SEQUENCES AND SERIES

Precalculus 1
Chapter 10

1
▪ This Slideshow was developed to accompany the textbook
▪ Precalculus
▪ By Richard Wright
▪ https://www.andrews.edu/~rwright/Precalculus-RLW/Text/TOC.html
▪ Some examples and diagrams are taken from the textbook.

Slides created by
Richard Wright, Andrews Academy
rwright@andrews.edu

2
3
10-01 SEQUENCES
In this section, you will:
• Write a sequence from a rule.
• Write an explicit rule for a sequence.
• Write a recursive rule for a sequence.
• Simplify factorial expressions.

3
10-01 SEQUENCES
▪ Sequence
▪ List of numbers following a rule
▪ 0, 3, 6, 9, 12 <− finite (ends)
▪ 0, 3, 6, 9, 12, … <− infinite (doesn’t end)

▪ 𝑛 = 1, 2, 3, 4, 5, … (term #) like x
▪ 𝑎𝑛 = 0, 3, 6, 9, 12, … (term value) like y

4
10-01 SEQUENCES
▪ Find the 1st 5 terms of 𝑎𝑛 = 5 + 2𝑛 −1 𝑛

𝑎1 = 5 + 2 1 −1 1 = 3
𝑎2 = 5 + 2 2 −1 2 = 9
𝑎3 = 5 + 2 3 −1 3 = −1
𝑎4 = 5 + 2 4 −1 4 = 13
𝑎5 = 5 + 2 5 −1 5 = −5

5
10-01 SEQUENCES
▪ Write the rule for the nth term. ▪ 2, -9, 28, -65, 126, …
1, 5, 9, 13, 17, …

𝑎𝑛 = 4𝑛 − 3
𝑛+1
𝑎𝑛 = −1 (𝑛3 + 1)

6
10-01 SEQUENCES
▪ Recursive Rules ▪ Find the first 5 terms.
▪ Use the value of one term to ▪ 𝑎1 = 6
find the next term. ▪ 𝑎𝑛 = 𝑎𝑛−1 + 1

▪ 𝑎𝑛 means current term

▪ 𝑎𝑛−1 means previous term

𝑎1 = 6
𝑎2 = 𝑎1 + 1 = 6 + 1 = 7
𝑎3 = 𝑎2 + 1 = 7 + 1 = 8
𝑎4 = 𝑎3 + 1 = 8 + 1 = 9
𝑎5 = 𝑎4 + 1 = 9 + 1 = 10

7
10-01 SEQUENCES
▪ Factorial (!) ▪ Simplify
9!
▪ Product of a whole number
3!7!
with all the whole numbers less
than it through 1.
▪ 6! = 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1
▪ 5! = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1
▪ 0! = 1

9!
3! 7!
9∙8∙7∙6∙5∙4∙3∙2∙1
3∙2∙1∙7∙6∙5∙4∙3∙2∙1
9∙8
3∙2∙1
72
6
12

8
10-01 SEQUENCES
▪ Simplify
𝑛+1 !
𝑛!

𝑛+1 !
𝑛!
𝑛+1 𝑛 𝑛−1 𝑛−2 ⋯
𝑛 𝑛−1 𝑛−2 ⋯
𝑛+1

9
10
10-02 SERIES
In this section, you will:
• Evaluate a summation.
• Write a series as a summation.

10
10-02 SERIES
▪ Series ▪ Summation Notation
▪ Sum of a sequence ▪ (Sigma Notation)

𝑛
▪ Sequence
෍ 𝑎𝑖 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛
▪ 2, 4, 6, 8
𝑖=1
▪ Series
▪2 + 4 + 6 + 8

