Section 1.3
The Multiplication Principle states that if experiment E1 has n1 possible
outcomes, and for each possible outcome of E1 experiment E2 has n2
possible outcomes, then the composite experiment E1E2 (performing E1
first and then E2) has n1n2 possible outcomes.
Each arrangement (ordering) of n distinguishable objects is called a
permutation, and the number of permutations of n distinguishable
objects is
(n)(n – 1)(n – 2)…(3)(2)(1) = n!
Each ordered arrangement of r out of n distinguishable objects is called
a permutation of n objects taken r at a time, and the number of possible
permutations of n objects taken r at a time is
When sampling without replacement, the number of possible ordered
samples of r objects taken from a set of n objects is
1. One box contains 4 red books and 3 blue books. A second box
contains 9 green books.
(a) Find the number of possible arrangements of all 9 green books on
a shelf, if these books can be distinguished from one another.
9! = 362880 arrangements
(b) Find the number of possible arrangements of 4 of the 9 green
books on a shelf, if these books can be distinguished from one
another.
(c) Suppose books of the same color can be distinguished from one
another. If books are selected at random and without replacement,
find the number of possible ordered samples of 2 books that could
be taken from the box of 4 red books and 3 blue books. Are these
samples equally likely?
Section 1.3
The Multiplication Principle states that if experiment E1 has n1 possible
outcomes, and for each possible outcome of E1 experiment E2 has n2
possible outcomes, then the composite experiment E1E2 (performing E1
first and then E2) has n1n2 possible outcomes.
Each arrangement (ordering) of n distinguishable objects is called a
permutation, and the number of permutations of n distinguishable
objects is
(n)(n – 1)(n – 2)…(3)(2)(1) = n!
Each ordered arrangement of r out of n distinguishable objects is called
a permutation of n objects taken r at a time, and the number of possible
permutations of n objects taken r at a time is
n!
(n)(n – 1)(n – 2)…(n – r + 1) = ——— = nPr
(n – r)!
When sampling without replacement, the number of possible ordered
samples of r objects taken from a set of n objects is
1. One box contains 4 red books and 3 blue books. A second box
contains 9 green books.
(a) Find the number of possible arrangements of all 9 green books on
a shelf, if these books can be distinguished from one another.
9! = 362880 arrangements
(b) Find the number of possible arrangements of 4 of the 9 green
books on a shelf, if these books can be distinguished from one
another.
9!
9P4 = ——— = 3024 arrangements
(9 – 4)!
(c) Suppose books of the same color can be distinguished from one
another. If books are selected at random and without replacement,
find the number of possible ordered samples of 2 books that could
be taken from the box of 4 red books and 3 blue books. Are these
samples equally likely?
Section 1.3
The Multiplication Principle states that if experiment E1 has n1 possible
outcomes, and for each possible outcome of E1 experiment E2 has n2
possible outcomes, then the composite experiment E1E2 (performing E1
first and then E2) has n1n2 possible outcomes.
Each arrangement (ordering) of n distinguishable objects is called a
permutation, and the number of permutations of n distinguishable
objects is
(n)(n – 1)(n – 2)…(3)(2)(1) = n!
Each ordered arrangement of r out of n distinguishable objects is called
a permutation of n objects taken r at a time, and the number of possible
permutations of n objects taken r at a time is
n!
(n)(n – 1)(n – 2)…(n – r + 1) = ——— = nPr
(n – r)!
When sampling without replacement, the number of possible ordered
samples of r objects taken from a set of n objects is
n!
(n)(n – 1)(n – 2)…(n – r + 1) = ——— = nPr
(n – r)!
1. One box contains 4 red books and 3 blue books. A second box
contains 9 green books.
(a) Find the number of possible arrangements of all 9 green books on
a shelf, if these books can be distinguished from one another.
9! = 362880 arrangements
(b) Find the number of possible arrangements of 4 of the 9 green
books on a shelf, if these books can be distinguished from one
another.
