Transcript Chapter 4

Chapter 4
Probability and Counting Rules
Section 4-1
Sample Spaces and
Probability
Learning Target
• Determine sample spaces and find the
probability of an event using classical
probability or empirical probability.
Basic Concepts
• Probability Experiment – a chance process that
leads to well-defined results called outcomes
• Outcome – the result of a single trial of a
probability experiment
• Sample Space – the set of all possible outcomes
of a probability experiment
– Examples: tossing a coin (head, tail), roll a die
(1,2,3,4,5,6), Answer a true/false question (true, false)
Sample Space for Rolling Two Dice
Die 2
Die 1
1
2
3
4
5
6
1
(1,1)
(2,1)
(3,1)
(4,1)
(5,1)
(6,1)
2
(1,2)
(2,2)
(3,2)
(4,2)
(5,2)
(6,2)
3
(1,3)
(2,3)
(3,3)
(4,3)
(5,3)
(6,3)
4
(1,4)
(2,4)
(3,4)
(4,4)
(5,4)
(6,4)
5
(1,5)
(2,5)
(3,5)
(4,5)
(5,5)
(6,5)
6
(1,6)
(2,6)
(3,6)
(4,6)
(5,6)
(6,6)
Cards in a Regular Deck of Cards
• 4 suits – spades, diamonds, hearts, clubs
• 13 of each suit
• There are 3 face cards in each suit – jack
queen, king
• 52 cards total
Gender of Children
• Find the sample space for the gender of the
children if a family has three children.
• How can this be done?
Tree diagram
• Device consisting of line segments emanating
from a starting point and also from the
outcome point. It is used to determine all
possible outcomes of a probability
experiment.
Outcomes
BBB
BBG
BGB
B
BGG
1st child
2nd child
3rd child
GBB
G
GBG
GGB
GGG
More Vocab
• Event – consists of a set of outcomes of a
probability experiment
• Simple Event – an event with one outcome
• Compound Event – event with more than one
outcome
– Example: the event of rolling an odd number on a
die
Classical Probability
• Uses sample spaces to determine the
numerical probability that an event will
happen
• Assumes that all outcomes are equally likely to
occur
• Formula – number of outcomes in E divided by
the total number of outcomes in the sample
space, 𝑃 𝐸 =
𝑛(𝐸)
𝑛(𝑆)
• Probabilities can be expressed as fractions,
decimals, or percents. Most problems will be
expressed as fractions or decimals. If the
problems starts in fractions the answer should
be a fraction. If the problem starts as a
decimal the answer should be a decimal.
• Fractions should always be reduced and
decimals rounded to two or three decimal
places.
Practice Problems
• A card is drawn at random from an ordinary deck
of cards. Find these probabilities.
–
–
–
–
–
Of getting a jack
Of getting a red ace
Of getting the 6 of clubs
Of getting a 3 or a diamond
Of getting a 3 or a 6
• If a family has three children, what is the
probability that two of the three children are
girls?
Solutions
•
•
•
•
•
•
4
52
2
52
1
52
16
52
8
52
3
8
=
=
=
=
1
13
1
26
4
13
2
13
4 Probability Rules
1. The probability of any event E is a number
(either a fraction or decimal) between and
including 0 and 1. This is denoted by 0 ≤
𝑃(𝐸) ≤ 1.
2. If an event E cannot occur (i.e., the event
contains no members in the sample space), its
probability is 0.
3. If an event E is certain, its probability is 1.
4. The sum of the probabilities of all the outcomes
in the sample space is 1.
Complementary Events
• The complement of an event E is the set of all
outcomes in the sample space that are not
included in the outcomes of event E. The
complement of E is denoted by E (read “E
bar”)
• Example: The event E of getting an odd
number is 1,3,5. The complement of E is
getting an even number (2,4,6).
Practice Problems
• Find the complement of each event.
