module 4: permutations, combinations, and...

Principles of Mathematics 12 Answer Key, Contents 121

Module 4

Module 4: Permutations, Combinations, and Probability

Section 1: Permutations and CombinationsLesson 1 Fundamental Counting Principle 123

Lesson 2 The Formula nPr and Factorial Notation 126

Lesson 3 Like Objects 131

Lesson 4 Combinations 134

Lesson 5 Binomial Theorem 139

Review 144Section 2: Probability

Lesson 1 Review of Probability Ideas 146

Lesson 2 The Probability Laws 150

Lesson 3 Two Special Cases 154

Lesson 4 Conditional Probability 156

Lesson 5 Practice Problems of Compound Events 160

Lesson 6 Probabilities Using Permutations and Combinations 166

Lesson 7 Binomial Probabilities 168

Review 170

122 Answer Key, Contents Principles of Mathematics 12

Module 4

Principles of Mathematics 12 Section 1, Answer key 123

Lesson 1

Answer Key

1. Ending in the digit 2: 312, 342, 352, 412, 432, 452, 512, 532,542Ending in the digit 4: 314, 324, 354, 514, 524, 534

2. a) __4__ · __1__ · __3__ = 12

b) __4__ · __4__ · __3__ = 48

or all possible numbers – numbers with 4 in middle = 60 – 12 = 48

c) __3__ · __3__ · __2__ = 18

d) __3__ · __4__ · __3__ = 36

e) Case 1: first digit is 1: __1__ · __3__ · __2__ = 6

Case 2: first digit is 2: __1__ · __3__ · __3__ = 9

Total number is: 6 + 9 = 15

3. 108 – 72 = 36 boys. Complementary problems.

4. a) The first digit cannot be a zero or it would be considered athree-digit number.__6__ · __6__ · __5__ · __4__ = 7200 re-enters the selection set after the first digit has beenselected.

b) Case 1: The last digit is 0.__6__ · __5__ · __4__ · __1__ = 120Case 2: The last digit is 2, 6, or 8.__5__ · __5__ · __4__ · __3__ = 300

The number of even four-digit numbers will be:120 + 300 = 420.

c) Case 1: The last digit is 0.

__3__ · __2__ · __1__ · __1__ = 6Case 2: The last digit is 2, 6, or 8.__2__ · __2__ · __1__ · __3__ = 12

The number of all even four-digit numbers will be:6 + 12 = 18.

Module 4

d) Case 1: The last digit is 0.

__6__ · __5__ · __4__ · __1__ = 120Case 2: The last digit is 5.__5__ · __5__ · __4__ · __1__ = 100

The number of four-digit numbers divisible by 5 will be:120 + 100 = 220.

5. a) We list the possible replacements under each blank.Case 1: The first digit is 7 or 8.__2__ · __6__ · __6__ · __6__ = 432{7, 8} {any digit}

Case 2: The first digit is 5 and second is 5, 7, or 8.__1__ · __3__ · __6__ · __6__ = 108{5} {5,7,8} {any digit}

Case 3: The first digit is 5 and second is 3.__1__ · __1__ · __2__ · __6__ = 12{5} {3} {7, 8} {any digit}

The total number is 432 + 108 + 12 = 552.

b) Case 1: The first digit is 7 or 8.

__2__ · __5__ · __4__ · __3__ = 120{7, 8} {losing one digit each time}

Case 2: The first digit is 5 and second is 7 or 8.__1__ · __2__ · __4__ · __3__ = 24{5} {7, 8} {any unused digit}

Case 3: The first digit is 5 and second is 3.__1__ · __1__ · __2__ · __3__ = 6{5} {3} {7,8} {7 or 8, 1, 2}

The total number is 120 + 24 + 6 = 150.

124 Section 1, Answer key Principles of Mathematics 12

Module 4

6. a) __5__ · __4__ = 20 b) __5__ · __5__ = 25

7. __23__ · __22__ · __21__ = 10 626

8. __5__ · __4__ · __3__ · __2__ · __1__ = 120

9. The first person must be a woman, the second a man, etc.

__4__ · __3__ · __3__ · __2__ · __2__ · __1__ · __1__ = 144w m w m w m w

10. The first person can be a woman or a man. After the firstperson they alternate._8__ · _4__ · _3__ · _3__ · _2__ · _2__ · _1__ · _1__ = 1152

11. Let the blanks always represent the smaller set, the lettersinstead of the mailboxes. There are two letters and fivemailboxes.a) __5__ · __4__ = 20 b) __5__ · __5__ = 25


Module 4

Lesson 2

Answer Key

1.

2.

3. a)

b)

c) n n

n nn n

n n

− =

− − =− + =

= = −

1 42

42 0

7 6 0

7 6

2

b g

b gb g or (reject)

n = 10

nn

+ ==

2 2018

2

( 3)! ( 3)( 2)!a) 3

( 2)! ( 2)!

( 7)! ( 7)( 6)!b) 7

( 6)! ( 6)

( 1)( 2)!c) ( 1)

( 2)!

7!( 2)! 7 6!( 2)( 1) ( 1)!d) 7( 2)( 1)

( 1)!6! ( 1)!6!

k k k kk k

n n n nn n

n n n n nn

r r r r r r r r r rr r

+ + += = ++ +

+ + += = ++ +

− − = −−

+ ⋅ + + ⋅ − ⋅= = + +− −

7! 7 6! 31! 31 30!a) 7 b) 316! 6! 30! 30!

c) 12(11) 132 d) 9(8)7 504

10 9 8 7 6 5e) 210 f) 154 3 2 1 2 1

g) 7! 5040 h) 6(2) 12

5!8! 5 4! 8 7!i) 6(24) 144 j) 5 8 404!7! 4!7!

⋅ ⋅= = = =

= =

⋅ ⋅ ⋅ ⋅= =⋅ ⋅ ⋅ ⋅= =

⋅ ⋅= = = ⋅ =


Module 4

d)

Note: d) can also be solved by a graphing calculator, inwhich case the intermediate steps do not appear.

e)

Note: In Questions 4 and 5, to reduce the factorials, scale thelarger factorial expression until it reaches the smallerexpression. For example, to reduce (r + 1)! with r! write(r + 1)! = (r + 1)r! and now compare to r!

4. a) b)

− ⋅− −

= =−

RHS

( 1)!( )!( 1)!

!LHS

( )! !

n n rn r r r

n rn r r

−− − − +

= =− − +

RHS

!( )( )( 1)! !( 1)

! LHS( 1)!( 1)!

n n rn r n r r r

nn r r

+ − − =−

+ − =

+ − =

+ − =

= − =

2

( 1)( )( 1)!30 0

( 1)!