11

i = index
1 = lower limit
n = upper limit

11
10-02 SERIES
5
▪ Find each sum
4
෍ 2 + 𝑘3
෍(4𝑖 + 1) 𝑘=2
𝑖=1

12

෍(4𝑖 + 1) = 4 1 + 1 + 4 2 + 1 + 4 3 + 1 + 4 4 + 1
𝑖=1
= 5 + 9 + 13 + 17 = 44

෍ (2 + 𝑘 3 ) = 2 + 23 + 2 + 33 + 2 + 43 + 2 + 53
𝑘=2
= 10 + 29 + 66 + 27 = 232

12
10-02 SERIES
∞
5
෍
10𝑛
𝑛=1

13

∞
5
෍
10𝑛
𝑛=1
Parital sums
5
= 0.5
101
5 5
1
+ 2 = 0.5 + 0.05 = 0.55
10 10
5 5 5
+ + = 0.5 + 0.05 + 0.005 = 0.555
101 102 103
5 5 5 5
+ + + = 0.5 + 0.05 + 0.005 + 0.0005 = 0.5555
101 102 103 104
Approaches
5
0.55555555555 … =
9

13
10-02 SERIES
▪ Shortcut formulas
𝑛

1 +1 + 1 + 1 + ⋯ = ෍1 = 𝑛
𝑛 𝑖=1
𝑛 𝑛+1
1 + 2 + 3 + 4 + ⋯ = ෍𝑖 =
2
𝑛 𝑖=1
𝑛 𝑛 + 1 2𝑛 + 1
1 + 4 + 9 + 16 + ⋯ = ෍ 𝑖 2 =
6
𝑖=1

14

14
10-02 SERIES
𝑛
𝑛2 𝑛 + 1 2
3
1 + 8 + 27 + 64 + ⋯ = ෍ 𝑖 =
4
𝑛 𝑖=1
4
𝑛 𝑛 + 1 2𝑛 + 1 3𝑛2 + 3𝑛 − 1
1 + 16 + 81 + 256 + ⋯ = ෍ 𝑖 =
30
𝑖=1 𝑛
5
𝑛2 𝑛 + 1 2
2𝑛2 + 2𝑛 − 1
1 + 32 + 243 + 1024 + ⋯ = ෍ 𝑖 =
12
𝑖=1

15

15
10-02 SERIES
▪ Evaluate
5

෍ 3𝑖 2 − 5𝑖
𝑖=1

16

5 5
2
෍ 3𝑖 − ෍ 5𝑖
𝑖=1 𝑖=1
𝑛 𝑛 + 1 2𝑛 + 1 𝑛 𝑛+1
=3 −5
6 2
5 5+1 2 5 +1 5 5+1
=3 −5
6 2
= 90

16
 10-03 ARITHMETIC
17
SEQUENCES AND
SERIES
In this section, you will:
• Write the explicit rule for an arithmetic sequence.
• Write the recursive rule for an arithmetic sequence.
• Evaluate the sum for an arithmetic series.

17
10-03 ARITHMETIC SEQUENCES AND
SERIES
▪ Arithmetic ▪ Rule for the nth term
▪ Common difference (d) ▪ 𝑎𝑛 = 𝑑𝑛 + 𝑐
▪ Where 𝑐 = 𝑎1 − 𝑑
▪ 3, 7, 11, 15, 19, …
▪ 𝑎𝑛 = 𝑎1 + 𝑛 − 1 𝑑

18

18
10-03 ARITHMETIC SEQUENCES AND
SERIES
▪ Find the rule for the nth term for 3, 7, 11, 15, 19, …

19

d = 7 – 3 = 4; 11 – 7 = 4, 15 – 11 = 4; 19 – 15 = 4
a1 = 3
𝑎𝑛 = 𝑎1 + 𝑛 − 1 𝑑
𝑎𝑛 = 3 + 𝑛 − 1 4
𝑎𝑛 = 3 + 4𝑛 − 4
𝑎𝑛 = 4𝑛 − 1

19
10-03 ARITHMETIC SEQUENCES AND
SERIES
▪ The 8th term of an arithmetic sequence is 25, and the 12th term is 41.
Write the rule for the nth term.

