000 x .05 + $8,000 x .01 +
$10,000 x .01
• = $0 + $300 + $320 + $300 + $80 + $100
= $1100
• The expected value of paying out is
$1100. In the long run, the average cost
of a claim is $1100
• Therefore, the company needs to charge
premiums of at least $1100
Amount of Claim Probability
$0
0.70
$2,000
0.15
$4,000
0.08
$6,000
0.05
$8,000
0.01
$10,000
0.01
Expected Value for Payoffs
•
•
•
•
But what does this mean?
A game is played using 1 die. If a 1,2, or 3 is rolled,
the player wins nothing
If a 4 or 5 is rolled, the player wins $3
If a 6 is rolled, the player wins $9
If there is a charge of $1 to play the game, what is the
game’s expected value?
Expected
Outcome Gain/Loss Probability Value
1,2, or 3
($1) 3/6
-3/6
4,5
$2
2/6
4/6
6
$8
1/6
8/6
Total: 9/6
This chart is how
you should calculate
expected values. It is
easy and well laid
out
Example
• E = (-$1) x (3/6) + ($2) x (2/6) + ($8) x (1/6)
• = (-$3 + $4 + $8)/6 = $9/6 = $1.50
• In the long run, a player can expect to win an average
of $1.50 for each game played
• This doesn’t mean a player will win $1.50 for a single
game
• Therefore (for example), if a player played 1000
games, he or she could expect to win about $1,500
• Obviously you may not win if you play just one
game. But “over time” you would expect to win…
Casino Games
• Unlike the dice game, casino games are set up for a
player to lose in the long run
• For example, Roullette:
 You bet $1 on a single number
 You can win $35, and you keep your $1 bet
 There are 38 slots –or 1 winning slot and 37 losing
• What is the expected value for betting $1 on the
number 20
Example
Expected
Outcome Gain/Loss Probability Value
Ball on 20
$35
1/38
35/38
Ball not on
20
($1)
37/38
-37/38
-2/38
• E = $35(1/38) + -$1(37/38) (you have a 1/38 probability
of winning $35 and you lose $1 37/38 times)
• = ($35-$37)/38 = -$2/38 = -$.05
• The expected value is about minus 5 cents
• In the long run, a player will lose about 5 cents per game
• If a player played 2000 games, they would lose about
$100
Example: Finding an Expected Value
At a raffle, 1500 tickets are sold at $2 each for four
prizes of $500, $250, $150, and $75. You buy one ticket
for 2 dollars. What is the expected value of your gain?
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Solution: Finding an Expected Value
• To find the gain for each prize, subtract
the price of the ticket from the prize:
 Your gain for the $500 prize is $500 – $2 = $498
 Your gain for the $250 prize is $250 – $2 = $248
 Your gain for the $150 prize is $150 – $2 = $148
 Your gain for the $75 prize is $75 – $2 = $73
• If you do not win a prize, your gain is $0 – $2 = –$2
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Solution: Finding an Expected Value
• Probability distribution for the possible gains
(outcomes)
Gain, x
P(x)
$498
1
1500
$248
1
1500
$148
1
1500
$73
1
1500
–$2
1496
1500
E (x )  xP (x )
1
1
1
1
1496
 $248 
 $148 
 $73 
 ($2) 
1500
1500
1500
1500
1500
 $1.35
 $498 
You can expect to lose an average of $1.35 for each ticket
you buy.
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Doing it with my chart
Outcome
Gain/Loss Probability
Expected Value
Value
Win $500
$498
1/1500
498*(1/1500)
0.33
Win $250
$248
1/1500
248*(1/1500)
0.17
Win $150
$148
1/1500
148*(1/1500)
0.10
Win $75
$73
1/1500
73*(1/1500)
0.05
Lose $2
-$2
1496/1500
-2*(1496/1500)
Sum:
Note: Gain/Lost is “List 1” and Probability is “List 2”
• Enter Lists into stats
• Go back into stats, then calculate, then 1-Var Stats
•Hit ENTER, and BEFORE you hit enter again, type L1, L2 (use the 2nd
button for each)
-1.99
-1.35
39
Section 4.1 Summary
• Distinguished between discrete random variables and
continuous random variables
• Constructed a discrete probability distribution and its
graph
• Determined if a distribution is a probability
distribution
• Found the mean, variance, and standard deviation of a
discrete probability distribution
• Found the expected value of a discrete probability
distribution
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Assignment
• Page 197-198 5-25 odd
Larson/Farber 5th ed
41
Assignment
• Page 197-198 5-25 odd
• Page 198 199
• 27-32 , 35-46
 Find the mean, variance, and standard deviation of
each chart
 Interpret the mean to the average reader
Larson/Farber 5th ed
42
Chapter 4 Quiz 1 (30 points, 5 points
each)
• There are 38 slots on a roulette wheel. 18 red, 18 black, and 2 green.