( 1) 30 0

30 0

( 6)( 5) 0

6 (reject) 5

n n nn

n n

n n

n n

n n

+ + − =

+ + − =

+ + − =

=

( 2)( 1) ( 1) 57(7)(6)(5)(4)3

( 2)( 1) ( 1) 19(3)(6)(5)(4)(7)3

( 2)( 1) ( 1) 18(19)(20)21

19 (match the numbers)

n n n n

n n n n

n n n n

n

+ + ⋅ ⋅ − − =−

⋅ ⋅+ + − =

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅+ + − =

( 2)( 1) ( 1)( 2)! 578!( 2)! 16

57 8 7!( 2)( 1) ( 1)

16

57 8 7 6 5 4 3 2 1( 2)( 1) ( 1)

16

n n n n nn

n n n n

n n n n


Module 4

c)

5.

By dividing both sides by 18! and expanding thedenominators, we get:

6. 5 2 5 3

5 5 100 2

6 1

5! 5!a) 20 b) 60

3! 2!

5! 100!c) 120 d) 9900

0! 98!

6!e) 6

5!

P P

P P

P

= = = =

= = = =

= =

( )( )( ) ( ) ( )( )

( )( ) ( )( )

=− − − − + +

=− − + +

+ + = − +==

2 2

1 118 17 16 ! ! 16 ! 2 1

1 118 17 2 1

3 2 306 3538 304

8

r r r r r r r r

r r r r

r r r rrr

( ) ( ) ( )=− − +18! 18!

18 ! ! 16 ! 2 !r r r r

LHS

! !( )! ( 1)! ( 1)( )!( 1)!

!( 1) !( 1)( )!( 1)!

!( 1) !( )!( 1)!

!( 1 )!( 1)!

( 1) !!( 1)!

( 1)! RHS!( 1)!

n nn r r r n r n r r

n n r n rn r n r r r

n n r n rr n r

n n r rr n r

n nr n r

nr n r

+− − − + − −

− + +=− + − −

− + +=− +

− + +=− +

+=− +

+= =− +


Module 4

7.

8.

9.

10.

n(n – 1)(n – 2) = 7(6)(5). By comparison n = 7.

11. The symbol 8P3 means the permutation of eight objectsthree at a time. The symbol 3P8 does not make sensebecause you cannot permute three objects eight at a timesince you do not have eight objects.

12.

13.

[Notice how the formula may be used for doing part ofthe question in part (b).]

c) All possible — the number that begins with G = 40 320 – 5 040 = 35 280

P

P

= = =

⎛ ⎞⋅ = = =⎜ ⎟⎝ ⎠

8 8

7 7

8!a) 8! 40 320

0!7!

b) 1 1 7! 5 0400!

( )( )− − =

=− + =−

− = = ≠ −

2 72 0! 72 9 8 0( 2)!

( 1) 9(8) 9 or 8

n nnor

n nn

n n n n

3 6 27( )

! 6!7

( 3)! 4!

( 1)( 2)( 3)! 7 6 5 4!( 3)! 4!

n P P

nn

n n n nn

=

= ⋅−

− − − ⋅ ⋅ ⋅=−

7 57!

2520 or 7 6 5 4 3 25202!

P = = ⋅ ⋅ ⋅ ⋅ =

5 550

5 120 4 2 1 120P = = = ⋅ ⋅ ⋅ ⋅ =!!

! or 5 3

12 5

8 4

10 3

12!a)

7!

8!b)

4!

10!c)

7!

P

P

P

=

=

=


Module 4

14.

c) All possible – number of words with no vowels = 60 – 6 = 54

( )3 33!d) 2 1 2 1 2 3! 120!

P ⎛ ⎞⋅ ⋅ = = =⎜ ⎟⎝ ⎠

5 35!

a) 602!

b) 3 2 1 6

P = =

⋅ ⋅ =

P

P

P

⎛ ⎞⋅ ⋅ = = =⎜ ⎟⎝ ⎠⎛ ⎞⋅ ⋅ = ⋅ = =⎜ ⎟⎝ ⎠⎛ ⎞⋅ ⋅ = = =⎜ ⎟⎝ ⎠

6 6

6 6

6 6

6!d) 1 1 1 1 6! 720

0!

6!e) 1 1 1 1 6! 720

0!6!

f) 2 1 2 1 2(6!) 14400!


Module 4

Lesson 3

Answer Key

1.

2.

3.

4. a) __3__ · __2__ · __2__ · __1__ · __1__ = 12

b) Since the odd numbers cannot change order, it followsthat you must divide by the number of ways of orderingthe odd numbers—by 3! = 6. Therefore, 12 ÷ 6 = 2.

5. a)

b) Case 1: if first digit is 3 Case 2: if first digit is 4 or 5

Total number of numbers is: 40 + 120 = 160.

c) Last digit must be 5.Case 1: if first digit is 3 Case 2: if first digit is 4

Total number of numbers is: 4 + 10 = 14.

15

3 21 10⋅ ⋅ =!

! !1 1

43

1 4⋅ ⋅ ⋅ =!!

26

3 2120⋅ =!

! !1 2

53

40⋅ ⋅ =!!

73 2

420!

! !=

9! 1 2603!2!4!

=

11!69 300

3!2!2!4!=

=

=

=

7!a) 1 260

2!2!

11!b) 1 663 200

2!2!3!

8!c) 3 360

3!2!


Module 4

6.

d) Exactly one R means that the first letter is an R and thesecond is not an R.

e) Exactly two Rs means that the first two letters are Rsand the third is not an R.

7. a) The odd digits cannot change order. Therefore,

b) The even digits cannot change order. Therefore,

8.

9. The ways of ordering six Rights and four Ups. Therefore,

10. To get to C you must order two Rs and three Us which is

Once at C, A must walk to B by ordering six Rs and four Us

which is ways.

By the Fundamental Principle, A can walk to B in 10(210) =2 100 ways.

10!210

6!4!=

5!10 ways.

3!2!=

106 4

210!

! !.=

12!1 995 840

5!2!=

73

840!!

.=

74

210!!

.=

4!1 1 4 960!

⋅ ⋅ ⋅ =

1 452

240⋅ ⋅ =!!

a)

b)

c)

162

360

1 150

120

1 1 140

24

⋅ =

⋅ ⋅ =

⋅ ⋅ ⋅ =

!!

!!

!!


Module 4

11.

12. a) The Ms cannot change order since they areindistinguishable. They are treated as a grouped objectwith no internal order changes. Therefore, 5! = 120 ways.

b) Case 1: if the order is MUM, then you have to order fiveobjects in 5! = 120 ways.Case 2: if the order is MMU and MMU is the last object,you have 4! = 24 ways.Case 3: if the order is MMU and MMU is not the lastobject, the object right after this triple cannot be an M oryou will repeat MMUM from Case 1. Therefore, there arethree choices for the letter right after MMU and thesethree objects can be ordered in 3! ways. Total numberfor case 3 is 3(3!) = 18 ways.Case 4 and Case 5 with order UMM are similar to Cases2 and 3. Therefore, 24 + 18 = 42 ways. Total of all possibilities with MUM together is: 120 + 24 + 18 + 24 + 18 = 204 ways.