20

𝑎𝑛 = 𝑎1 + 𝑛 − 1 𝑑
Fill in 𝑎8 = 25
25 = 𝑎1 + 8 − 1 𝑑
25 = 𝑎1 + 7𝑑
Fill in 𝑎12 = 41
41 = 𝑎1 + 12 − 1 𝑑
41 = 𝑎1 + 11𝑑
Solve system of eq.
−25 = −𝑎1 − 7𝑑
41 = 𝑎1 + 11𝑑
16 = 4𝑑
4=𝑑

25 = 𝑎1 + 7 4
𝑎1 = −3

𝑎𝑛 = −3 + 𝑛 − 1 4
𝑎𝑛 = 4𝑛 − 7

20
10-03 ARITHMETIC SEQUENCES AND
SERIES
▪ Recursive Rule for Arithmetic Sequences

▪ 𝑎1 = 𝑎1
▪ 𝑎𝑛 = 𝑎𝑛−1 + 𝑑

21

21
10-03 ARITHMETIC SEQUENCES AND
SERIES
▪ Arithmetic Series
▪ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

▪1 + 3+ 5+ 7+ 9
▪ 19 + 17 + 15 + 13 + 11
▪ 20 + 20 + 20 + 20 + 20 = 5 20 = 100

𝑛
▪ 𝑆𝑛 = 𝑎1 + 𝑎𝑛
2
22

22
10-03 ARITHMETIC SEQUENCES AND
SERIES
▪ Find the sum of the integers 1 to 57.

23

1 + 2 + 3 + 4 + ⋯ + 57
d = 1; a1 = 1; an = 57
𝑛
𝑆𝑛 = 𝑎 + 𝑎𝑛
2 1
57
𝑆57 = 1 + 57
2
57
𝑆57 = 58 = 1653
2

23
10-03 ARITHMETIC SEQUENCES AND
SERIES
▪ Find the 50th partial sum of the arithmetic sequence -6, -2, 2, 6, …

24

d = 4; a1 = -6
Find nth term to get an
𝑎𝑛 = 𝑎1 + 𝑛 − 1 𝑑
𝑎𝑛 = −6 + 𝑛 − 1 4
𝑎𝑛 = 4𝑛 − 10
𝑎50 = 4 50 − 10 = 190
Find the sum
𝑛
𝑆𝑛 = (𝑎 + 𝑎𝑛 )
2 1
50
𝑆50 = −6 + 190 = 4600
2

24
10-03 ARITHMETIC SEQUENCES AND
SERIES
▪ Evaluate
100

෍(3𝑖 + 2)
𝑖=1

25

Linear (in form of 𝑎𝑛 = 𝑑𝑛 + 𝑐) so arithmetic

𝑛
𝑆𝑛 = 𝑎 + 𝑎𝑛
2 1
100
𝑆100 = 3 1 + 2 + 3 100 + 2
2
= 50 5 + 302
= 15350

25
 10-04 GEOMETRIC
26
SEQUENCES AND
SERIES
In this section, you will:
• Write the explicit rule for a geometric sequence.
• Write the recursive rule for a geometric sequence.
• Evaluate the sum for a geometric series.

26
10-04 GEOMETRIC SEQUENCES AND SERIES
▪ Geometric 2
▪ Find the rule for 6, −2, , …
3
▪ Common ratio (r)

▪ 1, 3, 9, 27, 81, 243, …

▪ Rule for nth term

▪ 𝑎𝑛 = 𝑎1 𝑟 𝑛−1

27

2 1
𝑟=− =−
6 3
𝑎𝑛 = 𝑎1 𝑟 𝑛−1
𝑛−1
1
𝑎𝑛 = 6 −
3

27
10-04 GEOMETRIC SEQUENCES AND SERIES
▪ The 2nd term of a geometric sequence is -18, the 5th term is 2/3. Find the
rule for the nth term.

28

𝑎𝑛 = 𝑎1 𝑟 𝑛−1
Fill in 𝑎2 = −18
−18 = 𝑎1 𝑟 2−1
−18 = 𝑎1 𝑟
18
𝑎1 = −
𝑟
2
Fill in 𝑎5 = 3
2
= 𝑎1 𝑟 5−1
3
2
= 𝑎1 𝑟 4
3
Substitute
2 18
= − 𝑟4
3 𝑟
2
= −18𝑟 3
3
1
− = 𝑟3
27