• A normal bet is 1 dollar, and if you win you receive 36 dollars back (total)
1.
What is the expected value if you bet on #20 red? (use a chart)
2.
Using question # 1, if you bet 100 times, how much money would you expect to
win or lose?
3.
What is the expected value if you bet just on red? (use a chart)
4.
Using question # 3, if you bet 149 times, how much money would you expect to
win or lose?
5.
What is the expected value if you bet on green? (use a chart)
6.
Using question # 5, if you bet 520 times, how much money would you expect to
win or lose?
Expected
Outcome Gain/Loss Probability Value
This Chart
Larson/Farber 5th ed
Total:
43
Chapter 4 Quiz 2 (10 points, 5 points
each)
•
•
•
•
•
•
1.
2.
•
You pay 10 dollars to play a game. 5000 people play the game.
There are 5 grand prize winners. They win 500 dollars (they receive 500)
There are 10 1st prize winners. They win 50 dollars (they receive 50)
There are 15 2nd prize winners. They win 25 dollars (they receive 25)
There are 20 3rd prize winners. They win 10 dollars. (they receive 10)
Everyone else, obviously, is not a winner.
What is the expected value of this game?
If you played this game 100 times, how much money would you expect to win or
lose?
Create a chart to calculate this answer
Expected
Outcome Gain/Loss Probability Value
This Chart
Larson/Farber 5th ed
Total:
44
Section 4.2
Binomial Distributions
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Section 4.2 Objectives
• Determine if a probability experiment is a binomial
experiment
• Find binomial probabilities using the binomial
probability formula
• Find binomial probabilities using technology,
formulas, and a binomial probability table
• Graph a binomial distribution
• Find the mean, variance, and standard deviation of a
binomial probability distribution
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Binomial Experiments
1. The experiment is repeated for a fixed number of
trials, where each trial is independent of other trials.
2. There are only two possible outcomes of interest
for each trial. The outcomes can be classified as a
success (S) or as a failure (F).
3. The probability of a success P(S) is the same for each
trial.
4. The random variable x counts the number of
successful trials.
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Notation for Binomial Experiments
Symbol
n
p = P(S)
q = P(F)
x
Description
The number of times a trial is repeated
The probability of success in a single trial
The probability of failure in a single trial
(q = 1 – p)
The random variable represents a count of
the number of successes in n trials:
x = 0, 1, 2, 3, … , n.
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Example
•
•
•
•
•
•
•
From a standard deck of cards, pick a card
Note whether it is a club or not and replace
Repeat 5 times, so n=5
The outcome is classified as S = clubs, and F = not clubs
Probability of success: p=p(S)= ¼
Probability of failure: q=p(F) = ¾
The random variable x represents the number of clubs
selected during your test. So x can equal 0,1,2,3,4, or 5
• Note that X is a discrete random variable –we can list
(count) all possible outcomes for x
Example: Binomial Experiments
Decide whether the experiment is a binomial
experiment. If it is, specify the values of n, p, and q, and
list the possible values of the random variable x.
1. A certain surgical procedure has an 85% chance of
success. A doctor performs the procedure on eight
patients. The random variable represents the number
of successful surgeries.
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Solution: Binomial Experiments
Binomial Experiment
1. Each surgery represents a trial. There are eight
surgeries, and each one is independent of the others.
2. There are only two possible outcomes of interest for
each surgery: a success (S) or a failure (F).
3. The probability of a success, P(S), is 0.85 for each
surgery.
4. The random variable x counts the number of
successful surgeries.
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Solution: Binomial Experiments
Binomial Experiment
• n = 8 (number of trials)
• p = 0.85 (probability of success)
• q = 1 – p = 1 – 0.85 = 0.15 (probability of failure)
• x = 0, 1, 2, 3, 4, 5, 6, 7, 8 (number of successful
surgeries)
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Example: Binomial Experiments
Decide whether the experiment is a binomial
experiment. If it is, specify the values of n, p, and q, and
list the possible values of the random variable x.
