“Baseball is 90% mental. The other half
is physical.”
Yogi Berra
Probability
• Denoted by P(Event)
favorable outcomes
P( E ) 
total outcomes
This method for calculating probabilities is only appropriate
when the outcomes of the sample space are equally likely.
Law of Large Numbers
• As the number of repetitions of a
chance experiment increase, the
difference between the relative
frequency of occurrence for an
event and the true probability
approaches zero.
Basic Rules of Probability
Rule 1. Legitimate Values
For any event E,
The probability of
rain must be 110%!
0 < P(E) < 1
A probability
is a number
between 0 and 1.
Rule 2. Sample space
If S is the sample space,
P(S) = 1
“Something Has to Happen Rule”
The probability of the set of all possible outcomes
must be 1.
I’m 100% sure you
are going to have a
boy… or a girl.
Rule 3. Complement
For any event E,
P(E) + P(not E) = 1
P(E) = 1 – P(not E)
If the probability that you get to class
on time is .8, then the probability that
you do not get to class on time is .2.
Independent
• Two events are independent if knowing that
one will occur (or has occurred) does not
change the probability that the other occurs
– A randomly selected student is female - What is
the probability she plays soccer for SHS?
Independent
– A randomly selected student is female - What is
the probability she plays football for SHS?
Dependent
Rule 4. Multiplication
If two events A & B are independent,
P(A & B)  P(A)  P(B)
General rule:
P(A & B)  P(A)  P(B | A)
If the probability of rolling a 5 on a fair dice is 1/6, what is
the probability of rolling a 5 three times in a row?
P( three 5’s in a row) =
(1/6) x (1/6) x (1/6) = 1/216 or .004629
P( A  B) ?
P( A  B)
Yes
P( A)  P( B)
What does
this mean?
Independent?
Given a deck of
cards and a die,
one card is drawn
and the dice is
rolled. What is
the probability
that an ace is
drawn and an even
is rolled?
4 3 1
P(ace and even) = P(ace) * P(even) =

52 6 26
P( A  B) ?
P( A  B)
Yes
P( A)  P( B)
Independent?
No
P( A)  P( B | A)
P( A)  P( B | A)
P(heart and heart) =
13 12 1

52 51 17
Given a deck of
cards, two
cards are
drawn without
replacement.
What is the
probability that
they are both
hearts?
Ex 6) Suppose I will pick two cards from a standard
deck without replacement. What is the probability that
I select two spades?
Are the cards independent? NO
P(A & B) = P(A) · P(B|A)
Read “probability of B
given that A occurs”
P(Spade & Spade) = 1/4 · 12/51 = 1/17
The probability of getting a spade given
that a spade has already been drawn.
Rule 5. Addition
If two events E & F are disjoint,
P(E or F) = P(E) + P(F)
If the probability that a randomly selected
student is a junior (A) is .2 and the probability
that the student is a senior (B) is .5, what is
the probability that the student is either a
junior or a senior?
P(A υ B) = P(A) + P(B), if A and B are disjoint.
P(A υ B) = .2 + .5 = .7
• Two events that have no common outcomes
are said to be disjoint or mutually exclusive.
A and B are disjoint
events
Rule 5. Addition
If two events E & F are disjoint, P(E or F) = P(E) + P(F)
(General) If two events E & F are
not disjoint,
P(E or F) = P(E) + P(F) – P(E & F)
Probability of owning a MP3 player: .50
Probability of owning a computer: .90
So the probability of owning a
MP3 player or a computer is
1.40?
Not disjoint events!
P( E  F ) ?
P( E  F )
Yes
P( E )  P( F )
What does
this mean?
Mutually exclusive?
Given a deck of
cards, one card
is drawn.
What is the
probability that
it is a 3 or a
4?
P(3 or 4) = P(3) + P(4) =
4
4
2