28
 1
− =𝑟
3
Substitute
18
− = 𝑎1
1
−3
𝑎1 = 54

𝑛−1
1
𝑎𝑛 = 54 −
3

28
10-04 GEOMETRIC SEQUENCES AND SERIES
▪ Geometric Series ▪ Evaluate
7
1−𝑟 𝑛
▪ 𝑆𝑛 = 𝑎1 ෍ 2𝑛−1
1−𝑟
𝑛=1
𝑎1
▪ 𝑆∞ =
1−𝑟
▪ Where 𝑟 < 1

29

2𝑛−1 is in the form 𝑎1 𝑟 𝑛−1 where 𝑎1 = 1 and 𝑟 = 2

1 − 𝑟𝑛
𝑆𝑛 = 𝑎1
1−𝑟
1 − 27
𝑆7 = 1 = 127
1−2

29
10-04 GEOMETRIC SEQUENCES AND SERIES
∞ 𝑛
▪ Evaluate 1
5 + 0.5 + 0.05 + 0.005 + ⋯ ෍5
2
𝑛=0

30

1
𝑎1 = 5 and 𝑟 = 10
𝑎1
𝑆∞ =
1−𝑟
5
=
1
1 − 10
5 10 50
= =5 =
9 9 9
10

The lower limit is not 1, so calculate n = 0

0 ∞ 𝑛
1 1
5 +෍5
2 2
𝑛=1
The exponent is not n-1, so multiply and divide by ½
∞ 1 𝑛
1 2
5+ ෍5
2 1
𝑛=1 2

30
 ∞ 𝑛−1
5 1
5+ ෍
2 2
𝑛=1
5 1
𝑎1 = 2 and 𝑟 = 2
𝑎1
𝑆∞ =
1−𝑟
5
5+ 2
1
1−2
5
5+ 2
1
2
5 + 5 = 10

30
 10-05
31
MATHEMATICAL
INDUCTION
In this section, you will:
• Write a proof for a sum formula using mathematical induction.
• Prove other mathematical statements using mathematical induction.

31
10-05 MATHEMATICAL INDUCTION
▪ Proofs for sum formulas
▪ Show it works when n = 1
▪ Show it works for n + 1

▪ Steps
1. Show it works for n = 1
2. Assume formula works for 𝑛 = 𝑘
3. Show it works for 𝑛 = 𝑘 + 1
▪ If proving sum formula use 𝑆𝑘+1 = 𝑆𝑘 + 𝑎𝑘+1
32

32
10-05 MATHEMATICAL INDUCTION
▪ Prove 5 + 7 + 9 + 11 + 13 + ⋯ + 3 + 2𝑛 = 𝑛 𝑛 + 4

33

This is in the form 𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛 = 𝑆𝑛 , so

𝑎𝑛 = 3 + 2𝑛
𝑆𝑛 = 𝑛 𝑛 + 4
1. 𝑆1 = 1 1 + 4 = 5 = 𝑎1
2. Assume 𝑆𝑘 = 𝑘 𝑘 + 4
3. 𝑆𝑘+1 = 𝑆𝑘 + 𝑎𝑘+1
𝑘+1 𝑘+1 +4 = 𝑘 𝑘+4 + 3+2 𝑘+1
𝑘 + 1 𝑘 + 5 = 𝑘 2 + 4𝑘 + 3 + 2𝑘 + 2
𝑘 2 + 6𝑘 + 5 = 𝑘 2 + 6𝑘 + 5

33
10-05 MATHEMATICAL INDUCTION
𝑛 𝑛−1 𝑛+1
▪ Prove 1 1 − 1 + 2 2 − 1 + 3 3 − 1 + ⋯ + 𝑛 𝑛 − 1 =
3

34

This is in the form 𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛 = 𝑆𝑛 , so

