5
Frequency
4
3
2
1
0
Ace
2
3
4
5
6
7
Card
8
9
10
J
Q
K
What is the probability of picking an ace?
5
Frequency
4
3
2
1
0
Ace
2
3
4
5
6
7
Card
8
9
10
J
Q
K
Probability =
5
Frequency
4
3
2
1
0
Ace
2
3
4
5
6
7
Card
8
9
10
J
Q
K
What is the probability of picking an ace?
4 / 52 = .077 or 7.7 chances in 100
5
Frequency
4
3
2
1
0
Ace
2
3
4
5
6
7
Card
8
9
10
J
Q
K
Card
K (.077)
Q (.077)
J (.077)
10 (.077)
9 (.077)
8 (.077)
7 (.077)
6 (.077)
5 (.077)
4 (.077)
3 (.077)
2 (.077)
Ace (.077)
Frequency
Every card has the same probability of being
picked
5
4
3
2
1
0
Card
K (.077)
Q (.077)
J (.077)
10 (.077)
9 (.077)
8 (.077)
7 (.077)
6 (.077)
5 (.077)
4 (.077)
3 (.077)
2 (.077)
Ace (.077)
Frequency
What is the probability of getting a 10, J, Q, or
K?
5
4
3
2
1
0
Card
K (.077)
Q (.077)
J (.077)
10 (.077)
9 (.077)
8 (.077)
7 (.077)
6 (.077)
5 (.077)
4 (.077)
3 (.077)
2 (.077)
Ace (.077)
Frequency
(.077) + (.077) + (.077) + (.077) = .308
16 / 52 = .308
5
4
3
2
1
0
Card
K (.077)
Q (.077)
J (.077)
10 (.077)
9 (.077)
8 (.077)
7 (.077)
6 (.077)
5 (.077)
4 (.077)
3 (.077)
2 (.077)
Ace (.077)
Frequency
What is the probability of getting a 2 and then
after replacing the card getting a 3 ?
5
4
3
2
1
0
Card
K (.077)
Q (.077)
J (.077)
10 (.077)
9 (.077)
8 (.077)
7 (.077)
6 (.077)
5 (.077)
4 (.077)
3 (.077)
2 (.077)
Ace (.077)
Frequency
(.077) * (.077) = .0059
5
4
3
2
1
0
Card
K (.077)
Q (.077)
J (.077)
10 (.077)
9 (.077)
8 (.077)
7 (.077)
6 (.077)
5 (.077)
4 (.077)
3 (.077)
2 (.077)
Ace (.077)
Frequency
What is the probability that the two cards you
draw will be a black jack?
5
4
3
2
1
0
10 Card = (.077) + (.077) + (.077) + (.077) = .308
Ace after one card is removed = 4/51 = .078
(.308)*(.078) = .024
4
3
2
1
Card
K (.077)
Q (.077)
J (.077)
10 (.077)
9 (.077)
8 (.077)
7 (.077)
6 (.077)
5 (.077)
4 (.077)
3 (.077)
2 (.077)
0
Ace (.077)
Frequency
5
Practice
• What is the probability of rolling a “1” using
a six sided dice?
• What is the probability of rolling either a
“1” or a “2” with a six sided dice?
• What is the probability of rolling two “1’s”
using two six sided dice?
Practice
• What is the probability of rolling a “1” using
a six sided dice?
1 / 6 = .166
• What is the probability of rolling either a
“1” or a “2” with a six sided dice?
• What is the probability of rolling two “1’s”
using two six sided dice?
Practice
• What is the probability of rolling a “1” using
a six sided dice?
1 / 6 = .166
• What is the probability of rolling either a
“1” or a “2” with a six sided dice?
(.166) + (.166) = .332
• What is the probability of rolling two “1’s”
using two six sided dice?
Practice
• What is the probability of rolling a “1” using
a six sided dice?
1 / 6 = .166
• What is the probability of rolling either a
“1” or a “2” with a six sided dice?
(.166) + (.166) = .332
• What is the probability of rolling two “1’s”
using two six sided dice?
(.166)(.166) = .028
Cards
• What is the probability of drawing an ace?
• What is the probability of drawing another ace?
• What is the probability the next four cards you
draw will each be an ace?
• What is the probability that an ace will be in the
first four cards dealt?
Cards
•
•
•
•
•
What is the probability of drawing an ace?
4/52 = .0769
What is the probability of drawing another ace?
4/52 = .0769; 3/51 = .0588; .0769*.0588 = .0045
What is the probability the next four cards you
draw will each be an ace?