2. A jar contains five red marbles, nine blue marbles, and
six green marbles. You randomly select three marbles
from the jar, without replacement. The random
variable represents the number of red marbles.
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Solution: Binomial Experiments
Not a Binomial Experiment
• The probability of selecting a red marble on the first
trial is 5/20.
• Because the marble is not replaced, the probability of
success (red) for subsequent trials is no longer 5/20.
• The trials are not independent and the probability of
a success is not the same for each trial.
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Binomial Probability Formula
Binomial Probability Formula
• The probability of exactly x successes in n trials is
P( x)  n Cx p q
x
•
•
•
•
n x
n!

p x q n x
(n  x)! x !
This is really a mixture of a
combination problem (nCx) times a
probability of success times a
probability of failure we have “n
choose x” times probability of
success^number of successes x
probability of failure^number of
failures
n = number of trials
p = probability of success
q = 1 – p probability of failure
x = number of successes in n trials
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Example: Finding Binomial Probabilities
Microfracture knee surgery has a 75% chance of success
on patients with degenerative knees. The surgery is
performed on three patients. Find the probability of the
surgery being successful on exactly two patients.
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Solution: Finding Binomial Probabilities
Method 1: Draw a tree diagram and use the
Multiplication Rule
 9
P(2 successful surgeries)  3   0.422
 64 
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Solution: Finding Binomial Probabilities
Method 2: Binomial Probability Formula
n  3,
3
1
p  , q  1 p  , x  2
4
4
2
 3  1 
P(2 successful surgeries)  3 C2    
 4  4
3 2
2
 3  1 
3!

(3  2)!2!  4   4 
1
 9   1  27
 3    
 0.422
 16   4  64
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Solution: Finding Binomial Probabilities
• Using a Calculator:
• Binomial Distribution:
 2ndVarsBinomialPDF(N,P,S) –this solves for
1 point on the curve
 N = number of trials
 P = probability of success
 S = number of successes
• Therefore we type in: BinomialPDF(3,.75,2)
 = .421875
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Binomial Probability Distribution
Binomial Probability Distribution
• List the possible values of x with the corresponding probability
of each.
• Example: Binomial probability distribution for Microfracture
knee surgery: n = 3, p = 3
4
x
0
1
2
3
P(x)
0.016
0.141
0.422
0.422
 Use the binomial probability formula to find probabilities.
 Or: Use the calculator, and calculate each of these
successes, then create a distribution table
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Example
• A six sided die is rolled 3 times. Find the probability
of rolling exactly 1 six
• We could use the tree diagram, and the multiplication
rule
• Or we could use the calculator:
• 2nd vars (dist)binompdf then enter (in order):
 N trials, (3), then probability of success (1/6), then
number of successes (1)
 So you would have binompdf(3,1/6,1) (3 rolls, 1/6
probability of getting a 6, 1 success)
• =.34722222
Example: Constructing a Binomial
Distribution
In a survey, U.S. adults were asked to give reasons why
they liked texting on their cellular phones. Seven adults
who participated in the survey are randomly selected and
asked whether they like texting because it is quicker than
calling. Create a binomial
probability distribution for
the number of adults who
respond yes.
Hint: What percent is success?
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Solution: Constructing a Binomial
Distribution
• 56% of adults like texting because it is quicker than
Remember: This is for a chart with every possible
calling.
probability calculated
• n = 7, p = 0.56, q = 0.44, x = 0, 1, 2, 3, 4, 5, 6, 7
P(0) = 7C0(0.56)0(0.44)7 = 1(0.56)0(0.44)7 ≈ 0.0032
P(1) = 7C1(0.56)1(0.44)6 = 7(0.56)1(0.44)6 ≈ 0.0284
P(2) = 7C2(0.56)2(0.44)5 = 21(0.56)2(0.44)5 ≈ 0.1086
P(3) = 7C3(0.56)3(0.44)4 = 35(0.56)3(0.44)4 ≈ 0.2304
P(4) = 7C4(0.56)4(0.44)3 = 35(0.56)4(0.44)3 ≈ 0.2932
P(5) = 7C5(0.56)5(0.44)2 = 21(0.56)5(0.44)2 ≈ 0.2239
P(6) = 7C6(0.56)6(0.44)1 = 7(0.56)6(0.44)1 ≈ 0.0950
P(7) = 7C7(0.56)7(0.44)0 = 1(0.56)7(0.44)0 ≈ 0.0173
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Solution: Constructing a Binomial
Distribution
x
0
1
2
3
4
5
6
7
P(x)
0.0032
0.0284
0.1086
0.2304
0.2932
0.2239
0.0950
0.0173
All of the probabilities are between
0 and 1 and the sum of the
probabilities is 1.