𝑎𝑛 = 𝑛(𝑛 − 1)
𝑛 𝑛−1 𝑛+1
𝑆𝑛 =
3
1 1−1 1+1
1. 𝑆1 = 3
= 0 = 𝑎1
𝑘 𝑘−1 𝑘+1
2. Assume 𝑆𝑘 = 3
3. 𝑆𝑘+1 = 𝑆𝑘 + 𝑎𝑘+1
𝑘+1 𝑘+1 −1 𝑘+1 +1 𝑘 𝑘−1 𝑘+1
= + 𝑘+1 𝑘+1 −1
3 3
𝑘+1 𝑘 𝑘+2 𝑘3 − 𝑘
= + 𝑘2 + 𝑘
3 3
𝑘 3 + 3𝑘 2 + 2𝑘 𝑘 3 − 𝑘 3𝑘 2 3𝑘
= + +
3 3 3 3
𝑘 3 + 3𝑘 2 + 2𝑘 𝑘 3 + 3𝑘 2 + 2𝑘
=
3 3

34
10-05 MATHEMATICAL INDUCTION
▪ Prove 𝑛 + 1 ! > 2𝑛 where 𝑛 ≥ 2

35

1. 𝑛 = 2: 2 + 1 ! > 22
• 3! > 4
• 6>4
2. Assume 𝑘 + 1 ! > 2𝑘
3. Show 𝑛 = 𝑘 + 1: 𝑘 + 1 + 1 ! > 2𝑘+1
• 𝑘 + 2 ! > 2𝑘+1
• 𝑘 + 2 𝑘 + 1 ! > 2𝑘 ∙ 2
• 𝑘 + 2 > 2 always and 𝑘 + 1 ! > 2𝑘 was true from step 2, so whole
thing true.

35
10-05 MATHEMATICAL INDUCTION
▪ Prove 4 is a factor of 5𝑛 − 1

36

1. 𝑛 = 1: 51 − 1 = 4; 4 is a factor of 4
2. Assume 4 is a factor of 5𝑘 − 1
3. 𝑛 = 𝑘 + 1
• 5𝑘+1 − 1
• 5𝑘+1 − 5𝑘 + 5𝑘 − 1
• 5𝑘 ∙ 5 − 5𝑘 + 5𝑘 − 1
• 5𝑘 5 − 1 + 5𝑘 − 1
• 4 ∙ 5𝑘 + 5𝑘 − 1
• 4 is factor of both 4 ∙ 5𝑘 and (5𝑘 − 1)

36
37
10-06 BINOMIAL
THEOREM
In this section, you will:
• Evaluate combinations.
• Expand binomial expressions.

37
10-06 BINOMIAL THEOREM
(x + y)0 1
(x + y)1 1x 1y
(x + y)2 1x2 2xy 1y2
(x + y)3 1x3 3x2y 3xy2 1y3
(x + y)4 1x4 4x3y 6x2y2 4xy3 1y4

Properties
1. n + 1 terms 3. Sum of exponents of each term = n
2. Powers of x count down, y’s count up 4. Coefficients are symmetrical
𝑛𝐶𝑟
38

Rows are n starting at 0

Down to left diagonals are r starting at 0
nCr is the coefficient of the row and diagonal

38
10-06 BINOMIAL THEOREM
▪ Binomial theorem
𝑛
𝑛 11
𝑎+𝑏 = ෍ 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏𝑟 ▪
4
𝑟=0
𝑛!
▪ Where 𝑛𝐶𝑟 = 8
𝑛−𝑟 !𝑟! ▪
8
▪ Evaluate
4
▪ 9𝐶2 ▪
2

39

On TI graphing calc: MATH → PRB | nCr

9𝐶2
Type 9 nCr 2 = 36
11
= 11C4 = 330
4

8
= 8𝐶8 = 1
8

4
= 4𝐶2 = 6
2

39
10-06 BINOMIAL THEOREM
▪ Expand 𝑥 + 2 4

40

a = x; b = 2; n = 4
𝑛

෍ 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟
𝑟=0
4 0
4𝐶0 𝑥 2 + 4𝐶1 𝑥 2 + 4𝐶2 𝑥 2 22 + 4𝐶3 𝑥1 23 + 4𝐶4 𝑥 0 24
3 1