– Rolling a die and getting a 4
– Selecting a letter of the alphabet and getting a
vowel
– Selecting a month and getting a month that begins
with a J
– Selecting a day of the week and getting a weekday
Solutions
• Getting a 1,2,3,5,6
• Getting a consonant (assume y is a consonant)
• Getting February, March, April, May, August,
September, October, November, or December
• Getting Saturday or Sunday
Rule for Complementary Events
• 𝑃 𝐸 = 1 − 𝑃(𝐸) or 𝑃 𝐸 = 1 − 𝑃(𝐸) or
𝑃 𝐸 +𝑃 𝐸 =1
Empirical Probability
• Relies on actual experience to determine the
likelihood of outcomes
• Given a frequency distribution, the probability
of an event being in a given class is
𝑃 𝐸 =
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐𝑙𝑎𝑠𝑠
𝑡𝑜𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
=
𝑓
𝑛
Practice Problems
• In a sample of 50 people, 21 had type O
blood, 22 had type A blood, 5 had type B
blood, and 2 had type AB blood. Set up a
frequency distribution and find the following
probabilities.
– A person has type O blood
– A person has type A or type B blood
– A person has neither type A nor type O blood
– A person does not have type AB blood
Type
Frequency
A
22
B
5
AB
2
O
21
Solution
Total = 50
A. 𝑃 𝑂 =
𝑓
𝑛
B. 𝑃 𝐴 𝑜𝑟 𝐵
21
=
50
22
=
50
+
5
50
=
C. 𝑃 𝑛𝑒𝑖𝑡ℎ𝑒𝑟 𝐴 𝑛𝑜𝑟 𝑂 =
27
50
5
2
+
20
50
D. 𝑃 𝑛𝑜𝑡 𝐴𝐵 = 1 − 𝑃 𝐴𝐵 = 1
7
=
50
2
48
− =
50
50
=
24
25
Applying the
Concepts and
Exercise 4-1
Section 4-2
The Addition Rules
for Probability
Learning Target
• IWBAT find the probability of compound
events, using the addition rules.
Mutually Exclusive
• Two events are mutually exclusive if they
cannot happen at the same time.
• In other words they have no outcomes in
common.
• Example: getting a 4 and a 6 are mutually
exclusive.
Which ones are mutually exclusive?
a. Getting an odd number and getting an even
number
b. Getting a 3 and getting an odd number
c. Getting a 7 and a jack
d. Getting a club and getting a king
Addition Rule #1
• When two events are mutually
exclusive, the probability that
A or B will occur is
𝑃 𝐴 𝑜𝑟 𝐵 = 𝑃 𝐴 + 𝑃(𝐵)
Practice Problems
• A box contains 3 glazed doughnuts, 4 jelly
doughnuts, and 5 chocolate doughnuts. If a
person selects a doughnut at random, find the
probability that either is a glazed or chocolate
doughnut.
• At a political rally, there are 20 republicans, 13
democrats, and 6 independents. If a person is
selected at random, find the probability that he
or she is either a democrat or an independent.
Answers
• 𝑃 𝑔𝑙𝑎𝑧𝑒𝑑 𝑜𝑟 𝑐ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒 = 𝑃 𝑔𝑙𝑎𝑧𝑒𝑑 +
3
5
8
2
𝑃 𝑐ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒 =
+ = =
12
12
12
3
• 𝑃 𝑑𝑒𝑚𝑜𝑐𝑟𝑎𝑡 𝑜𝑟 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 =
𝑃 𝑑𝑒𝑚𝑜𝑐𝑟𝑎𝑡 + 𝑃 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 =
13
6
19
+ =
39
39
39
Addition Rule #2
• If A and B are not mutually
exclusive, then 𝑃 𝐴 𝑜𝑟 𝐵 =
𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 𝑎𝑛𝑑 𝐵)
Practice Problems
• In a hospital unit there are 8 nurses and 5
physicians; 7 nurses and 3 physicians are
females. If a staff person is selected, find the
probability that the subject is a nurse or a
male.
