section 1.3 each arrangement (ordering) of n distinguishable objects is called a permutation, and...

Section 1.3

Each arrangement (ordering) of n distinguishable objects is called a permutation, and the number of permutations of n distinguishable objects is

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

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 possible outcomes.n1n2

(n)(n – 1)(n – 2)…(3)(2)(1) = n!

1.

(a)

(b)

(c)

One box contains 4 red books and 3 blue books. A second box contains 9 green books.

Find the number of possible arrangements of all 9 green books on a shelf, if these books can be distinguished from one another.

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! = 362880 arrangements

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

Each arrangement (ordering) of n distinguishable objects is called a permutation, and the number of permutations of n distinguishable objects is

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 – 1)(n – 2)…(3)(2)(1) = n!

n!(n)(n – 1)(n – 2)…(n – r + 1) = ———

(n – r)!When sampling without replacement, the number of possible ordered samples of r objects taken from a set of n objects is

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 possible outcomes.n1n2

= nPr

1.

(a)

(b)

(c)

One box contains 4 red books and 3 blue books. A second box contains 9 green books.

Find the number of possible arrangements of all 9 green books on a shelf, if these books can be distinguished from one another.

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! = 362880 arrangements

9!9P4 = ——— = 3024 arrangements

(9 – 4)!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

Each arrangement (ordering) of n distinguishable objects is called a permutation, and the number of permutations of n distinguishable objects is

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 – 1)(n – 2)…(3)(2)(1) = n!

n!(n)(n – 1)(n – 2)…(n – r + 1) = ———

(n – r)!When sampling without replacement, the number of possible ordered samples of r objects taken from a set of n objects is

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 possible outcomes.

n!(n)(n – 1)(n – 2)…(n – r + 1) = ———

(n – r)!

n1n2

= nPr

= nPr

1.

(a)

(b)

(c)

One box contains 4 red books and 3 blue books. A second box contains 9 green books.

Find the number of possible arrangements of all 9 green books on a shelf, if these books can be distinguished from one another.

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! = 362880 arrangements

9!9P4 = ——— = 3024 arrangements

(9 – 4)!

7!7P2 = ——— = 42 samples

(7 – 2)!These samples are equally likely.

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?

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)

(e)

(f)

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. Are these samples equally likely?

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?

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?

72 = 49 samplesThese samples are equally likely.

7!———— = 21 samples2! (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

n!————r! (n – r)!

(number of ordered samples of r objects)—————————————————————– =(number of permutations of r distinguishable objects ) = nCr

n!————r! (n – r)! = nCr

(d)

(e)

(f)

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. Are these samples equally likely?

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?

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?

72 = 49 samplesThese samples are equally likely.

7!7C2 = ———— = 21 samples

2! (7 – 2)!These samples are equally likely.

9!9C3 = ———— = 84 samples

3! (9 – 3)!These samples are equally likely.

(g)

(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 can be distinguished from one another.

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! = 5040 arrangements

7!—— = 35 arrangements3! 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 #1Choose 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

16C9 .7C4 .

3C3 = 16!————— 9! (16 9)!

7!————4! (7 4)!

3!———— =3! (3 3)!

16!——— = 4004009! 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 #2Choose 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

16C3 .13C4 .

9C9 = 16!————— 3! (16 3)!

13!————— 4! (13 4)!

9!———— =9! (9 9)!

16!——— = 4004003! 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

n!————r! (n – r)!

(number of ordered samples of r objects)—————————————————————– =(number of permutations of r distinguishable objects )

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 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

n!————r! (n – r)!

n!—————r1! r2! … rk!

= nCr

= nCr

Number of Samples of Size r out of n Objects

WithReplacement

WithoutReplacement

Ordered

Unordered

nr

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?

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?

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?

165 = 1048576 samples These samples are equally likely.

16!16P5 = ———— = 524160 samples

(16 – 5)!These samples are equally likely.

(j)

(k)

(l)

Number of Samples of Size r out of n Objects

WithReplacement

WithoutReplacement

Ordered

Unordered

nr n!———(n – r)! = nPr

n!————r! (n – r)! = nCr

nr

=

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?

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?

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?

165 = 1048576 samples These samples are equally likely.

16!16P5 = ———— = 524160 samples

(16 – 5)!These samples are equally likely.

(j)

(k)

(l)

16 16!16C5 = = ————— = 4368 samples

5 5! (16 – 5)!These samples are equally likely.

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 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.

(m)

165

(93)(4)(3)

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

= 0.01152

(NOTE: When the calculations are difficult, it is generally acceptable to leave the answer is a “ready-to-calculate” format.)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.