1 ∙ 𝑥 ∙ 1 + 4 ∙ 𝑥 3 ∙ 2 + 6 ∙ 𝑥 2 ∙ 4 + 4 ∙ 𝑥 ∙ 8 + 1 ∙ 1 ∙ 16
4

𝑥 4 + 8𝑥 3 + 24𝑥 2 + 32𝑥 + 16

40
10-06 BINOMIAL THEOREM
▪ Expand 3 − 𝑥 2 5

41

a = 3; b = −𝑥 2 ; n = 5
𝑛

෍ 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟
𝑟=0
5 2 0 4 2 1
5𝐶0 3 −𝑥 + 5𝐶1 3 −𝑥 + 5𝐶2 33 −𝑥 2 2 + 5𝐶3 32 −𝑥 2 3 + 5𝐶4 31 −𝑥 2 4

+ 5𝐶5 30 −𝑥 2 5
1 ∙ 243 ∙ 1 + 5 ∙ 81 ∙ −𝑥 2 + 10 ∙ 27 ∙ 𝑥 4 + 10 ∙ 9 ∙ −𝑥 6 + 5 ∙ 3 ∙ 𝑥 8 + 1 ∙ 1
∙ −𝑥10
243 − 405𝑥 2 + 270𝑥 4 − 90𝑥 6 + 15𝑥 8 − 𝑥10

41
10-06 BINOMIAL THEOREM
▪ Find the coefficient of the term 𝑎 4 𝑏 7 in 2𝑎 − 3𝑏 11

42

a = 2a; b = -3b; n = 11
𝑛

෍ 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟
𝑟=0
Compare this to 𝑎4 𝑏 7 to see that r = 7
11−7
11𝐶7 2𝑎 −3𝑏 7
330 16𝑎4 −2187𝑏 7
−11,547,360𝑎4 𝑏 7
-11,547,360

42
43
10-07 COUNTING
PRINCIPLES
In this section, you will:
• Apply the fundamental counting principle
• Calculate permutations
• Calculate combinations

43
10-07 COUNTING PRINCIPLES
▪ Fundamental Counting Principle
▪ If events E1 and E2 occur in m1 and m2 ways, the number of ways both
events can occur is m1∙m2.

▪ A lock will open with the right choice of 3 numbers. How many different
sets of 3 numbers can you choose if each number is from 1 to 30
inclusive? (a) with repetition (b) without repetition

44

a. 30∙30∙30 = 27000
b. 30∙29∙28 = 24360

44
10-07 COUNTING PRINCIPLES
▪ How many license plates can be made if each is 2 letters follow by 4-
digits? (a) with repetition (b) without repetition

45

a. 26∙26∙10∙10∙10∙10 = 6,760,000
b. 26∙25∙10∙9∙8∙7 = 3,276,000

45
10-07 COUNTING PRINCIPLES
▪ Permutation
▪ Number of ways to order n objects taken r at a time
𝑛!
▪ 𝑛𝑃𝑟 =
𝑛−𝑟 !

▪ How many ways can 8 children line up in a row?

46

n = 8; r = 8
8!
8𝑃8 = = 8! = 40,320
8−8 !

46
10-07 COUNTING PRINCIPLES
▪ A club has 24 members, how many ways can 5 officers be selected?

47

Choose from 24 → n=24

Actually choosing 5 → r=5
24! 24 ∙ 23 ∙ 22 ∙ 21 ∙ 20 ∙ 19 ∙ 18 ∙ 17 ∙ 16 ⋯
24𝑃5 = =
24 − 5 ! 19 ∙ 18 ∙ 17 ∙ 16 ⋯
= 24 ∙ 23 ∙ 22 ∙ 21 ∙ 20
= 5,100,480

47
10-07 COUNTING PRINCIPLES
▪ Distinguishable Permutations
▪ What is some objects are exactly the same?
▪ ABB
▪ BAB and BAB are the same eventhough the B’s were switched

▪ We want the orders that look different (choosing all the objects)
𝑛!
𝑞1 ! ∙ 𝑞2 ! ∙ 𝑞3 ! ⋯
▪ Where n = number of objects; q = how many times each is repeated

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10-07 COUNTING PRINCIPLES
▪ How many distinguishable ways to order the letters in BANANA?

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n=6; q1=3 (A); q2=2 (N)

6!
= 60
3! ∙ 2!

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10-07 COUNTING PRINCIPLES
▪ Combinations
▪ Grouping of objects without order
▪ ABC is the same as BAC
𝑛!
𝑛𝐶𝑟 =
𝑛 − 𝑟 ! 𝑟!