13.

14. Since the ordering of the three selected objects isimmaterial, you should count it as if the order is importantand then cancel out all the orderings of three objects, i.e.,divide by 3! = 6.

Therefore, the answer is:

(This problem will be explained more fully in the nextlesson.)

5 3 6010 ways.

3! 6P = =

14!a) 151 351 200

2!4!3!2!

14!b) 30 270 240

3!5!2!2!

=

=

85 3

56!

! !=


Module 4

Lesson 4

Answer Key

1.

e) Use the nCr function on your calculator to get 190 578 024or cancel down the factorials as follows:

2. a) Both equal 120b) Both equal 10c) Both equal 210Each pair must be equal because if you select three personsout of 10 to take some place, you are simultaneouslyselecting seven persons out of 10 to leave behind. Everyselection has this dual action of selecting some objects andchoosing not to select the remaining objects.

3. The number of ways of choosing objects not to be selected isalso 120 = nCn–r.

4. 12C4 = 495

5. 5C3 · 8C4 · 5C3 = 10(70)10 = 7 000

6. Two students must be selected for every handshake. Theorder of selection is unimportant. Therefore, there are 30C2 = 435 handshakes.

7. a) You must select five cards in any order. 52C5 = 2 598 960 different poker hands.

b) There are four suits and there are 13 cards in each suit.Therefore, there are 4C1 · 13C5 = 5148 flushes.

( )( )( )( )( )120 119 118 117 116120!190 578 024

5!115! 5!= =

12!a) 11 8808!

12!b) 4958!4!

5!c) 5! 1200!

5!d) 10!5!

=

=

= =

=


Module 4

8. 49C6 = 13 983 816 choices

9. a) 9C6 = 84b) Case 1: with Dorothy or Oksana invited:

_2_ · 7C5 = 2(21) = 42Case 2: without Dorothy or Oksana: 7C6 = 7Total number of ways is: 42 + 7 = 49.

10. a) Each person can choose any hotel. 7C1 · 7C1 · 7C1 = 7(7)7 = 343 ways

b) Place the special people first.7C1 · 6C1 · 7C1 = 7(6)7 = 294 ways

c) All possible – the instance they all select this two-vacancy hotel = 343 – 1 = 342.

11. 6C5 · 5C3 = 6(10) = 60 ways

12. Case 1: five girls10C5 = 252

Case 2: four girls and one boy10C4 · 8C1 = 210(8) = 1 680

Case 3: three girls and two boys10C3 · 8C2 = 120(28) = 3 360 The number of choices with girls in the majority is: 252 + 1 680 + 3 360 = 5 292.

13. Tom and, hence, Roxolana, must be invited because if theyare not, that leaves only eight people, but Peter and Carolecannot both be invited. Therefore, the ambassador mustselect 6 more people.Case1: One of Peter or Carole is invited. 2C2 · 2C1 · 6C5 = 1(2)(6) = 12

Case 2: Neither Peter nor Carole is invited.2C2 · 6C6 = 1(1) = 1

Total number of ways to select her guests is: 12 + 1 = 13.


Module 4


Module 4

14. There are some repeated letters: two Es and two Ds.

Case 1: two Es and two Ds

Case 2: either two Es or two Ds _2_ · 6C2 = 2(15) = 30Case 3: no repeated letters 7C4 = 35 The only permutation so far is Case 1; therefore, the othercases must be multiplied by the number of orderings of fourletters, but remember 2 letters are the same ∴ number ofpermutations is 4!/2! = 12Total number of four-letter words is: 6 + 30(12) + 35(24) = 1206.

15. There are 13 possible pairs. Select two: 13C2 = 78.There are four cards of each denomination. Select two ofeach denomination selected above: 4C2 · 4C2 = 6(6) = 36.Select one more card from the remaining 11 denominations:44C1 = 44.The order in which the cards are dealt is unimportant.Therefore, the total number of ways of being dealt two pairsis 78(36)44 = 123 552.

16. a) Choose one of each: 10C1 · 8C1 = 10(8) = 80. b) Choose two of each: 10C2 · 8C2 = 45(28) = 1 260. c) Same as part (b) except they can switch partners:

1 260(2) = 2 520

17. To form a rectangle you need two vertical and twohorizontal lines. Selecting two of each: 7C2 · 4C2 = 21(6) = 126.

18. a) A chooses six and B gets the rest: 12C6 = 924 ways.b) Since unassigned, they can switch. Hence, the

subdivision is the same: 924 ÷ 2 = 462.c) A chooses four of 12; B chooses four of eight; C gets the

rest:12C4 · 8C4 = 495(70) = 34 650.

d) Any one of the three subdivisions in part (c) can beordered in 3! = 6 ways. Since unassigned, 34 650 ÷ 6 = 5 775 ways.

42 2

6!

! !=


Module 4

19. Count all possible lines which can be formed by joining anytwo points and subtract all the lines which are notdiagonals. The non-diagonal lines are on the perimeterforming the figure.Number of diagonals in a regular hexagon is: 6C2 – 6 = 15 – 6 = 9.Number of diagonals in a regular octagon is: 8C2 – 8 = 28 – 8 = 20. Number of diagonals in a regular n-gon is:

20. a) Jack and four others: __1__ · 8C4 = 70 ways.b) Jack is rejected: 8C5 = 56 ways.

9C5 = 126 and 70 + 56 =126. They are equal. Parts (a) and(b) cover all the possible ways of selecting five persons fromnine. Jack is on the committee or he is not; there is no otherpossibility.

21. You must select three points to form a triangle. The order ofselection is unimportant. Therefore, 7C3 = 35 such trianglesare possible.

22. 30C7 = 2 035 800 ways. They must choose five songs for themiddle part of their program: 28C5 = 98 280 ways.

( ) 2

2

1 3.

2 2n

n n n nC n n− −− = − =

23. The number of ways of choosing r out of n objects is thesame as choosing n – r out of n objects. The factorial proof is:

Therefore, nCn–r = nCr.

As explained in Question 2, every time you select r objectsout of n objects for some task, you are selecting n – r objectsthat will be left behind. Therefore, the count should be thesame.

( )

( )

n n r

n r

nCn n r n rn

n n r n rn

r n rn

n r rC

− =− − −

=− + −

=−

=−

=

!( ) !( )!

!!( )!

!!( )!

!( )! !