• .0769*.0588*.04*.02 = .000003
• What is the probability that an ace will be in the
first four cards dealt?
• .0769+.078+.08+.082 = .3169
Probability
.00
Event
will not
occur
1.00
Event
must
occur
Probability
• In this chapter we deal with discreet
variables
– i.e., a variable that has a limited number of
values
• Previously we discussed the probability of
continuous variables (Z –scores)
– It does not make sense to seek the probability
of a single score for a continuous variable
• Seek the probability of a range of scores
Key Terms
• Independent event
– When the occurrence of one event has no
effect on the occurrence of another event
• e.g., voting behavior, IQ, etc.
• Mutually exclusive
– When the occurrence of one even precludes
the occurrence of another event
• e.g., your year in the program, if you are in prosem
Key Terms
• Joint probability
– The probability of the co-occurrence of two or
more events
• The probability of rolling a one and a six
• p (1, 6)
• p (Blond, Blue)
Key Terms
• Conditional probabilities
– The probability that one event will occur given
that some other vent has occurred
• e.g., what is the probability a person will get into a
PhD program given that they attended Villanova
– p(Phd l Villa)
• e.g., what is the probability that a person will be a
millionaire given that they attended college
– p($$ l college)
Example
Owns a video Does not own
game
a video game
Total
No Children
10
35
45
Children
25
30
55
Total
35
65
100
What is the simple probability that a person will
own a video game?
Owns a video Does not own
game
a video game
Total
No Children
10
35
45
Children
25
30
55
Total
35
65
100
What is the simple probability that a person will
own a video game? 35 / 100 = .35
Owns a video Does not own
game
a video game
Total
No Children
10
35
45
Children
25
30
55
Total
35
65
100
What is the conditional probability of a person owning a
video game given that he or she has children?
p (video l child)
Owns a video Does not own
game
a video game
Total
No Children
10
35
45
Children
25
30
55
Total
35
65
100
What is the conditional probability of a person owning a
video game given that he or she has children?
25 / 55 = .45
Owns a video Does not own
game
a video game
Total
No Children
10
35
45
Children
25
30
55
Total
35
65
100
What is the joint probability that a person will own
a video game and have children? p(video, child)
Owns a video Does not own
game
a video game
Total
No Children
10
35
45
Children
25
30
55
Total
35
65
100
What is the joint probability that a person will own
a video game and have children? 25 / 100 = .25
Owns a video Does not own
game
a video game
Total
No Children
10
35
45
Children
25
30
55
Total
35
65
100
25 / 100 = .25
.35 * .55 = .19
Owns a video Does not own
game
a video game
Total
No Children
10
35
45
Children
25
30
55
Total
35
65
100
The multiplication rule assumes that the two events are
independent of each other – it does not work when there is
a relationship!
Owns a video Does not own
game
a video game
Total
No Children
10
35
45
Children
25
30
55
Total
35
65
100
Practice
Republican
Democrat
Total
Male
52
27
79
Female
18
65
83
Total
70
92
162
p (republican)
p (republican, male)
p (republican l male)
p(female)
p(female, republican)
p(male l republican)
Republican
Democrat
Total
Male
52
27
79
Female
18
65
83
Total
70
92
162
p (republican) = 70 / 162 = .43
p (republican, male) = 52 / 162 = .32
p (republican l male) = 52 / 79 = .66
Republican
Democrat
Total
Male
52
27
79
Female
18
65
83
Total
70
92
162
p(female) = 83 / 162 = .51
p(female, republican) = 18 / 162 = .11
p(male l republican) = 52 / 70 = .74
Republican
Democrat
Total
Male
52
27
79
Female
18
65
83
Total
70
92
162
Foot Race
• Three different people enter a “foot race”
• A, B, C
• How many different combinations are
there for these people to finish?
Foot Race
A, B, C
A, C, B
B, A, C
B, C, A
C, B, A
C, A, B
6 different permutations of these three names
taken three at a time
Foot Race
• Six different people enter a “foot race”
• A, B, C, D, E, F
• How many different permutations are there
for these people to finish?
Permutation
N!
N
 Pr
( N  r )!
Ingredients:
N = total number of events
r = number of events selected
Permutation
6!
 720
(6  6)!
Ingredients:
N = total number of events
r = number of events selected
A, B, C, D, E, F
Note: 0! = 1
Foot Race
• Six different people enter a “foot race”
• A, B, C, D, E, F
• How many different permutations are there
for these people to finish in the top three?
•
A, B, C
A, C, D
A, D, E
B, C, A
Permutation
N!
