STA 2023
Chapter 3 - Probability

Events, Sample Spaces, and Probability (3.1)
o Experiment – method of observation that leads to an outcome
o Sample Point – possibility of an experiment
o Sample Space – all possibilities of an experiment
 S represents sample space
o Venn Diagram – graphical display that contains all possible outcomes
o Example – One suit of a deck of cards
 S = {2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A}
 P(S) = 1  read “the probability that S occurs is 1”
1
 What is the probability that we draw a 3? P(3) =
13
3
 What is the probability that we draw a face card? P(F) =
13
 What is the probability that we do not draw a 3 or 4? P(Not 3 or 4) =
2
11
1
=
13
13
o Properties of a Sample Point
 0  P(sample point)  1 (probability of an outcome is between 0 and 1)
 P(S) =  P(sample point) = 1 (probability of the sample space is 1)
o Example – Exercise 3.2 – page 117
a. What is this type of diagram called? Venn Diagram
b. Suppose the sample points are equally likely. Find P(A) and P(B). P(A)
= P(4) + P(5) + P(6) = .10 + .10 + .10 = .30, while P(B) = P(6) + P(7) =
.10 + .10 = .20.
c. Suppose P(1) = P(2) = P(3) = P(4) = P(5) = .05, and P(6) = P(7) = P(8) =
P(9) = P(10) = .15. Find P(A) and P(B). P(A) = P(4) + P(5) + P(6) = .05
+ .05 + .15 = .25, while P(B) = P(6) + P(7) = .15 + .15 = .30.

Unions and Intersections (3.2)
o Union – (AB) occurs if A occurs, B occurs, or both occur
 (AB) is read “A union B” or “A or B”
o Intersection – (AB) occurs if A and B both occur
 (AB) is read “A intersect B” or “A and B”
o Example – Standard deck of playing cards – draw one card
 What is the probability that we draw a card that is red or black? P(RB)
= P(card is red or black or both) = 1
 What is the probability that we draw a card that is red and black? P(RB)
= P(card is red and black) = 0 (no cards are red AND black)
 What is the probability that we draw a card that is red and a king?
2
P(RK) = P(card is a red king) =
52
 What is the probability that we draw a card that is a diamond and a 9?
1
P(D 9) = P(card is a nine of diamonds) =
52
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STA 2023
Chapter 3 - Probability

What is the probability that we draw a card that is a diamond or a 9?
16
P(D 9) = P(card is a nine or a diamond or both) =
52

Complementary Events (3.3)
o Complement – AC occurs if A does not occur
 AC is read “A complement”
o Properties of Complementary Events
 P(A) + P(AC) = 1
 P(A) = 1 – P(AC) or P(AC) = 1 – P(A)
o Example – Standard Deck of Cards
26
 Find P(RC). P(RC) =
52
39
 Find P(DC). P(DC) =
52
48
 Find P(4C). P(4C) =
52
 Find P[(45) C]. P[(45) C] = 1
46
 Find P[(FR) C]. P[(FR) C] =
52
51
 Find P[(4S) C]. P[(4S) C] =
52

The Additive Rule and Mutually Exclusive Events (3.4)
o Additive Rule of Probability – useful for solving “or” probabilities
 P(AB) = P(A) + P(B) – P(AB)
o Mutually Exclusive – A and B are mutually exclusive if P(AB) = 0
o Probability of Union of Two Mutually Exclusive Events
 P(AB) = P(A) + P(B)
o Example – Offensive Efficiency of a Football Team on a drive, given that P(FG)
= .35 and P(TD) = .20.
 What is the probability that a team scores on an offensive possession?
P(Scores) = P(FG TD) = P(FG) + P(TD) – P(FG TD) = .35 + .20 – 0
= .55

Conditional Probability (3.5)
o Conditional Rule of Probability
P( A  B)
 P(A|B) =
P( B)
 P(A|B) is read “conditional probability that even A will occur given that
event B has already occurred”
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STA 2023
Chapter 3 - Probability
o Example – Kobe Bryant’s field goal shooting statistics
 Given that Kobe Bryant has scored a field goal, what is the probability that
it was a three-pointer? Assume P(FG) = .452, P(2FG) = .406, and
.046
P(3  FG)
P(3FG) = .046. P(3|FG) =
=
= .102
.452
P( FG)