(n)

(16)(15)(14)(13)(12)

(9)(8)(7)(4)(3)

6048 samples with the first 3 books selected green, the 4th book selected red, and the 5th book selected blue

= 0.01154

16!16P5 = ———— = (16)(15)(14)(13)(12) = 524160

(16 – 5)!equally likely samples of any 5 books 9!

9P3 = ————(4)(3) = (9)(8)(7)(4)(3) = (9 – 3)!

(NOTE: When the calculations are difficult, it is generally acceptable to leave the answer is a “ready-to-calculate” format.)

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.

(o)

1008 samples with the first 3 books selected green, the 4th book selected red, and the 5th book selected blue

= 0.23077

16!16C5 = ————— = (16)(14)(13) = 4368

5! (16 – 5)!equally likely samples of any 5 books

9!9C3 = ————(4)(3) =

3! (9 – 3)!

(NOTE: When the calculations are difficult, it is generally acceptable to leave the answer is a “ready-to-calculate” format.)

16

9

4

1

3

1

9

3

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?

(p)

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 ways. Therefore, these samples are not equally likely.10To 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}.

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?

(p)

20 20!20C5 = = ————— = 15504 samples

5 5! (20 – 5)!

The number of possible samples is

These samples are not equally likely.

Number of Samples of Size r out of n Objects

WithReplacement

WithoutReplacement

Ordered

Unordered

nr n!———(n – r)! = nPr

(n + r – 1)!———— r! (n – 1)!

n!————r! (n – r)! = nCr

nr

=

2.

(a)

(b)

An urn contains 7 red chips, 8 blue chips, and 9 white chips. Find the probability of

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.

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.

(24)(23)(22)(21)(20)(19)(18)(17)

(7)(8)(7)(6)(9)(8)(7)(6)or

24!—16!

7!—6!

8!—5!

9!—5!

248

(7)(83)(94)

selecting one red chip, three blue chips, and four white chips if eight chips are selected at random and without replacement.

selecting one red chip, three blue chips, and four white chips if eight chips are selected at random and with replacement.

(c)

(d)

24

8

7

1

8

3

9

4

or24!—16!

7!—6!

8!—5!

9!—5!8!

1! 3! 4!

248

(7)(83)(94)8!

1! 3! 4!

(e)

(f)

selecting all white chips if five chips are selected at random and without replacement.

(24)(23)(22)(21)(20)

(9)(8)(7)(6)(5)or

24!—19!

9!—4!

or24

5

9

5

selecting all white chips if five chips are selected at random and with replacement.

245

95

(g)

(h)

exactly four white chips if seven chips are selected at random and without replacement.

24

7

9

4

15

3

exactly four white chips if seven chips are selected at random and with replacement.

247

(94)(153)7!

3! 4!

(i)

(j)

selecting no red chips if three chips are selected at random and without replacement.

selecting no red chips if three chips are selected at random and with replacement.

24

3

17

3

243

173

24

5

16

5

1 –

(k)

(l)

selecting at least one blue chip if five chips are selected at random and without replacement.

selecting at least one blue chip if five chips are selected at random and with replacement.

245

165

1 –

(m)

(n)

selecting exactly two red chips if six chips are selected at random and without replacement.

selecting exactly two red chips if six chips are selected at random and with replacement.

24

6

7

2

17

4

246

(72)(174)6!

2! 4!

(o)

(p)

selecting more than 20 chips in order to get the fifth blue chip if chips are selected at random and without replacement.

selecting more than 20 chips in order to get the fifth blue chip if chips are selected at random and with replacement.

24

20

8

4

16

16

2420

1620 + (8)(1619)20!

1! 19!+ (82)(1618)

20!

2! 18!

+ (83)(1617)20!

3! 17!+ (84)(1616)

20!

4! 16!

(q)

(r)

selecting a white chip on the third try if chips are selected at random and without replacement.

selecting a white chip on the third try if chips are selected at random and with replacement.

9 3— = —24 8

9 3— = —24 8

3.

(a)

(b)

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.Two chips are randomly selected without replacement and the random variable X = "sum of the observed integers" is recorded. Find each of the following:P(X < 4) =

P(X > 2) =

P(X = 2) + P(X = 3) = 1— + 5

2— = 5

3— 5

1 – P(X = 2) = 11 – — = 5

4— 5

Two chips are randomly selected with replacement and the random variable X = "sum of the observed integers" is recorded. Find each of the following:P(X < 4) =

P(X > 2) =

P(X = 2) + P(X = 3) = 1— + 4

1— = 3

7—12

1 – P(X = 2) = 11 – — = 4

3— 4