N
 Pr
( N  r )!
Ingredients:
N = total number of events
r = number of events selected
Permutation
6!
 120
(6  3)!
Ingredients:
N = total number of events
r = number of events selected
Foot Race
• Six different people enter a “foot race”
• If a person only needs to finish in the top
three to qualify for the next race (i.e., we
don’t care about the order) how many
different outcomes are there?
Combinations
N!
N
 Cr
r!( N  r )!
Ingredients:
N = total number of events
r = number of events selected
Combinations
6!
 20
3!(6  3)!
Ingredients:
N = total number of events
r = number of events selected
Note:
• Use Permutation when ORDER matters
• Use Combination when ORDER does not
matter
Practice
• There are three different prizes
– 1st $1,00
– 2nd $500
– 3rd $100
• There are eight finalist in a drawing who
are going to be awarded these prizes.
• A person can only win one prize
• How many different ways are there to
award these prizes?
Practice
8!
 336
(8  3)!
336 ways of awarding the three different prizes
Practice
• There are three prizes (each is worth
$200)
• There are eight finalist in a drawing who
are going to be awarded these prizes.
• A person can only win one prize
• How many different ways are there to
award these prizes?
Combinations
8!
 56
3!(8  3)!
56 different ways to award these prizes
Practice
• A shirt comes in four sizes and six colors.
One also has the choice of a dragon,
alligator, or no emblem on the pocket.
How many different kinds of shirts can you
order?
Practice
• A shirt comes in four sizes and six colors.
One also has the choice of a dragon,
alligator, or no emblem on the pocket.
How many different kinds of shirts can you
order?
• 4*6*3 = 72
• Don’t make it hard on yourself!
Practice
• In a California Governor race there were
135 candidates. The state created ballots
that would list candidates in different
orders. How many different types of
ballots did the state need to create?
Practice
135!

(135 135)!
2.6904727073180495e+230
Or
26,904,727,073,180,495,000,000,000,000,
000,000,000,000,000,000,000,000,000,000
,000,000,000,000,000,000,000,000,000,00
0,000,000,000,000,000,000,000,000,000,0
00,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000
,000,000,000,000,000,000,000,000,000,00
0,000,000,000,000,000,000,000,000,000
Bonus Points
•
Suppose you’re on a game show and you’re given the choice of three
doors. Behind one door is a car; behind the others, goats. The car and the
goats were placed randomly behind the doors before the show. The rules of
the game show are as follows: After you have chosen a door, the door
remains closed for the time being. The game show host, Monty Hall, who
knows what is behind the doors, now has to open one of the two remaining
doors, and the door he opens must have a goat behind it. If both remaining
doors have goats behind them, he chooses one randomly. After Monty Hall
opens a door with a goat, he will ask you to decide whether you want to stay
with your first choice or to switch to the last remaining door. Imagine that
you chose Door 1 and the host opens Door 3, which has a goat. He then
asks you “Do you want to switch to Door Number 2?” Is it to your advantage
to change your choice?
• http://www.nytimes.com/2008/04/08/scienc
e/08monty.html?_r=1
You pick #1
Door 1
Door 2
Door 3
GAME 1
AUTO
GOAT
GOAT
GAME 2
GOAT
AUTO
GOAT
GAME 3
GOAT
GOAT
AUTO
GAME 4
AUTO
GOAT
GOAT
GAME 5
GOAT
AUTO
GOAT
GAME 6
GOAT
GOAT
AUTO
Results
Switch and you
lose.
Switch and you
win.
Switch and you
win.
Stay and you
win.
Stay and you
lose.
Stay and you
lose.
Practice
• The probability of winning “Blingoo” is .30
• What is the probability that you will win 20
of the next 30 games of Blingoo ?
• Note: previous probability methods do not
work for this question
Binomial Distribution
• Used with situations in which each of a
number of independent trials results in one
of two mutually exclusive outcomes
Binomial Distribution
N!
X (NX )
p( X ) 
p q
X !( N  X )!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Game of Chance
• The probability of winning “Blingoo” is .30
• What is the probability that you will win 20
of the next 30 games of Blingoo ?
• Note: previous probability methods do not
work for this question
Binomial Distribution
N!
X (NX )
p( X ) 
p q
X !( N  X )!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
N!
X (NX )
p( X ) 
.30 q
X !( N  X )!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
N!
X
(NX )
p( X ) 
.30 .70
X !( N  X )!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
30!
X
( 30  X )
p( X ) 
.30 .70
X !(30  X )!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
30!
20
( 30  20 )
p( X ) 
.30 .70
20!(30  20)!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
30!