9!
9P4 = ——— = 3024 arrangements
(9 – 4)!
(c) Suppose books of the same color can be distinguished from one
another. If books are selected at random and without replacement,
find the number of possible ordered samples of 2 books that could
be taken from the box of 4 red books and 3 blue books. Are these
7!
samples equally likely?
7P2 = ——— = 42 samples
(7 – 2)!
These samples are equally likely.
When sampling with replacement, the number of possible ordered
samples of r objects taken from a set of n objects is nr
When sampling without replacement, the number of possible unordered
samples of r objects taken from a set of n objects is
Each unordered subset of r out of n distinguishable objects is called a
combination of n objects taken r at a time, and the number of possible
combinations of n objects taken r at a time is
The number of distinguishable permutations of n objects, where r objects
are of one type and n – r objects are of another type, is
The number of distinguishable permutations of n objects, where r1
objects are of type 1, r2 objects are of type 2, …, and rk objects are of
type k (See Text Exercise 1.3-18.), is
(d) Suppose books of the same color can be distinguished from one
another. If books are selected at random and with replacement,
find the number of possible ordered samples of 2 books that
could be taken from the box of 4 red books and 3 blue books.
72 = 49 samples
Are these samples equally likely?
These samples are equally likely.
(e) Suppose books of the same color can be distinguished from one
another. If books are selected at random and without replacement,
find the number of possible unordered samples of 2 books that
could be taken from the box of 4 red books and 3 blue books. Are
these samples equally likely?
7!
———— = 21 samples These samples are equally likely.
2! (7 – 2)!
(f) Suppose books of the same color can be distinguished from one
another. Find the number of possible combinations of the 9 green
books taken 3 at a time. Are these samples equally likely?
When sampling with replacement, the number of possible ordered
samples of r objects taken from a set of n objects is nr
When sampling without replacement, the number of possible unordered
samples of r objects taken from a set of n objects is
(number of ordered samples of r objects)
n!
—————————————————————– = ———— = C
(number of permutations of r distinguishable objects ) r! (n – r)! n r
Each unordered subset of r out of n distinguishable objects is called a
combination of n objects taken r at a time, and the number of possible
n!
combinations of n objects taken r at a time is ————
= C
r! (n – r)! n r
The number of distinguishable permutations of n objects, where r objects
are of one type and n – r objects are of another type, is
The number of distinguishable permutations of n objects, where r1
objects are of type 1, r2 objects are of type 2, …, and rk objects are of
type k (See Text Exercise 1.3-18.), is
(d) Suppose books of the same color can be distinguished from one
another. If books are selected at random and with replacement,
find the number of possible ordered samples of 2 books that
could be taken from the box of 4 red books and 3 blue books.
72 = 49 samples
Are these samples equally likely?
These samples are equally likely.
(e) Suppose books of the same color can be distinguished from one
another. If books are selected at random and without replacement,
find the number of possible unordered samples of 2 books that
could be taken from the box of 4 red books and 3 blue books. Are
these samples equally likely?
7!
7C2 = ———— = 21 samples These samples are equally likely.
2! (7 – 2)!
(f) Suppose books of the same color can be distinguished from one
another. Find the number of possible combinations of the 9 green
books taken 3 at a time. Are these samples equally likely?
9!
9C3 = ———— = 84 samples
3! (9 – 3)!
These samples are equally likely.
(g) Find the number of possible arrangements of the 4 red books and 3
blue books on a shelf, if books of the same color can be
distinguished from one another.
7! = 5040 arrangements
(h) Find the number of possible arrangements of the 4 red books and 3
blue books on a shelf, if books of the same color cannot be
distinguished from one another.
7!
—— = 35 arrangements
3! 4!