52 52 13
P( E  F ) ?
P( E  F )
Yes
P( E )  P( F )
Mutually exclusive?
No
P( E )  P( F )  P( E  F )
Given a deck of
cards, one card
is drawn.
What is the
probability that
it is an ace or
a red card?
P(ace or red) = P(ace) + P(red) – P(ace and red) =
4
26
+
52
52
2
28
=
52
52
P( E  F ) ?
P( E  F )
Mutually exclusive?
No
Yes
P( E )  P( F )
P( E )  P( F )  P( E  F )
Independent?
Yes
P( E)  P( F )
Ex 5)
If P(A) = 0.45, P(B) = 0.35, and A &
B are independent, find P(A or B).
Is A & B disjoint?
NO, independent events cannot be disjoint
If A &
are –disjoint,
are they
P(A or B) = P(A)
+B
P(B)
P(A & B)
Disjoint
events are
independent?
Disjoint events
doIf not happen at the same
dependent!
How can you
independent,
time.
find the
P(A or B) =So,
.45if +A multiply
.35
.45(.35)
=
0.6425
occurs, can B occur?
probability of
A & B?
In a class, there are 12 boys made up of 8 Seniors and
4 Juniors . There are also 8 girls, made up of 3 Seniors
and 5 Juniors. Find the probability of choosing a boy or
a Senior.
Note that choosing a boy and choosing a Senior are not
disjoint (they can occur simultaneously).
P(boy or a senior) = P(Boy) + P(Senior) – P(Senior boy)
= 12  11  8  15  3
20
20
20
20
4
Boy
Girl
Total
Senior
8
3
11
Junior
4
5
9
Total
12
8
20
Rule 6. At least one
The probability that at least one
outcome happens is 1 minus the
probability that no outcomes
happen.
P(at least 1) = 1 – P(none)
For a sales promotion the manufacturer
places winning symbols under the caps
of 10% of all Dr. Pepper bottles. You
buy a six-pack. What is the probability
that you win something?
P(at least one winning symbol) =
1 – P(no winning symbols)
1 - .96 = .4686
Suppose that 40% of cars in Fort Smith are
manufactured in the United States, 30% in Japan,
10% in Germany, and 20% in other countries.
If cars are selected at random, what is
the probability that it is not US made?
P(not US made) = 1 – P(US made) = 1 - .4 = .6
Suppose that 40% of cars in Fort Smith are
manufactured in the United States, 30% in Japan,
10% in Germany, and 20% in other countries.
If cars are selected at random, what is
the probability that it is made in Japan
or Germany?
P(Japanese or German) = P(Japanese) + P(German)
= .3 + .1 = .4
Suppose that 40% of cars in Fort Smith are
manufactured in the United States, 30% in Japan,
10% in Germany, and 20% in other countries.
If cars are selected at random, what is
the probability that you see two in a row
from Japan?
P(2 Japanese in a row) = P(Japanese) and
P(Japanese) = P(J) x P(J) = .3 x .3 = .09
Suppose that 40% of cars in Fort Smith are
manufactured in the United States, 30% in Japan,
10% in Germany, and 20% in other countries.
If cars are selected at random, what is
the probability that none of three cars
came from Germany?
P(no Germany in three) = P(not G) x P(not G) x
P(not G) = .9 x .9 x .9 = .729
Suppose that 40% of cars in Fort Smith are
manufactured in the United States, 30% in Japan,
10% in Germany, and 20% in other countries.
If cars are selected at random, what is
the probability that at least one of three
cars is US made?
P(at least one US in three) = 1 – P(no US in three)
= 1 – (.6)(.6)(.6) = .784
Suppose that 40% of cars in Fort Smith are
manufactured in the United States, 30% in Japan,
10% in Germany, and 20% in other countries.
If cars are selected at random, what is
the probability that the first Japanese
car is the fourth one you choose?
P(first J is the fourth car) = P(not J) x P(not J) x
P(not J) x P(J) = (.7)3 (.3) = .1029
Watch out for:
• probabilities that don’t add up to 1
• don’t add probabilities of events if
they are not disjoint
• don’t multiply probabilities of events
if they are not independent
• don’t confuse disjoint and independent
Rule 7: Conditional Probability
• A probability that takes into
account a given condition
P(A  B)
P(B | A) 
P(A)
P(and)
P(B | A) 
P(given)
In a class, there are 12 boys made up of 8 Seniors and
4 Juniors . There are also 8 girls, made up of 3 Seniors
and 5 Juniors. Find the probability of choosing a boy
given that he is a Senior.
Boy
Girl
Total
Senior
8
3
11
Junior
4
5
9
Total
12
8
20
P(Boy|Senior) = 8/11
P(Senior|Boy) = 8/12 = 2/3
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
12) What is the probability that the driver is a
student?
195
P (Student ) 
359
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
13) What is the probability that the driver drives a
European car?
45
P (European ) 
359
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
14) What is the probability that the driver drives
an American or Asian car?
212  102
P (American or Asian ) 
359
Disjoint?
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
15) What is the probability that the driver is staff
or drives an Asian car?
164  102  47
P (Staff or Asian ) 
359
Disjoint?
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
16) What is the probability that the driver is staff
and drives an Asian car?
47
P (Staff and Asian ) 
359
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
17) If the driver is a student, what is the
probability that they drive an American car?
107
P (American |Student ) 
195
Condition
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
18) What is the probability that the driver is a
student if the driver drives a European car?
33
P (Student |European ) 
45
Condition
Definition of Independent Events
Two events E and F are independent
if and only if
P(F | E) = P(F) or P(E | F) = P(E)
EXAMPLE
Illustrating Independent Events
The probability a randomly selected murder victim is
male is 0.7515. The probability a randomly selected
murder victim is male given that they are less than
18 years old is 0.6751.
Since P(male) = 0.7515 and
P(male | < 18 years old) = 0.6751,
the events “male” and “less than 18 years old” are
not independent. In fact, knowing the victim is less
than 18 years old decreases the probability that the
victim is male.
I draw one
card and look
at it. I tell
you it is red.
What is the
probability it is
a heart?
P( heart | red) =
13
P(heart and red) 52 1


26 2
P(red)
52
Are “red card”
and “spade”
mutually
exclusive? Are
they
independent?
A red card
can’t be a
spade so they
ARE mutually
exclusive
Are “red card”
and “ace”
mutually
exclusive? Are
they
independent?
2 aces are
red cards so
they are NOT
mutually
exclusive
Are “face card”
and “king”
mutually
exclusive? Are
they
independent?
Kings are
Face cards so
they are NOT
mutually
exclusive
“Slump? I ain’t in no slump. I
just ain’t hittin.”
Yogi Berra