Answer
Staff
Females
Males
Total
Nurses
7
1
8
Physicians
3
2
5
Total
10
3
13
𝑃 𝑛𝑢𝑟𝑠𝑒 𝑜𝑟 𝑚𝑎𝑙𝑒 = 𝑃 𝑛𝑢𝑟𝑠𝑒 + 𝑃 𝑚𝑎𝑙𝑒 − 𝑃 𝑛𝑢𝑟𝑠𝑒 𝑎𝑛𝑑 𝑚𝑎𝑙𝑒
8
3
1
10
=
+
−
=
13 13 13 13
For 3 events
• Mutually Exclusive
– 𝑃 𝐴 𝑜𝑟 𝐵 𝑜𝑟 𝐶 = 𝑃 𝐴 + 𝑃 𝐵 + 𝑃 𝐶
• Not Mutually Exclusive
– 𝑃 𝐴 𝑜𝑟 𝐵 𝑜𝑟 𝐶 = 𝑃 𝐴 + 𝑃 𝐵 + 𝑃 𝐶 −
𝑃 𝐴 𝑎𝑛𝑑 𝐵 − 𝑃 𝐴 𝑎𝑛𝑑 𝐶 − 𝑃 𝐵 𝑎𝑛𝑑 𝐶 +
𝑃(𝐴 𝑎𝑛𝑑 𝐵 𝑎𝑛𝑑 𝐶)
Venn Diagrams
P(A)
P(B)
P(A and B)
Mutually Exclusive
P(A)
P(B)
Not Mutually Exclusive
Exercises 4-2
1-25 odd and #8
• Graded for correct answer
– 2, 6, 10, 14, 18, 20, 24, 26
Section 4-3
The Multiplication Rules and
Conditional Probability
Learning Target
• IWBAT find the probability of compound
events, using the multiplication rule.
Multiplication Rules
• The multiplication rules are used to find the
probability of events that happen in
sequence.
• For example, when you toss a coin and roll a
die, you can find the probability of flipping a
head and rolling a 4.
• The events are independent since the
outcome of the first event does not effect the
second.
Rule #1
• When two events are
independent, the probability of
both occurring is
𝑃 𝐴 𝑎𝑛𝑑 𝐵 = 𝑃 𝐴 ∙ 𝑃(𝐵)
Examples
• A coin is flipped and a die is rolled. Find the
probability of getting a head on the coin and a
4 on the die.
• A card is drawn from a deck and replaced;
then a second card is drawn. Find the
probability of getting a queen then an ace.
More Examples
• An urn contains 3 red marbles, 2 blue marbles,
and 5 white marbles. A marble is selected and
its color noted. Then it is replaced. A second
ball is selected and its color noted. Find the
probability of each of these.
– Selecting 2 blue marbles
– Selecting 1 blue marble then 1 white marble
– Selecting 1 red marble then 1 white marble
Solutions
• 𝑃 ℎ𝑒𝑎𝑑 𝑎𝑛𝑑 4 = 𝑃 ℎ𝑒𝑎𝑑 ∙ 𝑃 4 =
1 1
∙
2 6
• 𝑃 𝑞𝑢𝑒𝑒𝑛 𝑎𝑛𝑑 𝑎𝑐𝑒 = 𝑃 𝑞𝑢𝑒𝑒𝑛 ∙ 𝑃 𝑎𝑐𝑒
4
52
• 𝑃
• 𝑃
• 𝑃
=
16
2704
=
1
169
2
2
4
1
𝑏𝑙𝑢𝑒 𝑎𝑛𝑑 𝑏𝑙𝑢𝑒 =
∙ =
=
10 10
100
25
2
5
10
1
𝑏𝑙𝑢𝑒 𝑎𝑛𝑑 𝑤ℎ𝑖𝑡𝑒 = ∙ =
=
10 10
100
10
3
2
6
3
𝑟𝑒𝑑 𝑎𝑛𝑑 𝑏𝑙𝑢𝑒 = ∙ =
=
10 10
100
50
1
=
12
4
= ∙
52
The multiplication rule can be
extended to three or more events by
using the formula…
𝑃 𝐴 𝑎𝑛𝑑 𝐵 𝑎𝑛𝑑 𝐶 𝑎𝑛𝑑 … 𝑎𝑛𝑑 𝐾
= 𝑃 𝐴 ∙ 𝑃 𝐵 ∙ 𝑃 𝐶 ∙ ⋯ ∙ 𝑃(𝐾)
Dependent Events
When the outcome or occurrence of the first
event effects the outcome or the occurrence of
the second event in such a way that a
probability is changed, the events are said to be
dependent.