Module 4

Lesson 5

Answer Key

1.

c) 271

2 172

2 173

2 174

2 1

75

2 176

2 1 1

128 448 672 560 280 84 14 1

7 6 5 2 4 3 3 4

2 5 6 7

7 6 5 4 3 2

x x x x x

x x

x x x x x x x

b g b g b g b g b g

b g b g

+FHGIKJ +

FHGIKJ +

FHGIKJ +

FHGIKJ

+FHGIKJ +

FHGIKJ +

= + + + + + + +

b) x x y x y x y

x y x y x y x y

x y y

x x y x y x y x y x yx y x y

9 8 1 7 2 6 3

5 4 4 5 3 6 2 7

8 9

9 8 7 2 6 3 5 4 4 5

3 6 2 7

91

92

93

94

95

96

97

98

9 36 84 126 126

84 36

+FHGIKJ − +

FHGIKJ − +

FHGIKJ − +

FHGIKJ − +

FHGIKJ − +

FHGIKJ − +

FHGIKJ −

+FHGIKJ − + −

= − + − + − +− +

b g b g b g

b g b g b g b g

b g b g

9 8 9xy y−

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞+⎜ ⎟

⎝ ⎠

= + + + + + +

+ + +

10 9 8 2 7 3 6 4

5 5 4 6 3 7 2 8

9 10

10 9 8 2 7 3 6 4 5 5

4 6 3 7 2 8

10 10 10 10a)

1 2 3 4

10 10 10 105 6 7 8

109

10 45 120 210 252

210 120 45 10

a a b a b a b a b

a b a b a b a b

ab b

a a b a b a b a b a b

a b a b a b +9 10ab b

Since and are both equal to 1 they will not be stated

in the expansions.

n nn0

FHGIKJFHGIKJ


Module 4

2.

3.

4. 74 4

235

6416 35

4

7 4 4 3

4

FHGIKJFHGIKJ −FHGIKJ =FHGIKJFHGIKJ =

−yy

yy y

86

288 6 6 2 6FHGIKJ =−a b a b

( ) ( ) ( )

212 11 10

12 10 8

22 26 5 4

6 6 5 7 4 8

12 121 1a) . . .1 2

12 66 . . .

6 6b) 3 3 3 . . .

1 3 2 3

729 486 135 . . .

x x xx x

x x x

x xax ax ax

a x a x a x

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜+ ⎟ − + ⎟ − +⎟ ⎟⎜ ⎜⎜ ⎜⎟ ⎟⎟ ⎟⎟ ⎟⎜ ⎜⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠= − +

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎟ ⎟⎟ ⎟⎜ ⎜⎜ ⎜⎟ ⎟+ ⎟ + ⎟ +⎜ ⎜⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎜ ⎜⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠= + +

7 6 5 2

7 5 3

7 72 2c) . . .2 1 2 2 2

7 21 . . .128 32 8

x x xx x

x x x

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜⎟ ⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜ ⎜+ ⎟ − + ⎟ − +⎟ ⎟ ⎟ ⎟ ⎟⎜ ⎜⎜ ⎜ ⎜ ⎜ ⎜⎟ ⎟⎟ ⎟ ⎟ ⎟ ⎟⎟ ⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜ ⎜⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

= − +

f) 231

21 3

22

1 18

12 6 13 22 3

2 3+FHGIKJFHGIKJ +FHGIKJFHGIKJ + FHGIKJ = + + +

x x x x x x

e) 341

33 4

23

3 43

33 3

81 324 486 32481

3 4 3 3 3 22

33 4

12 8 44

a aa

aa

aa a

a a aa

d i d i d i d i+FHGIKJFHGIKJ +FHGIKJFHGIKJ +FHGIKJFHGIKJ + FHGIKJ

= + + + +

d) 251

252

253

2

54

2

32 80 80 40 10

2 5 2 4 2 3 2 2 2 3

2 4 5

10 9 8 7 6 5

x x x x x x x

x x x

x x x x x x

d i d i b g d i b g d i b g

d ib g b g

+FHGIKJ − +

FHGIKJ − +

FHGIKJ −

+FHGIKJ − + −

= − + − + −


Module 4

5.

x x xx

x or x x xx

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

=

4 12 4 112

24 10 48 42 36

2

1e) The pattern of the exponent of is : ( ) , ( ) ,

1( ) . . . , , , . . . . The exponent is

decreasing by 6. To reach the constant te rm, you need48

8 decreases. The requested term is 6

xx

⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

84 4

2

the ninth

12 1 495term: (2 ) .

8 2 16

( ) ( )

( )

x x xx

x x x xx

x x

10 95 53

285 50 42 34

3

2 50

1d) The pattern of the exponent of is: , ,

1. . . or , , , . . . . The exponent is

decreasing by 8. Therefore, to reach fr om ,48

you need 6 decreases. The re8

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

=

x xx

4 652

3

quested term is the

10 2seventh term: 840 .

6 2

⎛ ⎞⎛ ⎞ ⎛ ⎞⎟⎟⎜⎜ ⎟⎜⎟⎟ − =⎟⎜⎜ ⎜⎟⎟ ⎟⎟⎜ ⎜⎜ ⎟ ⎟⎜ ⎜ ⎝ ⎠⎝ ⎠⎝ ⎠

( ) ( )

x x x x x xx x x

x x x

14 13 4 12 8

14 17 20

212 4 20

c) The pattern of the exponent of is: , , . . .

or , , , . . . . The requested term is the thi rd term

14which is: 2 372 736 .

2

⎛ ⎞⎟⎜ ⎟ − =⎜ ⎟⎜ ⎟⎜⎝ ⎠

( )xx

66

b) Since there are 13 terms, the middle t erm will be

12 1the seventh term : 2 924

6 2

⎛ ⎞ ⎛ ⎞⎟⎜ ⎟⎜⎟ − =⎟⎜ ⎜⎟ ⎟⎟⎜⎜ ⎟⎜ ⎝ ⎠⎝ ⎠

( ) ( )11 5 5 6 5 6 511a) 2 462 64 29 568

5y x y x y x−⎛ ⎞

= =⎜ ⎟⎝ ⎠


Module 4

from

requested

term

third

6.

7. The middle term will be the fifth term:

84

313

7044F

HGIKJ −FHG

IKJ =x

xb g .

x x xx x

xx

x x x x

x xx x

−

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞+ − + − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠= − + −

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

= −

23 11 3 10 3 9

33 8

33 29 25 21

73 11 7 4 12

14 7

11 111 1(2 ) (2 ) (2 )

1 4 2 4

11 1(2 )

3 4

2 048 2 816 1 760 660

11 1 1The eighth term is (2 ) 330(2 )

7 4 2

3301 024

x x= −5 5165.

512

( ) ( )r

rr

r r

ra ar a r

9 29 39

2

9 91 3h) 3 16 2

−−−⎛ ⎞ ⎛ ⎞⎛ ⎞⎟ ⎟⎜ ⎜⎟⎜⎟ − = − ⎟⋅ ⋅⎟⎜ ⎜⎜⎟ ⎟⎟⎟⎜⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠⎝ ⎠ ⎝ ⎠

63 18 18 183

6

9 84( 8) 224g) ( 2) 84 ( 8)

3 3 3 729 243x x x x⎛ ⎞ ⎛ ⎞⎛ ⎞ − −− = − = =⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠

( )

( ) ( )

x x x x x x

x x x

x x x

211 10 2 9 2

11 12 13

38 2 14

f) The pattern of the exponent of is: , , . . .

or , , , . . . . The requested term is the fou rth

11term which is: 2 42 240 .