(i) Find the number of possible arrangements of the 9 green books, 4
red books, and 3 blue books on a shelf, if books of the same color
cannot be distinguished from one another. Do this two different
ways by applying the multiplication principle corresponding to
each of the following sets of instructions for arranging the books:
Instruction Set #1
Choose the 9 positions for green books.
Choose the 4 positions for red books.
Choose the 3 positions for blue books.
The number of possible arrangements is
16!
7!
3!
.
.
7C4
3C3 = ————— ———— ———— =
16C9
9! (16  9)! 4! (7  4)! 3! (3  3)!
16!
——— = 400400
9! 4! 3!
(i) Find the number of possible arrangements of the 9 green books, 4
red books, and 3 blue books on a shelf, if books of the same color
cannot be distinguished from one another. Do this two different
ways by applying the multiplication principle corresponding to
each of the following sets of instructions for arranging the books:
Instruction Set #2
Choose the 3 positions for blue books.
Choose the 4 positions for red books.
Choose the 9 positions for green books.
The number of possible arrangements is
16!
13!
9!
.
.
13C4
9C9 = ————— ————— ———— =
16C3
3! (16  3)! 4! (13  4)! 9! (9  9)!
16!
——— = 400400
3! 4! 9!
When sampling with replacement, the number of possible ordered
samples of r objects taken from a set of n objects is nr
When sampling without replacement, the number of possible unordered
samples of r objects taken from a set of n objects is
(number of ordered samples of r objects)
n!
—————————————————————– = ———— = C
(number of permutations of r distinguishable objects ) r! (n – r)! n r
Each unordered subset of r out of n distinguishable objects is called a
combination of n objects taken r at a time, and the number of possible
n!
combinations of n objects taken r at a time is ————
= C
r! (n – r)! n r
The number of distinguishable permutations of n objects, where r objects
n!
are of one type and n – r objects are of another type, is
————
r! (n – r)!
The number of distinguishable permutations of n objects, where r1
objects are of type 1, r2 objects are of type 2, …, and rk objects are of
n!
type k (See Text Exercise 1.3-18.), is
—————
r1! r2! … rk!
Number of Samples of Size r out of n Objects
With
Replacement
Ordered
Unordered
nr
Without
Replacement
(j) Suppose books of the same color can be distinguished from one
another. If books are selected at random and with replacement,
find the number of possible ordered samples of 5 books that could
be taken from all 16 books. Are these samples equally likely?
165 = 1048576 samples
These samples are equally likely.
(k) Suppose books of the same color can be distinguished from one
another. If books are selected at random and without replacement,
find the number of possible ordered samples of 5 books that could
be taken from all 16 books. Are these samples equally likely?
16!
16P5 = ———— = 524160 samples These samples are equally likely.
(16 – 5)!
(l) Suppose books of the same color can be distinguished from one
another. If books are selected at random and without replacement,
find the number of possible unordered samples of 5 books that
could be taken from all 16 books. Are these samples equally
likely?
Number of Samples of Size r out of n Objects
With
Replacement
Ordered
Unordered
nr
Without
Replacement
n!
——— = nPr
(n – r)!
n!
———— = nCr =
r! (n – r)!
n
r
(j) Suppose books of the same color can be distinguished from one
another. If books are selected at random and with replacement,
find the number of possible ordered samples of 5 books that could
be taken from all 16 books. Are these samples equally likely?
165 = 1048576 samples
These samples are equally likely.
(k) Suppose books of the same color can be distinguished from one
another. If books are selected at random and without replacement,
find the number of possible ordered samples of 5 books that could
be taken from all 16 books. Are these samples equally likely?
16!
16P5 = ———— = 524160 samples These samples are equally likely.
(16 – 5)!
(l) Suppose books of the same color can be distinguished from one
another. If books are selected at random and without replacement,
find the number of possible unordered samples of 5 books that
could be taken from all 16 books. Are these samples equally
likely?