When situations involve not replacing the item
that was selected first, the events are
dependent.
Conditional Probability
• The conditional probability of an event B in
relationship to an event A is the probability
that event B occurs after event A has already
occurred.
Rule #2
• When two events are dependent, the
probability of both occurring is
𝑃 𝐴 𝑎𝑛𝑑 𝐵 = 𝑃(𝐴) ∙ 𝑃(𝐵 𝐴)
Means that B happens
given that A happened first.
(Conditional Probability)
Examples
• Three cards are drawn from an ordinary deck
and not replaced. Find the probability of these
events.
– Getting 3 jacks
– Getting an ace, a king, and a queen in order
– Getting a club, a spade, and a heart in order
– Getting 3 clubs
Solutions
• 𝑃 3 𝑗𝑎𝑐𝑘𝑠 =
4
3
2
∙ ∙
52 51 50
• 𝑃 𝑎𝑐𝑒 𝑎𝑛𝑑 𝑘𝑖𝑛𝑔 𝑎𝑛𝑑
64
132600
=
8
16575
24
1
=
=
132600
5525
4
4
4
𝑞𝑢𝑒𝑒𝑛 = ∙ ∙
52 51 50
• 𝑃 𝑐𝑙𝑢𝑏 𝑎𝑛𝑑 𝑠𝑝𝑎𝑑𝑒 𝑎𝑛𝑑 ℎ𝑒𝑎𝑟𝑡 =
2197
132600
• 𝑃 3
169
=
10200
13 12 11
𝑐𝑙𝑢𝑏𝑠 = ∙ ∙
52 51 50
=
1716
132600
=
13 13 13
∙ ∙
52 51 50
=
11
850
=
Finding Conditional Probability
The probability that the second event B occurs
given that the first event A has occurred can be
found by dividing the probability that both
events occurred by the probability that the first
event has occurred. The formula is
𝑃(𝐴 𝑎𝑛𝑑 𝐵)
𝑃 𝐵𝐴 =
𝑃(𝐴)
Example
• A box contains black chips and white chips. A
person selects 2 chips without replacement. If
the probability of selecting a black chip and a
15
white chip is , and the probability of
56
3
is
8
selecting a black chip on the first draw
,
find the probability of selecting the white chip
on the second draw, given that the first chip
selected was a black chip.
Solution
•𝑃 𝑊 𝐵
15 56
3 8
=
𝑃(𝐵 𝑎𝑛𝑑 𝑊)
=
=
𝑃(𝐵)
15
3
15 8
5
÷ = ∙ =
56
8
56 3
7
Example 2
• A recent survey asked 100 people if they thought women
in the armed forces should be permitted to participate in
combat. The results of the survey are shown.
Gender
Yes
No
Total
Male
32
18
50
Female
8
42
50
Total
40
60
100
• Find the probability
– The respondent answered yes, given that the respondent
was female.
– The respondent was a male, given that the respondent
answered no.
Solutions
• Before you start, determine a variable to
represent each outcome.
Let
M=respondent was male
F=respondent was female
Y=respondent answered yes
N=respondent answered no
a. Respondent answered yes given that
the respondent was female.
The problem is to find P(Y F). The rule states
𝑃 𝑌𝐹 =
𝑃(𝐹 𝑎𝑛𝑑 𝑌)
.
𝑃(𝐹)
P(F and Y) is
P(F) is
8
.
100
50
.
100
Then P(Y
8 100
F)=
50 100
=
4
25
b. Respondent was male, given the
respondent answered no.
The problem is to find P(M N).