3

⎛ ⎞⎟⎜ ⎟ =⎜ ⎟⎜ ⎟⎜⎝ ⎠


Module 4

fourth

8. In the expansion of every second term is the

opposite of every second term of In the sum,

these even position terms will produce a zero. Every oddterm will be twice as large since each binomial iscontributing the same value. Therefore, the answer will betwice every odd term:

9. In the binomial expansion, if the exponent of x is:a) x2 then the exponent of 1 is n – 2 since the sum of the

exponents of any term is nb) x3 then the exponent of 1 is n – 3 According to the problem:

10. The terms are:

To approximate (1.001)12, set x = 0.001. The approximationis: 1 + 0.012 + 0.000066 + 0.00000022 = 1.01206622.

12 11 10 2 9 3

2 3

12 12 121 1 1 1

1 2 3

1 12 66 220 . . .

x x x

x x x

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

+ + + +

Therefore,

Cancelling like terms yields

Hence, n – 2 = 12, or n = 14.

( ) ( )

43 2

! !4 .

3! 3 ! 2! 2 !

1 14 .

3 2

n nn nn n

n n

n

⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜⎟= ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜− −⎝ ⎠ ⎝ ⎠

=− −

=−

2 272

2 174

2 176

2 1

2 128 672 1 280 1 14 1

2 128 672 672 280 560 280 14 4242 14

28 644 2548 2188

7 52

34 6

2 3

2 3 2

3 2

+FHGIKJ + +

FHGIKJ + +

FHGIKJ +

LNM

OQP

= + + + + + +

= + + + + + + + ++

= + + +

x x x

x x x

x x x x xxx x x

e j e j e j

b g b g b g[

]

2 17

+ +xe j .2 1

7− +xe j


Module 4

Review

Answer Key

1.

2. The first and last person must be a girl. Hence,_4_ · _3_ · _3_ · _2_ · _2_ · _1_ · _1_ = 144 or 4!3! = 144.

3.

4.

(The 3! ways of ordering the three even digits are cancelledout. )

5. The three calculus and four algebra books can be selected in5C3 · 7C4 = 10(35) = 350 ways. These seven books can beordered in 7! = 5 040 ways. Therefore, the number of waysof selecting and then ordering these books on a shelf is:350(5 040) = 1 764 000 ways.

6. Three points must be selected to form a triangle. Hence,10C3 = 120.

7. x x x

x x x

x x xx x

x x x

x x x

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= + + +

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞+ − + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − + +

⎛ ⎞ ⎛ ⎞ ⎛+ − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

8 7 6 2

8 7 6

210 9 8

10 8 6

6 5 2

8 8 8a) (2 ) (2 ) (3) (2 ) (3) . . .

0 1 2

256 3 072 16 128 . . .

10 10 101 1b) ( ) ( ) ( ) . . .

0 1 2

10 45 . . .

6 6 6c) ( ) ( ) ( )

0 1 2x x

x x x

⎞− +⎜ ⎟

⎝ ⎠

= − + +

4 2 2

6 7 8

( ) ( ) . . .

6 15 . . .

73

840!!=

=

=

=

8!a) 20 160

2!

8!b) 5 040

2!2!2!

10!c) 50 400

3!3!2!

10 4104 6

210C = =!! !


Module 4

8.

9. The pattern of the exponent of x is:

x8 + x7(x2) + x6(x2)2 + . . . = x8 + x9 + x10 + . . ., the exponentincreases by 1.

Therefore, you require the sixth term:

10.

11.

There are 10 + 16 + 8 + 3 = 37 possible paths.

O

1 1

2 1

2 3 1

5 X 1

5 5 1 1

10 6 2 1

10 10 8 3

⋅ ⋅ ⋅⋅ = ⋅⋅ ⋅ ⋅

= ⋅

=

6! 7! 6 5 7 6 54!2! 4!3! 2 1 3 2 1

15 35

525 ways

3 2 5 3 2 5 138(2 ) ( ) 56(2 ) ( ) 448 .

5x x x x x⎛ ⎞

= =⎜ ⎟⎝ ⎠

84

708 4 5 1 4 4FHGIKJ − =− −a b a bb g


Module 4

Lesson 1

Answer Key

1.

2. a)

b)

c)

d)(1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1)(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2)(1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3)(1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4)(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)(1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6)

H H H T

T H T T

1 2 34 5 6

H T

1 3 1a) b)6 6 22 1 3 1c) d)6 3 6 24 2 3 1e) f)6 3 6 26 2 1g) 1 (certainty) h)6 6 30i) 0 (impossible)6

=

= =

= =

= =

=


Module 4

e)

3.

4.

5.3a)8

5b)8

5c)8

3 5 3d) or 1 (odd) 18 8 8

4 1e)8 2

P− = − =

=

( )

( )

( )

3 1 4 1a) b)12 4 12 3

3 45 7c) d)12 12 12

3 4 7 5 7e) or (not white) 1 (white) 112 12 12 12

3 5 8 2 4 8f) or (not blue) 1 (blue) 112 12 3 12 12

P P

P P

= =

+=

+ ⎛ ⎞⎟⎜= = − = − =⎟⎜ ⎟⎟⎜⎝ ⎠

+ ⎛ ⎞⎟⎜= = = − = − =⎟⎜ ⎟⎟⎜⎝ ⎠

( )

= =

+=

= =

= = =

4 1 13 1a) b)

52 13 52 4

13 31 4c) d)

52 52 13

12 3 32 8e) f)

52 13 52 13

26 2 1 2 1g) h)

52 4 2 52 26

HHH HHT

THH HTT

TTH TTT

HTH

THT


Module 4

6.

7. A tree diagram may be the best way to solve this problem.

Using the sample space provided by the results above:

8.760 152a)

1475 295

425 17b)1475 59

600 24 35c) 1 (Senior 2) 1 11475 59 59

250 10d)1475 59

P

=

=

− = − = − =

=

a)

b)

14

24

12

=

First child Second child Result

girl GG

girl

boy GB

girl BG

boy

boy BB

3 1a)

120 40

1 39b) 1 (key does open the door) 1

40 40P

=

⎛ ⎞− = − =⎜ ⎟⎝ ⎠


Module 4

9.

10. a) b)

c) d)

e) f)

g) h)

i)

136

636

16

936

14

536

136

0

936

14

636

16

636

16

=

=

= =

=

1a)26

3b)26

5c) (if "y" is not considered a vowel)26

21d)26


Module 4

Lesson 2

Answer Key

1.