(l) Suppose books of the same color can be distinguished from one
another. If books are selected at random and without replacement,
find the number of possible unordered samples of 5 books that
could be taken from all 16 books. Are these samples equally
likely?
16C5
=
16
5
=
16!
————— =
5! (16 – 5)!
4368 samples
These samples are equally likely.
(m) Suppose books of the same color can be distinguished from one
another. If 5 books are selected at random and with replacement,
find the probability that the first 3 books selected are green, the 4th
book selected is red, and the 5th book selected is blue.
165 = 1048576 equally likely samples of any 5 books
(93)(4)(3) = 8748 samples with the first 3 books selected green,
the 4th book selected red, and the 5th book selected blue
(93)(4)(3)
165
= 0.01152
(NOTE: When the calculations are difficult, it is generally
acceptable to leave the answer is a “ready-to-calculate” format.)
(n) Suppose books of the same color can be distinguished from one
another. If 5 books are selected at random and without
replacement, find the probability that the first 3 books selected are
green, the 4th book selected is red, and the 5th book selected is
blue.
(n) Suppose books of the same color can be distinguished from one
another. If 5 books are selected at random and without
replacement, find the probability that the first 3 books selected are
green, the 4th book selected is red, and the 5th book selected is
blue. 16!
16P5 = ———— = (16)(15)(14)(13)(12) = 524160 equally likely
(16 – 5)!
samples of any 5
books
9!
9P3 = ————(4)(3) = (9)(8)(7)(4)(3) =
(9 – 3)!
6048 samples with the first 3 books selected green,
the 4th book selected red, and the 5th book
selected blue
(9)(8)(7)(4)(3)
= 0.01154
(16)(15)(14)(13)(12)
(NOTE: When the calculations are difficult, it is generally
acceptable to leave the answer is a “ready-to-calculate” format.)
(o) Suppose books of the same color can be distinguished from one
another. If 5 books are selected at random and without
replacement, find the probability that 3 books are green, 1 book is
red, and 1 book is blue.
16!
16C5 = ————— = (16)(14)(13) = 4368 equally likely samples of
5! (16 – 5)!
any 5 books
9!
9C3 = ————(4)(3) = 1008 samples with the first 3 books selected
3! (9 – 3)!
green, the 4th book selected red, and the 5th
book selected blue
9
4
3
3
1
1
= 0.23077
16
9
(NOTE: When the calculations are difficult,
it is generally acceptable to leave the
answer is a “ready-to-calculate” format.)
(p) Suppose books of the same color can be distinguished from one
another. If books are selected at random and with replacement,
find the number of possible unordered samples of 5 books that
could be taken from all 16 books. Are these samples equally
likely?
If we label the books b1, b2, b3, …, b16, then we can observe that
the unordered sample {b1, b1, b1, b1, b1} can occur in only one
way, whereas the unordered sample {b1, b1, b1, b2, b2} can occur
in 10 ways. Therefore, these samples are not equally likely.
To count the number of possible samples, we can represent each
possible sample with a sequence of 15 slashes (/) and 5 zeros (0).
The number of times b1 occurs in the sample is designated by the
number of zeros to the left of the first slash, the number of times
b2 occurs in the sample is designated by the number of zeros to
the left of the second slash and to the right of the first slash, etc.
00000 / / / / / / / / / / / / / / / represents the sample {b1, b1, b1, b1, b1}.
000 / 00 / / / / / / / / / / / / / / represents the sample {b1, b1, b1, b2, b2}.
/ / 0 / / 00 / / / / 00 / / / / / / / represents the sample {b3, b5, b5, b9, b9}.
(p) Suppose books of the same color can be distinguished from one
another. If books are selected at random and with replacement,
find the number of possible unordered samples of 5 books that
could be taken from all 16 books. Are these samples equally
likely?
The number of possible samples is
20C5
=
20
5
=
20!
————— = 15504 samples
5! (20 – 5)!
These samples are not equally likely.