The Multiplicative Rule and Independent Events (3.6)
o Multiplicative Rule of Probability – useful for solving “and” probabilities
 P(AB) = P(A) P(B|A) or P(AB) = P(B) P(A|B)
o Tree Diagram
o Independent Events
 If events A and B are independent then P(AB) = P(A) P(B)
 The converse is also true
 Mutually exclusive events are dependent events (converse not necessarily
true)
 Looking only at a Venn diagram, we can determine whether or not two
events are mutually exclusive, however, we cannot determine whether or
not two events are independent
o Example – Standard Deck of Cards
13
 Are the events of drawing a heart and a 4 independent? P(H) =
, P(4)
52
4
1
= , so P(H)*P(4) = = P(H 4). Yes, they are independent.
52
52
 Are the events of drawing a heart and a 4 mutually exclusive? Since
1
P(H 4) =  0, so they are not mutually exclusive.
52
13
 Are the events of drawing a spade and diamond independent? P(S) =
,
52
13
1
P(D) = , so P(S)*P(D) =  0 = P(S D), so they are not
16
52
independent.
 Are the events of drawing a spade and diamond independent? Since P(S
D) = 0, the events are mutually exclusive. Furthermore, since the events
are mutually exclusive, we know that the events are also dependent, which
agrees with our previous answer.
o Example – Independent Events
 Suppose that the living members of Pink Floyd (Roger, David, Richard,
and Nick) decide to do a reunion tour, and because of their highly volatile
personalities, the probability that each member quits the band after any
show is as follows: P(Roger quits) = .50, P(David quits) = .40, P(Richard
quits) = .20, and P(Nick quits) = .10. Assuming that the band members
act independently of one another, find the probability that all members of
the band quit immediately after the next show. We are trying to find
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STA 2023
Chapter 3 - Probability
P(Roger quits  David quits  Richard quits  Nick quits). Since they
are independent this is equivalent to P(Roger quits)*P(David
quits)*P(Richard quits)*P(Nick quits) = (.50)(.40)(.20)(.10) = .004.

Random Sampling (3.7)
o Random Sample – when all elements of a population have equal probability of
being selected
o Factorial – n! = n(n-1)…1
 0! = 1
 n! is read “n factorial”
o Example – Using factorials
 Calculate 6!. 6! = 6*5*4*3*2*1 = 720
 Calculate 10!. 10! = 10*9*8*7*6! = 10*9*8*7*720 = 3,628,800
o Combination – how many ways we can select n objects from a set of N (N > n)
N
N!
   =
 n  n!( N  n)!

Alternative Notation:
N Cn
or C nN
N
  is read “N choose n”
n
o Example – Basic Combinations
 How many ways can we choose two marbles from a set of five? Here,
 N  5
5!
5!
N=5 and n=2, so   =   =
=
= 10.
 n   2  2!(5  2)! 2!*3!
 How many possible Florida lottery combinations are there? Here, N=53
 N   53 
53!
53!
and n=6, so   =   =
=
= 22,957,480.
 n   6  6!(53  6)! 6!*47!


Some Counting Rules (3.8)
o Permutation – how many ways can we select and order n objects from a set of N
N!
 PnN =
( N  n)!
N
 Pn is read “N permute n”
o Example – Executive Board for a campus organization
 A campus organization consisting of 12 members wishes to select 4
members for the positions of President, Vice President, Secretary, and
Treasurer. In how many possible ways can this be accomplished? Here,
12!
12!
we have N=12 and n=4, so P412 =
=
= 11,880.
(12  4)! 8!
o Multiplicative Rule – how many ways we can select one element from each of k
sets where ni objects are in set i
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STA 2023
Chapter 3 - Probability
 Number of ways = n1*n2*…*nk
o Example – Eating
 Suppose you are out to a nice dinner and you are to choose one appetizer
(of 3 possible), one main course (of 5 possible), one drink (of 4
possible), and one dessert (of 3 possible). How many possible
combinations can be formed? Using the multiplicative rule, the number
of combinations that can be formed is 3*5*4*3 = 180.
o Partitions Rule – number of ways to sort all elements of a set of N objects into k
groups, where ni objects are in set I
N!
 Number of ways =
where n1 + n2 + … + nk = N
n1!*n2 !*... * nk !
o Example – Offensive Football Team Positions
 Suppose I have a team of 15 players in which I need to choose 11 of
them to start on offense and further choose 1 quarterback, 2 running
backs, 3 wide receivers, and 5 linemen. How many different starting
15!
lineups are possible? Using the partitions rule, we have
=
1!*2!*3!*5!*4!
37,837,800 possible lineups.
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