2.

3.

Note: In Question 3, after one red marble is withdrawn,three reds remain in a total of nine marbles. Hence, thesecond multiplied term is

4.

Note: In this question, since the first marble is replacedbefore the second is drawn, the probability of selecting thesecond red marble is not affected by the selection of the firstred marble. When P(red2⎢red1) = P(red2), we say that theevent red2 is independent of the event red1. When the firstmarble is replaced before the second marble is drawn, theevents red1 and red2 are independent events.

This is not the case when the first marble is not replacedbefore the second marble is drawn as in Question 3.

1 2 14 4 2

(red ) (red |red ) = 10 10 25

⋅ ⋅ =P P

39 .

1 2 14 3 12 2

(red ) (red |red )10 9 90 15

⋅ = ⋅ = =P P

4 26 2 28 7(queen) (red) (red queen)

52 52 52 52 13P P P+ − = + − = =

3 4 1 6 3a) (1) (shaded) (1 and shaded)8 8 8 8 4

1 3 1 1b) or (1) (shaded|1)8 8 3 8

3 5c) 1 (1) 18 8

4 1d)8 2

P P P

P P

P

+ − = + − = =

⋅ = ⋅ =

− = − =

=


Module 4

5.

P(divisible by 3 and greater than 7) = P(divisible by 3) · P(sum > 7⎢divisible by 3)

6. 1 – P(student gets an A) = 1 – 0.12 = 0.88

7. P(Bombers win this year or next year) = P(Bombers win thisyear) + P(Bombers win next year) – P(Bombers win this yearand next year) = 0.85 + 0.60 – 0.50 = 0.95

8. P(student plays for the Lions or basketball) = P(studentplays for the Lions) + P(student plays basketball) –P(student plays for the Lions and basketball)= 0.08 + 0.07 – 0.02 = 0.13

9. A tree diagram could be used to produce the followingsample space:

In this sample space only one point has three heads.

Therefore, P(HHH) = 18

.

HHH HHT HTH THH HTT THT TTH TTT

=+ + +

⋅+

+ + +=

2 5 4 136

4 12 5 4 1

536

b g b gb g

Sums

Divisible by 3 Diagonals

sum is greater than 7

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12


Module 4

You can also use the Product Rule (twice):P(HHH) = P(H1) · P((H2H3)⎢H1) = P(H1) · P(H2⎢H1) · P(H3⎢(H2 H1)

=

Each result of flipping the coin is independent of the previous result(s), i.e., the coin does not remember, or care,what happened on any of the previous result(s).

Therefore, P(HHH) =

10. a) P(king1) · P(king2⎢king1) =

(independent events)

b) P(king1) · P(king2⎢king1) =

(dependent events, since one of the kings has been removed before the second card is draw)

11.

No other event is possible. You can select only a red marble or a blue marble.

12. a)

( ) 6 1sum 5

36 6P < = =

Sums 1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

a)

b)

c)

310

710

1010

1=

452

351

1221

⋅ =

452

452

1169

⋅ =

18

.

12

12

12

⋅ ⋅ .


Module 4

b)

c)

( )5 6 5 4(5 sum 9)

36 9P

+ +< < = =

Sums 1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

P < = =10 5(5 or sum 5)

36 18

Sums 1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12


Module 4

Lesson 3

Answer Key

1. Since the events are mutually exclusive,

2. P(even) + P(prime) – P(even and prime) =

3. Since these events are independent, P(even and heads) =

P(even) · P(heads) =

4. Since no ties are assumed, the events are mutually

exclusive, P(2 or 3) = P(2) + P(3) =

5.

6. Since the first card is replaced before the second card isdrawn the events are independent.

7. a)

b)

452

351

1221

1252

1151

11221

⋅ =

⋅ =

a)

b)

452

452

1169

1252

1252

9169

⋅ =

⋅ =

5 115 3

=

110

110

15

+ = .

36

12

14

⋅ = .

3 3 1 56 6 6 6+ − =

1 1 1(5 or 4) = (5) + (4) = .

6 6 3P P P + =


Module 4

8. Since the cards are not replaced between successive drawsthe events are dependent.

9. Since the cards are replaced between successive draws theevents are independent.

c) Since the two cards cannot be spades and diamondssimultaneously, the events are mutually exclusive.

10. Assuming Harry and Henry are working independently, itfollows that the events are independent.

P(both solve the problem) =

P(at least one solves the problem) = P(Harry solves it) + P(Henry solves it) – P(both)

= + − =12

13

16

23

12

13

16

⋅ =

13 13 13 13 1(SS or DD)

52 52 52 52 8P = ⋅ + ⋅ =

a)

b)

152

152

12704

1352

1352

116

⋅ =

⋅ =

⋅ ⋅ =

⋅ ⋅ = =

⋅ ⋅ =

4 3 2 1a)

52 51 50 5525

26 25 24 24 2b)

52 51 50 204 17

52 12 11 22c)

52 51 50 425


Module 4

Lesson 4

Answer Key

1.

2.

3.

4. P

PP

=

⋅= =

(penny shows heads|nickel shows heads)

(penny shows heads and nickel shows (nickel shows heads)

1 112 2

1 22

(France|Mexico)

(France and Mexico)(Mexico)

0.40 20.60 3

P

PP

=

= =

P

PP

=

= =

(Janice develops lung cancer|Janice smok es)

(Janice develops lung cancer and Janice smokes)(Janice smokes)

4815

3 910

(Andrea passes|Andrea studied)(Andrea passes and Andrea studied)

(Andrea studied)17

682015 7516

PP

P=

= =


Module 4

smokes

heads)

5. P(heads) = The answers are equal since the occurrence of

heads on either coin is independent of the occurrence ofheads on a particular coin. Just because the nickel fell headsdoes not have any effect on the probability of the pennyfalling heads.

6. a) since there are two even shaded numbers and a

total of four shaded regions.

b)

7. The sample space changes from

because both children cannot be girls.

Therefore, P(BB⎜not GG) =

8. a)

b) P(even) =

c) The new sample space is:

or use the Venn diagram with the reduced sample space:

P(less than 5⎜even) = 23

.

(less than 5 and even)(less than 5|even)

(even)2

263 36

PPP

=

= =

2 4 6

3 16 2=

1 3 5 2 4 6

13

.

BB BG GB GG BB BG GB to

(shaded and even)(even|shaded)

(shaded)2

2 184 4 28

PPP

=

= = =

24

12

= ,

12

.


Module 4

9. a)

b)

i)

ii)

Note: The answer can be deduced very quickly by

using the reduced sample space.

10. a) Since not white, the number of points in the samplespace changes from 12 to 8.

Hence, P(blue⎜not white) =

or

b) Not white reduces the sample space to 8. Therefore,

P(not red⎜not white) = Notice that this answer is the

same as part (a), since not white and not red impliesblue.

38

.