20
( 30  20 )
p( X ) 
.30 .70
20!(30  20)!
p = .000029
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
What does this mean?
• p = .000029
• This is the probability that you would win
EXACTLY 20 out of 30 games of Blingoo
Game of Chance
• Playing perfect black jack – the probability
of winning a hand is .498
• What is the probability that you will win 8
of the next 10 games of blackjack?
Binomial Distribution
N!
X (NX )
p( X ) 
p q
X !( N  X )!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
10!
8
(10 8 )
p( X ) 
.498 .502
8!(10  8)!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
10!
8
(10 8 )
p( X ) 
.498 .502
8!(10  8)!
p = .0429
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Excel
Binomial Distribution
What is this doing?
Its combining together what you have learned so far!
One way to fit our 8 wins would be (joint probability):
W, W, W, W, W, W, W, W, L, L =
(.498)(.498)(.498)(.498)(.498)(.498)(.498)(.498)(.502)(.502)=
(.4988)(.5022)=.00095
pX q(N-X)
Binomial Distribution
N!
X (NX )
p( X ) 
p q
X !( N  X )!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
N!
X (NX )
p( X ) 
p q
X !( N  X )!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
Other ways to fit our question
W, L, L, W, W, W, W, W
L, W, W, W, W, L, W, W
W, W, W, L, W, W, W, L
L, L, W, W, W, W, W, W
W, L, W, L, W, W, W, W
W, W, L, W, W, W, L, W
Binomial Distribution
Other ways to fit our question
W, L, L, W, W, W, W, W = .00095
L, W, W, W, W, L, W, W = .00095
W, W, W, L, W, W, W, L = .00095
L, L, W, W, W, W, W, W = .00095
W, L, W, L, W, W, W, W = .00095
W, W, L, W, W, W, L, W = .00095
Each combination has the same probability – but how many
combinations are there?
Combinations
N!
N
 Cr
r!( N  r )!
Ingredients:
N = total number of events
r = number of events selected
45 different combinations
Binomial Distribution
N!
X (NX )
p( X ) 
p q
X !( N  X )!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
• Any combination would work
• . 00095+ 00095+ 00095+ 00095+ 00095+
00095+ 00095+ 00095+. . . . . . 00095
• Or 45 * . 00095 = .04
Binomial Distribution
N!
X (NX )
p( X ) 
p q
X !( N  X )!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Practice
• You bought a ticket for a fire department
lottery and your brother has bought two
tickets. You just read that 1000 tickets
were sold.
– a) What is the probability you will win the
grand prize?
– b) What is the probability that your brother will
win?
– c) What is the probably that you or your
bother will win?
5.2
• A) 1/1000 = .001
• B)2/1000 = .002
• C) .001 + .002 = .003
Practice
• Assume the same situation at before
except only a total of 10 tickets were sold
and there are two prizes.
– a) Given that you didn’t win the first prize, what is the
probability you will win the second prize?
– b) What is the probability that your borther will win the
first prize and you will win the second prize?
– c) What is the probability that you will win the first
prize and your brother will win the second prize?
– d) What is the probability that the two of you will win
the first and second prizes?
5.3
•
•
•
•
A) 1/9 = .111
B) 2/10 * 1/9 = (.20)*(.111) = .022
C) 1/10 * 2/9 = (.10)*(.22) = .022
D) .022 + .022 = .044
Practice
• In some homes a mother’s behavior
seems to be independent of her baby's,
and vice versa. If the mother looks at her
child a total of 2 hours each day, and the
baby looks at the mother a total of 3 hours
each day, and if they really do behave
independently, what is the probability that
they will look at each other at the same
time?
5.8
• 2/24 = .083
• 3/24 = .125
• .083*.125 = .01
Practice
• Abe ice-cream shot has six different
flavors of ice cream, and you can order
any combination of any number of them
(but only one scoop of each flavor). How
many different ice-cream cone
combinations could they truthfully
advertise (note, we don’t care about the
order of the scoops and an empty cone
doesn’t count).
5.29
6!
6
1!(1  6)!
6!
 15
2!(2  6)!
6!
 20
3!(3  6)!
6!
 15
4!(6  4)!
6!
6
5!(6  5)!
6!
1
6!(6  6)!
6 + 15 + 20 +15 + 6 + 1 = 63
Extra Brownie Points!
• Lottery
• To Win:
• choose the 5 winnings numbers
– from 1 to 49
• AND
• Choose the "Powerball" number
– from 1 to 42
• What is the probability you will win?
– Use combinations to answer this question