Number of Samples of Size r out of n Objects
With
Replacement
Ordered
Unordered
nr
(n + r – 1)!
————
r! (n – 1)!
Without
Replacement
n!
——— = nPr
(n – r)!
n!
———— = nCr =
r! (n – r)!
n
r
2. An urn contains 7 red chips, 8 blue chips, and 9 white chips. Find
the probability of
(a) selecting a red chip on the first draw, blue chips on the second,
third, and fourth draws, and white chips on the remaining draws
if eight chips are selected at random and without replacement.
7! 8! 9!
— — —
6! 5! 5!
(7)(8)(7)(6)(9)(8)(7)(6)
or
24!
(24)(23)(22)(21)(20)(19)(18)(17)
—
16!
(b) selecting a red chip on the first draw, blue chips on the second,
third, and fourth draws, and white chips on the remaining draws
if eight chips are selected at random and with replacement.
(7)(83)(94)
248
(c) selecting one red chip, three blue chips, and four white chips if
eight chips are selected at random and without replacement.
7
8
9
1
3
4
8!
or
1! 3! 4!
24
8
7! 8! 9!
— — —
6! 5! 5!
24!
—
16!
(d) selecting one red chip, three blue chips, and four white chips if
eight chips are selected at random and with replacement.
8!
(7)(83)(94)
1! 3! 4!
248
(e) selecting all white chips if five chips are selected at random and
without replacement.
9
9!
—
5
4!
(9)(8)(7)(6)(5)
or
or
24!
(24)(23)(22)(21)(20)
24
—
19!
5
(f) selecting all white chips if five chips are selected at random and
with replacement.
95
245
(g) exactly four white chips if seven chips are selected at random and
without replacement.
9
15
4
3
24
7
(h) exactly four white chips if seven chips are selected at random and
with replacement.
7!
(94)(153)
3! 4!
247
(i) selecting no red chips if three chips are selected at random and
without replacement.
17
3
24
3
(j) selecting no red chips if three chips are selected at random and
with replacement.
173
243
(k) selecting at least one blue chip if five chips are selected at
random and without replacement.
16
5
1 –
24
5
(l) selecting at least one blue chip if five chips are selected at
random and with replacement.
1
–
165
245
(m) selecting exactly two red chips if six chips are selected at random
and without replacement.
7
17
2
4
24
6
(n) selecting exactly two red chips if six chips are selected at random
and with replacement.
6!
(72)(174)
2! 4!
246
(o) selecting more than 20 chips in order to get the fifth blue chip if
chips are selected at random and without replacement.
8
16
4
16
24
20
(p) selecting more than 20 chips in order to get the fifth blue chip if
chips are selected at random and with replacement.
1620
+
20!
1! 19!
(8)(1619)
+
20!
(82)(1618)
2! 18!
20!
+
3! 17!
2420
(83)(1617)
+
20!
4! 16!
(84)(1616)
(q) selecting a white chip on the third try if chips are selected at
random and without replacement.
9
3
—=—
24 8
(r) selecting a white chip on the third try if chips are selected at
random and with replacement.
9
3
—=—
24 8
An urn contains one red chip labeled with the integer 1, two blue
chips labeled distinctively with the integers 1 and 2, three white
chips labeled distinctively with the integers 1, 2, and 3.
(a) Two chips are randomly selected without replacement and the
random variable X = "sum of the observed integers" is recorded.
Find each of the following:
1
2
3
P(X < 4) = P(X = 2) + P(X = 3) = — + — = —
5
5
5
1
4
P(X > 2) = 1 – P(X = 2) = 1 – — = —
5
5
(b) Two chips are randomly selected with replacement and the
random variable X = "sum of the observed integers" is recorded.
Find each of the following:
1
1
7
P(X < 4) = P(X = 2) + P(X = 3) = — + — = —
4
3
12
1
3
P(X > 2) = 1 – P(X = 2) = 1 – — = —
4
4
3.