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Example: Finding Binomial Probabilities
Using Technology
The results of a recent survey indicate that 67% of U.S.
adults consider air conditioning a necessity. If you
randomly select 100 adults, what is the probability that
exactly 75 adults consider air conditioning a necessity?
Use a technology tool to find the probability. (Source:
Opinion Research Corporation)
Solution:
• Binomial with n = 100, p = 0.67, x = 75
• Solve on the calculator
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Solution: Finding Binomial Probabilities
Using Technology
From the displays, you can see that the probability that
exactly 75 adults consider air conditioning a necessity
is about 0.02.
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Chapter 4 Quiz 2 (5 points each)
• A game is played using 1 die. If a 1,3, or 5 is rolled, the player wins
nothing. If a 2 or 4 is rolled, the player gets $5 back. If a 6 is rolled, the
player gets $8 back. There is a charge of $1 to play the game.
Outcome
1,3, or 5
Gain/Loss
Probability
Expected Value
1/6
7/6
$4
1.
2.
3.
4.
Make this chart, and fill in all missing numbers
What is the game’s expected value?
What is the mean of this distribution chart?
Over time, the person running this game will either make or lose money.
How much money will they expect to make or lose total, if 500 games were
played?
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Example: Finding Binomial Probabilities
A survey indicates that 41% of women in the U.S.
consider reading their favorite leisure-time activity. You
randomly select four U.S. women and ask them if
reading is their favorite leisure-time activity. Find the
probability that at least two of them respond yes.
Solution:
• n = 4, p = 0.41, q = 0.59
• At least two means two or more.
• Find the sum of P(2), P(3), and P(4).
• You can solve each on the calculator and add
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Solution: Finding Binomial Probabilities
P(2) = 4C2(0.41)2(0.59)2 = 6(0.41)2(0.59)2 ≈ 0.351094
P(3) = 4C3(0.41)3(0.59)1 = 4(0.41)3(0.59)1 ≈ 0.162654
P(4) = 4C4(0.41)4(0.59)0 = 1(0.41)4(0.59)0 ≈ 0.028258
P(x ≥ 2) = P(2) + P(3) + P(4)
≈ 0.351094 + 0.162654 + 0.028258
≈ 0.542
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Solution Using the Calculator
• There is a way to calculate cumulative probabilities
• In this case, we choose 2nd Vars binomcdf(
 This stands for binomial cumulative (binompdf
stands for binomial point –when you want a
specific value at one point on the curve)
• This solves “up to” problems
 In other words –We want to know the probability
up to and including a certain number of successes
• There’s only one problem: This particular problem
isn’t an “up to” problem, but a “more than and
including problem”
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Solving With a Calculator (Continued)
• Enter 2nd Vars Binomcdf(N,P,S)
 N = number of trials (4 in this example)
 P = probability (.41 in this example)
 S = number of successes (in this case just 1)
• Enter Binomcdf(4,.41,1)
 We get .457
• This is not the solution
-what now?
• This is the COMPLEMENT of our solution –we were
searching for 2 or more successes. We calculated up to and
including 1 success
• To find our solution, we simply take 1-.457 and we get .542,
which is our solution
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Example: Finding Binomial Probabilities
Using a Table
About ten percent of workers (16 years and over) in the
United States commute to their jobs by carpooling. You
randomly select eight workers. What is the probability
that exactly four of them carpool to work? Use a table to
find the probability. (Source: American Community Survey)
Solution:
• Binomial with n = 8, p = 0.10, x = 4
• Solve this on the calculator
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Solution: Finding Binomial Probabilities
Using a Table
• A portion of Table 2 is shown
The probability that exactly four of the eight workers
carpool to work is 0.005.