𝑃(𝑁 𝑎𝑛𝑑 𝑀) 18 100
3
𝑃 𝑀𝑁 =
=
=
𝑃(𝑁)
60 100 10
Probabilities for “at least”
• The multiplication rule can be used along with
the complement rule to simplify problems
involving “at least”.
Example #1
• A game is played by drawing 4 cards from an ordinary
deck and replacing each card after it is drawn. Find
the probability that at least one ace is drawn.
• Solution: Since the outcome is looking for 1 ace, 2
aces, 3 aces, or 4 aces, it would be easier to find the
probability that 0 aces have been drawn and then
subtracting from 1.
12 12 12 12 20736
𝑃 0 𝑎𝑐𝑒𝑠 =
∙
∙
∙
=
13 13 13 13 28561
20736
7825
𝑃 𝑤𝑖𝑛𝑛𝑖𝑛𝑔 = 1 − 𝑃 0 𝑎𝑐𝑒𝑠 = 1 −
=
= .27
28561 28561
Example #2
• A coin is tossed five times. Find the probability of
getting at least one tail.
P(at least 1 tail) = 1 – P(all heads)
1 5
1
𝑃 𝑎𝑙𝑙 ℎ𝑒𝑎𝑑𝑠 = ( ) =
2
32
1
31
𝑃 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 1 𝑡𝑎𝑖𝑙 = 1 −
=
32 32
Applying the
Concepts and
Exercises 4-3
Section 4-4
Counting Rules
Learning Target
• IWBAT find the total number of outcomes in a
sequence of events, using the fundamental
counting rule.
• IWBAT find the number of ways that r objects
can be selected from n objects, using the
permutation rule.
• IWBAT find the number of ways that r objects
can be selected from n objects without regard
to order, using the combination rule.
The Fundamental Counting Rule
• In a sequence of n events in which the first
one has 𝑘1 possibilities and the second event
has 𝑘2 possibilities and the third has 𝑘3 and so
forth, the total number of possibilities in the
sequence will be 𝑘1 ∙ 𝑘2 ∙ 𝑘3 ⋯ 𝑘𝑛
• These problems will include repetitions and
no repetitions.
Example
• A paint manufacturer wishes to manufacture
several different paints. The categories include
Color
Red, blue, white, black, green, brown, yellow
Type
Latex, oil
Texture
Flat, semigloss, high gloss
Use
Outdoor, indoor
• How many different kinds of paint can be
made if you can select one color, one type,
one texture, and one use?
• Solution: 7 × 2 × 3 × 2 = 84
Factorial Notation
• For any counting n
𝑛! = 𝑛(𝑛 − 1)(𝑛 − 2) ⋯ 1
0! = 1
Example:
5! = 5 × 4 × 3 × 2 × 1 = 120
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3628800
Note: You can find the factorial key under math
PRB
4
Permutation
• A permutation is an arrangement of n objects
in a specific order.
• The arrangement of n objects in a specific
order using r objects is called a permutation of
n objects taking r objects at a time. It is
𝑛!
written as 𝑛𝑃𝑟 and the formula is 𝑛𝑃𝑟 =
(𝑛−𝑟)!
Example
• A television news director wants to use 3 news
stories on an evening show. One story will be the
lead story, one will be the second story and the
last will be the closing story. If the director has a
total of 8 stories to choose from, how many
possible ways can the program be set up?
• Solution: 𝑛𝑃𝑟 = 8𝑃3 =
8!
(8−3)!
= 336
• Note: You can use your calculator to do the
calculations. Type in number for n, then Math
PRB nPr, then number for r.
Example 2
• A school musical director can select 2 musical
plays to present next year. One will be
presented in the fall, the other in the spring. If
she has 9 to pick from, how many different
possibilities are there?
• Solution: 9𝑃2 =
9!
(9−2)!
=
9!
7!
= 72
Combinations
• A selection of distinct objects without regard
to order is called a combination.
• The number of combinations of r objects
selected from n objects is denoted by nCr and
𝑛!
is given by the formula 𝑛𝐶𝑟 =
(𝑛−𝑟)!𝑟!
