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Introduction to Probability
and Statistics
Twelfth Edition
Chapter 4
Probability and Probability
Distributions
Some graphic screen captures from Seeing Statistics ®
Some images © 2001-(current year) www.arttoday.com
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.
Example
• Toss a fair coin twice. What is the
probability of observing at least one
head (Event A)?
S={HH, HT, TH, TT}
A={HH, HT, TH}
1st Coin
2nd Coin
Ei
P(Ei)
H
HH
1/4
T
HT
H
TH
1/4 = P(A)
1/4 = P(HH) + P(HT) + P(TH)
H
T
T
TT
P(at least 1 head)
1/4 = 1/4 + 1/4 + 1/4 = 3/4
#A
P ( A) 
Copyright ©2006 Brooks/Cole
#S
A division of Thomson Learning, Inc.
Example
• A bowl contains three M&Ms®, two reds, one blue.
A child selects two M&Ms at random. What is the
probability that exactly two reds (Event A)?
r1
1st M&M
r1
r2
2nd M&M
b
b
r1b
P(Ei)
1/6
r2
r1r2
1/6
b
r2b
r1
r2r1
1/6 = P(r1r2) + P(r2r1)
1/6 = 1/6 +1/6 = 1/3
br1
1/6
br2
1/6
r1
b
r2
Ei
r2
A={r1r2, r2r1}
P(A)
#A
P
(
A
)

Copyright ©2006 Brooks/Cole
# S Inc.
A division of Thomson Learning,
Counting Rules
• If the simple events in an experiment are
equally likely, we can calculate
# A number of simple events in A
P( A) 

# S total number of simple events
• We can use counting rules to find #A
and #S.
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.
Counting
• How many ways from A to C?
32=6
• How many ways from A to D?
3  2  2 = 12
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.
The mn Rule
• If an experiment is performed in two stages,
with m ways to accomplish the first stage and
n ways to accomplish the second stage, then
there are mn ways to accomplish the
experiment.
• This rule is easily extended to k stages, with
the number of ways equal to
n1 n2 n3 … nk
Example: Toss two coins. The total number of
simple events is:
22=4
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.
m
Examples
m
Example: Toss three coins. The total number of
simple events is:
222=8
Example: Toss two dice. The total number of
simple events is:
6  6 = 36
Example: Two M&Ms are drawn from a dish
containing two red and two blue candies. The total
number of simple events is:
4  3 = 12
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.
Permutations
Example: How many 3-digit lock combinations
can we make by using 3 different numbers
among 1, 2, 3, 4 and 5?
5(4)(3)  60
The order of the choice is
important!
The number of ways you can arrange n
distinct objects, taking them r at a time is
n!
n
Pr 
(n  r )!
where n! n(n  1)( n  2)...( 2)(1) and 0! 1.
5!
5(4)(3)( 2)(1)
P 

 60
©2006 Brooks/Cole
(5  3)! Copyright
2(1)
A division of Thomson Learning, Inc.
5
3
Example
Example: A lock consists of five parts and
can be assembled in any order. A quality
control engineer wants to test each order for
efficiency of assembly. How many orders are
there?
The order of the choice is
important!
5!
P   5(4)(3)( 2)(1)  120
0!
5
5
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.
Example
• How many ways to select a student
committee of 3 members: chair, vice chair,
and secretary out of 8 students?
8!
P 
(8  3)!
(8)(7)(6)(5)( 4)(3)( 2)(1)

5(4)(3)( 2)(1)
 8(7)(6)  336
8
3
The order of the choice is
important! ---- Permutation
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.
Example
• How many ways to select a student
committee of 3 members out of 8 students?
• (Don’t assign chair, vice chair and
secretary).
The order of the choice is
NOT important! 
Combination
C
8
3
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.
Combinations
• The number of distinct combinations of n
distinct objects that can be formed,
taking them r at a time is n
n!
Cr 
r!(n  r )!
Example: Three members of a 5-person committee must
be chosen to form a subcommittee. How many different
subcommittees could be formed?
The order of
the choice is
not important!
5!
5(4)(3)( 2)1 5(4)
C 


 10
3!(5  3)! 3(2)(1)( 2)1 (2)1
5
3
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.
Example
• How many ways to select a student
committee of 3 members out of 8 students?
• (Don’t assign chair, vice chair and
secretary).
8!
8
C3 
The order of the choice is
NOT important! 
Combination
3!(8  3)!
8(7)( 6)(5)( 4)(3)( 2)(1)

3( 2)(1)5(4)(3)( 2)(1)
8(7)( 6)

 56
3( 2)(1)
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.
Question
• A box contains six M&Ms®, 4 reds
and 3 blues. A child selects three M&Ms at
random.
• What is the probability that exactly one is red
(Event A) ?
r1
r2
r3
r4
b1
b2
b3
• Simple Events and sample space S:
{r1r2r3, r1r2b1, r2b1b2…... }
• Simple events in event A:
{r1b1b2, r1b2b3, r2b1b2……}
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.
Solution
• Choose 3 MMs out of 7. (Total number of
ways, i.e. size of sample space S)
The order of
the choice is
not important!
7! 7(6)(5)
C 

 35
3!4! 3(2)(1)
7
3
• Event A: one red, two blues
Choose
one red
Choose
Two
Blues
4!
C 
4
4  3 = 12 ways to
1!3!
4
1
3!
C 
3
2!1!
3
2
#A
P( A) 
#S
12

35
choose 1 red and 2
greens ( mn Rule)
Copyright ©2006 Brooks/Cole
A division of Thomson Learning, Inc.