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Example: Graphing a Binomial
Distribution
Sixty percent of households in the U.S. own a video
game console. You randomly select six households and
ask each if they own a video game console. Construct a
probability distribution for the random variable x. Then
graph the distribution. (Source: Deloitte, LLP)
Solution:
• n = 6, p = 0.6, q = 0.4
• Find the probability for each value of x
• Solve each solution on the calculator and then
graph by hand
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Solution: Graphing a Binomial
Distribution
x
0
1
2
3
4
5
6
P(x)
0.004
0.037
0.138
0.276
0.311
0.187
0.047
Histogram:
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Mean, Variance, and Standard Deviation
• Mean: μ = np
• Variance: σ2 = npq
• Standard Deviation:   npq
These are population parameters
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Example: Finding the Mean, Variance,
and Standard Deviation
In Pittsburgh, Pennsylvania, about 56% of the days in a
year are cloudy. Find the mean, variance, and standard
deviation for the number of cloudy days during the
month of June. Interpret the results and determine any
unusual values. (Source: National Climatic Data Center)
Solution: n = 30, p = 0.56, q = 0.44
Mean: μ = np = 30∙0.56 = 16.8
Variance: σ2 = npq = 30∙0.56∙0.44 ≈ 7.4
Standard Deviation:   npq  30  0.56  0.44  2.7
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Solution: Finding the Mean, Variance, and
Standard Deviation
μ = 16.8 σ2 ≈ 7.4
σ ≈ 2.7
• On average, there are 16.8 cloudy days during the
month of June.
• The standard deviation is about 2.7 days.
• Values that are more than two standard deviations
from the mean are considered unusual.
 16.8 – 2(2.7) =11.4, a June with 11 cloudy days
or fewer would be unusual.
 16.8 + 2(2.7) = 22.2, a June with 23 cloudy days
or more would also be unusual.
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Section 4.2 Summary
• Determined if a probability experiment is a binomial
experiment
• Found binomial probabilities using the binomial
probability formula
• Found binomial probabilities using technology and a
binomial table
• Graphed a binomial distribution
• Found the mean, variance, and standard deviation of
a binomial probability distribution
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Assignment
• Page 211-214 3-6, 9-26, 27,29
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Chapter 4 Quiz 3
• Do problem 30 on page 214
• Do parts a, b and c (skip part d)
• 30 points –ten points for each
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Section 4.3
More Discrete Probability
Distributions
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Section 4.3 Objectives
• Find probabilities using the geometric distribution
• Find probabilities using the Poisson distribution
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Geometric Distribution
• A discrete probability distribution.
• Satisfies the following conditions
 A trial is repeated until a success occurs.
 The repeated trials are independent of each other.
 The probability of success p is constant for each
trial.
 x represents the number of the trial in which the
first success occurs.
• The probability that the first success will occur on trial
x is P(x) = p(q)x – 1, where q = 1 – p.
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Example: Geometric Distribution
Basketball player LeBron James makes a free throw
shot about 74% of the time. Find the probability that the
first free throw shot LeBron makes occurs on the third
or fourth attempt.
Solution:
• P(shot made on third or fourth attempt) = P(3) + P(4)
• Geometric with p = 0.74, q = 0.26, x = 3
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Solution: Geometric Distribution
• P(3) = 0.74(0.26)3–1 = 0.050024
• P(4) = 0.74(0.26)4–1 ≈ 0.013006
P (shot made on third or fourth attempt)
= P(3) + P(4)
≈ 0.050024 + 0.013006
≈ 0.063
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Geometric Probability on the Calculator
• 2nd Vars Geometpdf (p,x) –this solves for one
specific probability, or point on the curve
• P = Probability of Success
• X = number of trials until success
• If you want to solve a cumulative probability:
• 2nd Vars Geometcdf (p,x) –this solves for
cumulative probability – “up to”
• P = Probability of Success
• X = number of trials you want to solve “up to”
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Geometric Probability on the Calculator
Basketball player LeBron James makes a free throw
shot about 74% of the time. Find the probability that the
first free throw shot LeBron makes occurs on the third
or fourth attempt.