(blue|not white)

(blue and not white) (not white)

3128

1238

P

PP

=

=

=

38

.

34

(prime and odd)(prime|odd)

(odd)3

384 48

PPP

=

= =

4 1(odd)

8 2P = =

1 3 5 7

2 4 6 8

1 3 5 7

2 4 6 8


Module 4

11. a) Reduced sample space is:

P(even⎜greater than 40) =

b) In the reduced sample space: 10 12 14 . . . 46 48 50,there are 21 points of which five are greater than 40.

Therefore, P(greater than 40⎜even) =

c) The sample space is:

This reduced sample space contains four prime numbers.

Therefore, P(prime⎜between 20 and 40) =

12. a) P(black⎜coin fell heads) = P(black|Box I) =

b) P(white⎜coin fell tails) = P(white|Box II) = 23

12

419

.

21 22 23 24 25 26 27 28 29

30 31 32 33 34 35 36 37 38 39

521

.

5 110 2

=

41 42 43 44 45 46 47 48 49 50


Module 4

Lesson 5

Answer Key

1. An alternate method is to use a tree diagram.

Therefore, P(B) =

(that is, all the branches that end in black).

2.

P(Improper seal) = 0.002 + 0.0018 + 0.002 = 0.0058

3.

P(BG or GB) =1 1 14 4 2+ =

Boy

Boy

Girl

Boy

Girl

Girl

= 14

= 14

12

12

12

12

12

12

Improper = 0.002I

Proper

Improper = 0.0018II

Proper

Improper = 0.002III

Proper

0.010.99

0.994

0.996

0.006

0.004

0.3

0.2

0.5

+ =1 1 53 12 12

Black

Box I

Red

Black

Box II

Red

12

46

26

16

56

12

13

P= ←

16

P= ←

112

P= ←

512

P= ←

(Box I and black)

(Box II and black)

(Box II and red)

(Box I and red)


Module 4

4.

P(both defective) =

5.

a) P(RR) =

b) P(same colour) = P(RR) + P(WW) + P(BB)

= + + =115

215

115

415

115

Red

Red White

Blue

Red

White White

Blue

Red

Blue White

Blue

= 115

= 215

= 115

29

29

39

310

310

410

145

defective =

defective

~ defective

defective

~ defective

~ defective

19

89

29

79

810

210

145


Module 4

6. a)

b) P(ends in heads) = P(H1) + P(T1H2) + P(T1T2H3)

orP(ends in heads) = 1 – P(does not ends in heads)

c) P(ends in tails) =

7.

P(exactly two hits) = P(HHM) + P(HMH) + P(MHH) = 0.063 + 0.063 + 0.063 = 0.189

Hit

Hit

~ Hit = 0.063

HitHit = 0.063

~ Hit

~ Hit

Hit = 0.063

Hit

~ Hit

~ HitHit

~ Hit

~ Hit

0.3

0.3

0.3

0.3

0.3

0.3

0.30.7

0.7

0.7

0.7

0.7

0.7

0.7

18

1 71

8 8= − =

1 1 1 72 4 8 8

= + + =

HHH HHT

THH HTT

TTH TTT

HTH

THT


Module 4

8.

P(undetected) = 0.0125

9. Note: H represents hit and M represents miss.

a) P(HH) = P(H by Lady) × P(H by Sharp) =

b) P(HM or MH) =

c) P(at least one hit) = 1 – P(MM) =

10.

P(win) = 0.2 + 0.35 = 0.55

win = 0.2

snow

~ win = 0.3

win = 0.35

~ snow

~ win = 0.15

0.3

0.7

0.4

0.5

0.5

0.6

115

25

2325

− ⋅ =

45

25

15

35

1125

⋅ + ⋅ =

45

35

1225

⋅ =

Detected

Detected

~ Detected

~ Detected = 0.01250.25

0.05 0.75

0.95


Module 4

11. P(7 before 6) = P[7 or (not (6 or 7) and 7) or (not (6 or 7) and not (6 or 7) and 7) or (not (6 or 7) and not (6 or 7) and not (6 or 7) and 7) or . . .]

= P(7) + P(not (6 or 7) and 7) + P(not (6 or 7) and not (6 or 7) and 7) + P(not (6 or 7) and not (6 or 7) and not (6 or 7) and 7) + . . .

(from two-dice table)

This is a sum of an infinite geometric sequence with

12. P(at least one match) = 1 – P(no match)

P(at least one match) = 1 – 0.933712 = 0.066288

13. P(TTT) = P(T) · P(T) · P(T) =

(not tails all 3 times)1 (all tails)

11

878

PP= −

= −

=

12

12

12

18

⋅ ⋅ =

25 times

364 364 364 364 3641 . . .365 365 365 365 365

⎛ ⎞= − ⋅ ⋅ ⋅ ⋅⎜ ⎟⎝ ⎠

125 with 36 1

6 6636 36(7 before 6)

25 11 11136 36

tr Sr

P

= =−

= = =−

= + ⋅ + ⋅ ⋅ +

⋅ ⋅ ⋅ +

636

2536

636

2536

2536

636

2536

2536

2536

636

. . .


Module 4

14. P(at least one club) = 1 – P(no clubs)

You have to find the value of n so that

Using logarithms:

Since n is an integer, it follows that n = 3. Therefore, atleast three cards must be selected.

15. P(at least two were born the same month) = 1 – P(all born on different months)

This is an extremely likely event!

12 11 10 11 . . . 1 0.000053723 0.999946

12 12 12 12⎛ ⎞= − ⋅ ⋅ ⋅ ⋅ = − =⎜ ⎟⎝ ⎠

112

34

05 075030103 0124939

2 409

− > FHGIKJ

>− > −

<

n

nn

n

log( . ) log( . ). .

.

134

12

− FHGIKJ >

n

.

3 3 3 3 31 . . . 14 4 4 4 4

n⎛ ⎞ ⎛ ⎞= − ⋅ ⋅ ⋅ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


Module 4

Lesson 6

Answer Key

1. a) P( first and second correct) =

b) P( first, second, and third correct) =

2. Number of ways of placing the red or green books at the endis: __2__ · __3__ · __2__ · __1__ · __1__ = 12.

Number of ways of placing five books on a shelf is: 5! = 120.

P(red and green books at the end) =

3. P(all three are in descending ages) =

4. Treat the men as “like” objects.

P(women are in descending order of age) =

5. P(3 clubs and 2 diamonds) =

6. a)

(You must chose three males and two females.)

b)

c) P(at least one female) = 1 – P(all males)

= − = − =1 1134

3334

10 5

18 5

CC

P

P

C C C C C CC C C

=

⋅ ⋅ ⋅= + +

= + + = = ≈

8 3 10 2 8 4 10 1 8 5 10 0

18 5 18 5 18 5

(majority are female)

(3F and 2M) or (4F and 1M ) or (5F)

56(45) 70(10) 56(1) 3276 13 0.38248568 8568 8568 8568 34

C CC⋅ = ≈10 3 8 2

18 5

200.392157

51

C CC⋅ =13 3 13 2

52 5

0.008583

=

7!13!4!