▪ There are 31 students. How many different groups of 4 can be made?

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31𝐶4 = 31,465

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10-07 COUNTING PRINCIPLES
▪ You are forming a 10-person committee from 9 women and 12 men. How
many different committees if 5 women and 5 men?

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Order is not important → combination

Combination of women × combination of men

9𝐶5 ∙ 12𝐶5 = 99,792

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52
10-08 PROBABILITY
In this section, you will:
• Calculate simple probability
• Calculate the probability of compound events
• Calculate the probability of multiple events
• Calculate the complement of an event

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10-08 PROBABILITY
▪ Probability
▪ Number from 0 to 1 indicating how likely something is to happen.
▪ 0 = never happens
▪ 1 = always happens

𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑃 𝐴 =
𝑡𝑜𝑡𝑎𝑙 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

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10-08 PROBABILITY
▪ A box contains 3 red marbles, 5 black marbles, and 2 yellow marbles. If a
marble is selected at random, what is the probability of choosing yellow?

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𝑦𝑒𝑙𝑙𝑜𝑤 2 1
𝑃 𝑌 = = =
𝑡𝑜𝑡𝑎𝑙 10 5

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10-08 PROBABILITY
▪ 2 dice are rolled, what is the probability that the sum is 8?

11 21 31 41 51 61
12 22 32 42 52 62
13 23 33 43 53 63
14 24 34 44 54 64
15 25 35 45 55 65
16 26 36 46 56 66

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8′ 𝑠 5
𝑃 8 = =
𝑡𝑜𝑡𝑎𝑙 36

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10-08 PROBABILITY
▪ Compound Events
▪ 1 event with 2 accepted outcomes

▪𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵

▪ If 𝑃 𝐴 ∩ 𝐵 = 0, then called mutually

exclusive

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10-08 PROBABILITY
▪ You draw one card from a standard 52-card deck. What is the probability
of drawing a heart or red?

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𝑃 ℎ𝑒𝑎𝑟𝑡 ∪ 𝑟𝑒𝑑 = 𝑃 ℎ𝑒𝑎𝑟𝑡 + 𝑃 𝑟𝑒𝑑 − 𝑃 ℎ𝑒𝑎𝑟𝑡 ∩ 𝑟𝑒𝑑

13 26 13 26 1
= + − = =
52 52 52 52 2

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10-08 PROBABILITY
▪ Multiple Events
▪ Two events with 2 outcomes
▪ Independent – Event A does not affect event B
▪ 𝑃 𝐴 𝑎𝑛𝑑 𝐵 = 𝑃 𝐴 ∙ 𝑃 𝐵

▪ Dependent – Event A does affect event B

▪ 𝑃 𝐴 𝑎𝑛𝑑 𝐵 = 𝑃 𝐴 ∙ 𝑃 𝐵 𝐴
▪ Where 𝑃(𝐵|𝐴) is the probability that B occurs given that A already
occured
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10-08 PROBABILITY
▪ You draw 2 cards from a standard 52-card deck. What is the probability
you draw a heart and a red? (a) with replacement (b) without
replacement

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13 56 1
a. 𝑃 ℎ𝑒𝑎𝑟𝑡 𝑎𝑛𝑑 𝑟𝑒𝑑 = 𝑃 ℎ𝑒𝑎𝑟𝑡 ∙ 𝑃 𝑟𝑒𝑑 = 52 ∙ 52 = 8 = 0.125
13 25
b. 𝑃 ℎ𝑒𝑎𝑟𝑡 𝑎𝑛𝑑 𝑟𝑒𝑑 = 𝑃 ℎ𝑒𝑎𝑟𝑡 ∙ 𝑃 𝑟𝑒𝑑 ℎ𝑒𝑎𝑟𝑡 = 52 ∙ 52 = 0.123

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10-08 PROBABILITY
▪ Complement
▪ Opposite

▪𝑃 𝐴 = 1 − 𝑃 𝐴

▪ At least one P(n ≥ 1) is easier with the complement never 𝑃(0)

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