On the calculator:
•2nd Vars Geometpdf (p,x)
•2nd Vars Geometpdf (.74,3) =.0500
•2nd Vars Geometpdf (.74,4) = .013
•Add them and get .063
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Example
• From previous experience you know that the probability
of making a sale on any given phone call is .23
• Find the probability that the sale will occur on the fourth
or fifth call
• Find the probability for the 4th call, then find the
probability for the 5th call, then add them
• P(x) = P*(q) x-1
• p(4) = .23*.77^3 = .105
• p(5) = .23*.77^4 = .08
• So p(sale on 4 or 5 call) = .105 + .08 = .186
Example
• Find the probability that you will make a sale before the 4th
call
• You will need to find the p(1) + p(2) + p(3) Then sum them
• 2nd VARSgeometpdf(p,x) –do this 3 times with X as 1,
then 2, then 3 (remember the probability of success is .23)
• You should get .543467
• You can also do a cumulative probability: Geometcdf(.23,3)
and you will get the same answer
• Notice: This type of distribution is discrete (x number of
trials is easily counted) but we could go on forever –where x
is any large number. Therefore, the sum of the probabilities
will approach 1, but won’t ever be exactly 1
• Therefore, you can NOT calculate the complement to this
Poisson Distribution
• So far, in binomial experiments we have focused on the
probability of success for any given number of trials
• What if we want to know the probability that a specific
event occurs a specific number of times within a given
amount of time or space?
• (In other words, we may not be focused on success this
time, just that something occurred)
• For example, what is the probability that an employee will
take 15 days sick leave within a year?
• To calculate these types of problems, we use the Poisson
Distribution
Poisson Distribution
• Siméon-Denis Poisson (21 June 1781 – 25 April
1840), was a French mathematician, geometer, and
physicist.
• He was first published at the age of 18, and was
considered a giant in his field until he passed away
• One of his most famous theories is called “The
Poisson Distribution”
Poisson Distribution
Poisson distribution
• A discrete probability distribution.
• Satisfies the following conditions
 The experiment consists of counting the number of
times x an event occurs in a given interval. The
interval can be an interval of time, area, or volume.
 The probability of the event occurring is the same for
each interval.
 The number of occurrences in one interval is
independent of the number of occurrences in other
intervals.
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Poisson Distribution
Poisson distribution
• Conditions continued:
 The probability of exactly x occurrences in an interval
is
x 

e
P (x ) 
x!
where e is an irrational number ≈ 2.71828 and μ is
the mean number of occurrences per interval unit.
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The Constant e (FYI)
• The mathematical constant e is the unique real
number such that the value of the derivative
(slope of the tangent line) of the function f(x) =
ex at the point x = 0 is exactly 1.
• The number e is of eminent importance in
mathematics, alongside 0, 1, π and i. Besides
being abstract objects, all five of these numbers
play important and recurring roles across
mathematics, and, coincidentally are the five
constants appearing in one formulation of
Euler's identity.
• The number e is irrational; it is not a ratio of
integers. Furthermore, it is transcendental; it is
not a root of any non-zero polynomial with
rational coefficients. The numerical value of e
truncated to 20 decimal places is
• 2.71828182845904523536….
Example: Poisson Distribution
The mean number of accidents per month at a certain
intersection is 3. What is the probability that in any
given month four accidents will occur at this
intersection?
Solution:
• Poisson with x = 4, μ = 3
34(2.71828)3
P (4) 
 0.168
4!
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On a Calculator
• The mean number of accidents per month at a certain
intersection is 3. What is the probability that in any given
month four accidents will occur at this intersection?
• On a calculator:
• 2ndVARS
• Poissonpdf(μ,x)
• Where μ is the mean and x is the number of occurrences
• Thus: poissonpdf(3,4) = .168
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Cumulative Poisson
• What is the probability that more than 4 accidents will occur in any given
month at the intersection?
• 2 ways:
• Use Poissonpdf(μ,x) and solve for each probability
• Thus we solve P(0), P(1), P(2), P(3) and P(4)
• Add all these probabilities = .81526
• Subtract this answer from 1 .1847
• This new answer will be all the probabilities more than P(4)
• 2nd way:
• Use Poissoncdf(μ,x) and solve for up to and including 4 accidents
• Poissoncdf(3,4) = .81526
• Subract from 1 = .1847
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Assignment
• Page 222 1-24 (skip 9 and 10)
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Section 4.3 Summary
• Found probabilities using the geometric distribution
• Found probabilities using the Poisson distribution
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Chapter 4 Quiz 4 (on your own)
• Do problem 25 on page 224
1. Part A) Solve for a binomial point distribution.
 On your paper, write what you used for: n,p,s (5 pts)
 Show the probability of success (5 pts)
2. Part B) Solve for a Poisson point distribution. You
must calculate the mean to solve this problem.
 On your paper, write your equation and what you
calculated for the mean (5 pts)
 Show the probability of success (5 pts)
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