7! 243!

1 13! 6

=

12 1120 10

=

3 3

9 3

6 1504 84

= =PP

PP

= =2 2

9 2

2 172 36


Module 4

7. All possible permutations:

All counts when Ns are together:

P(Ns are together) =

8. All possible permutations:

a) P(As together) =

b) P(As and Ns together) =

c) P(Ns not together) = 1 – P(Ns together)

9. If the boy and his girlfriend are on the committee, then onemore boy, out of the remaining four boys, and two more girls,out of the remaining five girls, are required to complete thecommittee.

P( both are on the committee) =

10. Number of ways of selecting five cards: 52C5 = 2 598 960

a) P(exactly 4 hearts and 1 non-heart)

b) P(at least 4 hearts) = P(4 hearts) + P(5 hearts)

= 0.010729 +

= 0.010729 + 0.000495 = 0.011224 [= 11/980]

c) P(no hearts) = 39 5 575 757 21090.221534 or 2 598 960 2 598 960 9520

C = =

13 5

2 598 960C

C C ⎡ ⎤⋅= = = =⎢ ⎥⎣ ⎦

13 4 39 1 27 885 1430.0107292 598 960 2 598 960 13 328

C CC C

⋅ = =⋅

4 1 5 2

5 2 6 3

4(10) 1200 5

5!20 23!1 1

60 60 3= − = − =

3! 160 10

=

4!12 12!

60 60 5= =

6!60

3!2!=

2 520 110 080 4

=

7!2520

2!=

8!10 080

2!2!=


Module 4

Lesson 7

Answer Key

1.

2. P(4 threes and 3 non-threes) =

3. a) P(3H and 2T) =

b) P(at least 3H) = P(3H or 4H or 5H) =

c) P(at most 3H) = P(3H or 2H or 1H or 0H) =

4. (b) + (c) = The reason is that P(3 heads) =

is contained in both cases. If this count is removed from this

sum, i.e., the sum of all possibilities

reappears.

5. P(6 hits in 10 trials) =6 410! 3 2 0.250823

6!4! 5 5⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2116

516

1616

1− = = ,

516

12

1316

2116

1+ = ≠ .

5 5 5 1 26 1316 16 32 32 32 16

+ + + = =

4 1 5 05 5! 1 1 5! 1 1 116 4!1! 2 2 5!0! 2 2 2

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

3 25! 1 1 53!2! 2 2 16

⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

4 37! 1 5 0.0156294!3! 6 6

⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )

1 3

3 1

4 0

4! 1 1 1a) (1H and 3T)1!3! 2 2 4

4! 1 1 1b) (3H and 1T)3!1! 2 2 4

1 4! 1 1 1 1 5c) (3H and 1T) or (4H)4 4!0! 2 2 4 16 16

P

P

P

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


Module 4

6.

7. P(2T and 3H) =

The results are the same, since any toss of a coin is independent of the previous tosses.

8.

9.

10.

20

5 15

1 2(correct) . Therefore, (~correct) .3 3

2a) 0.00033

20! 1 2b) 0.1457035!15! 3 3

c) P(10) + P(11) + P(12) + P(20) = 1-bino mcdf (20,1/3,9) = 0.9190

= =

⎛ ⎞ =⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

P P

P P

P

= =

⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞− = − =⎜ ⎟⎝ ⎠

4 1

5

5 3(black) . Therefore, (~black) .8 8

5! 5 3a) 0.2861024!1! 8 8

5b) 1 (5B) 1 0.9046338

P P

P

= = =

⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= + = =⎜ ⎟⎝ ⎠

4 1

5

13 1 3(spade) . Therefore, (~spade) .52 4 4

5! 1 3 15a)4!1! 4 4 1024

15 1 16 1b) (4 spade or 5 spade)1024 4 1024 64

2 35! 1 1 52!3! 2 2 16

⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 2

0 3

6 1 5(7) . Therefore, (~7) .36 6 6

3! 1 5 25a)

1!2! 6 6 72

3! 1 5 125 91b) 1 (no 7s) 1 10!3! 6 6 216 216

125c) (no 7s) , as part (b) above216

P P

P

P

= = =

⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞− = − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=


Module 4

Review

Answer Key

1. P(king) + P(black) – P(black king) =

2. a) P(white1)P(white2⎜white1) =

b) P(white1)P(white2⎜white1) =

3. Using the 36 points of the sample space of rolling two diceP(8) + P(9) – 0 (since events are mutually exclusive) =

4. A binomial probability:

Note:

5. P(first student does not achieve an A)

P(second student does not achieve an A)= (1 – 0.12) (1 – 0.12) = (0.88)(0.88) = 0.7744 (These are independent events.)

6.

7. P(penny shows heads|nickel showed heads) = P(penny

shows heads) =

The events are independent since the nickel result has noeffect on how the penny will land.

12

(wins and practiced)(wins|practiced) =

(practiced)4

4 10 8157 15 7 21

10

PPP

= = ⋅ =

⎛ ⎞⎜ ⎟⎝ ⎠

5 3

5means

3C

53

12

12

10132

516

2 3FHGIKJFHGIKJFHGIKJ = FHG

IKJ =

5 4 9 136 36 36 4

+ = =

5 5 259 9 81⋅ =

5 4 59 8 18⋅ =

4 26 2 28 752 52 52 52 13

+ − = =


Module 4

8. a) P(blue|not white) = since the sample space changes

from 10 to 7 points.

b) P(not red|not white) = P(blue|not white) =

as in part (a).

9. P(first box)P(red|first box) + P(second box)P(red|second

box) =

10. a)

b) P(ends in tails) =

11. P(2 defective and 1 non-defective) =

12.

13.

14. P(majority of girls) = P(4 girls and 0 boys) +

P(3 girls and 1 boy) =

15. Binomial probability. 104

25

35

0 25084 6F

HGIKJFHGIKJFHGIKJ = .

( )⋅+ = + =5 4 5 3 4 1

9 4 9 4

10 45 5126 126 14

C C CC C

2 4 48

135

b g ! !!

=

8 29

29

! !!=

2 2 18 1

20 3

1(18) 31140 190

C CC⋅ = =

14

heads

heads

tails

tails

12

12

12

12

12

12

12

14

FHGIKJ =

12

12

14

FHGIKJ =

1 5 1 5 252 9 2 6 36⋅ + ⋅ =

27

,

27

,


Module 4

Assume that A and T are one letter and permute the remaining8 letters as 8! A and T could be on either side of one another, soyou must multiply by 2 (or 2! which is actually 2) and divide bythe total number of ways for 9 letters (9!).

module 4: permutations, combinations, and...

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