Example
• A newspaper editor has received 8 books to
review. He decides that he can use 3 reviews
in his newspaper. How many different ways
can these 3 reviews be selected?
• Solution: 𝑛𝐶𝑟 = 8𝐶3 =
8!
(8−3)!3!
=
8!
5!3!
= 56
• Note: You can use your calculator for
calculations. Type in the number for n then
math PRB nCr, then the number for r.
Example 2
• In a club there are 7 women and 5 men. A
committee of 3 women and 2 men is to be
chosen. How many different possibilities are
there?
• Solution: 7𝐶3 ∙ 5𝐶2 =
7!
5!
∙
7−3 !3! 5−2 !2!
= 350
• Note: Because there are two events and they
have to occur together you must multiply the
events. Therefore, you will have to use rules from
previous sections.
Exercises 4-4
1-47 odd
Section 4-5
Probability and
Counting Rules
Learning Target
• IWBAT find the probability of an event using
the counting rules.
Example 1
Find the probability of getting four aces when five
cards are drawn from an ordinary deck of cards.
Steps:
1. Find the total number of ways to draw 5 cards
from a deck. (52C5 = 2598960)
2. Find the number of ways to draw 4 aces and 1
other card. (1 way to get four aces then 48 ways
to get the other card, therefore there are 48
ways)
3. Find the probability of drawing 4 aces if 5 cards
48
1
are drawn. (P 4aces =
=
2598960
54145
Example 2
A box contains 24 transistors, 4 of which are
defective. If 4 are sold at random, find the
following probabilities.
a. Exactly 2 are defective
b. None are defective
c. All are defective
d. At least one is defective
A. Exactly 2 are defective
Steps:
1. Find the number of ways we can draw 4
transistors. (24C4=10626)
2. Find the ways that 2 are defective and 2 are not
defective. (4C2 and 20C2)
3. 𝑃 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 2 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 =
190
1771
𝐶∙ 𝐶
𝐶
24 4
4 2 20 2
=
1140
10626
=
B. None are defective
Steps:
1. Find number of ways to draw 4 transistors
(10626)
2. Find number of ways to choose no defective.
(20C4)
3. 𝑃 𝑛𝑜𝑛𝑒 𝑎𝑟𝑒 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 =
𝐶
𝐶
24 4
20 4
=
4845
10626
=
1615
3542
C. All are defective
Steps:
1. Number of ways to draw any 4 (10626)
2. Number of ways to draw 4 defective (4C4)
3. 𝑃 𝑎𝑙𝑙 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 =
𝐶
𝐶
24 4
4 4
=
1
10626
D. At least 1 is defective
Steps:
1. Number of ways to draw any 4. (10626)
2. Number of ways to draw no defective. (20C4)
3. 𝑃 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 1 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 = 1 −
1615
1927
𝑃 𝑛𝑜 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 = 1 −
=
3542
3542
Example 3
A store has 6 TV graphic magazines and 8
newstime magazines on the counter. If two
customers purchased a magazine, find the
probability that one of each magazine was
purchased.
Solution:
48
6𝐶1 ∙ 8𝐶1
𝑃 1 𝑇𝑉 𝑎𝑛𝑑 1 𝑁𝑒𝑤𝑠𝑡𝑖𝑚𝑒 =
=
91
14𝐶2
Example 4
A combination lock consists of the 26 letters of the
alphabet. If a 3-letter combination is needed, find
the probability that the combination will consist of
the letter ABC in that order. The same letter can be
used more than once. (Note: a combination lock is
actually a permutation lock.)
Solution:
1
1
𝑃 𝐴𝐵𝐶 = 3 =
26
17576
Example 5
There are 8 married couples in a tennis club. If 1
man and 1 woman are selected at random to
plan the summer tournament, find the
probability that they are married to each other.
Solution:
8
1
𝑃 𝑚𝑎𝑟𝑟𝑖𝑒𝑑 𝑡𝑜 𝑒𝑎𝑐ℎ 𝑜𝑡ℎ𝑒𝑟 = 2 =
8
8
Exercises 4-5
1-17 odd
Wednesday