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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
G
2nd child
3rd child
GBB
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
 That probability that is supposed to happen Theoretical
 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
aJ
 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
 Probability that happens in an experiment –
Experimental
 Given a frequency distribution, the probability of an
event being in a given class is
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐𝑙𝑎𝑠𝑠
𝑓
𝑃 𝐸 = 𝑡𝑜𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 = 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
𝑓
21
A. 𝑃 𝑂 = 𝑛 = 50
B. 𝑃 𝐴 𝑜𝑟 𝐵 =
22
5
+
50
50
=
27
50
5
2
7
C. 𝑃 𝑛𝑒𝑖𝑡ℎ𝑒𝑟 𝐴 𝑛𝑜𝑟 𝑂 = 20 + 50 = 50
2
48
24
D. 𝑃 𝑛𝑜𝑡 𝐴𝐵 = 1 − 𝑃 𝐴𝐵 = 1 − 50 = 50 = 25
Law of Large Numbers
 The larger the number of trials, the closer the empirical
probability gets to the classical probability.
Subjective Probability
 An educated guess regarding the chance that an
event will occur
Applying the
Concepts and
Exercises 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 + 𝑃 𝑐ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒 =
5
12
8
2
= 12 = 3
 𝑃 𝑑𝑒𝑚𝑜𝑐𝑟𝑎𝑡 𝑜𝑟 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 = 𝑃 𝑑𝑒𝑚𝑜𝑐𝑟𝑎𝑡 +
13
6
19
𝑃 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 =
+
=
39
39
39
3
+
12
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)
Mutually
Exclusive
P(A and
B)
P(A)
Not Mutually
Exclusive
P(B)
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
=
16
2704
=
1
169
2
2
4
1
∙ =
=
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
 𝑃 𝑏𝑙𝑢𝑒 𝑎𝑛𝑑 𝑏𝑙𝑢𝑒 =
𝑃
𝑃
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
Means that B happens
=given
𝑃(𝐴)
𝑃(𝐵 𝐴)
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
4
3
2
24
1
 𝑃 3 𝑗𝑎𝑐𝑘𝑠 = 52 ∙ 51 ∙ 50 = 132600 = 5525
4
4
4
64
8
 𝑃 𝑎𝑐𝑒 𝑎𝑛𝑑 𝑘𝑖𝑛𝑔 𝑎𝑛𝑑 𝑞𝑢𝑒𝑒𝑛 = 52 ∙ 51 ∙ 50 = 132600 = 16575
13 13 13
2197
169
 𝑃 𝑐𝑙𝑢𝑏 𝑎𝑛𝑑 𝑠𝑝𝑎𝑑𝑒 𝑎𝑛𝑑 ℎ𝑒𝑎𝑟𝑡 = 52 ∙ 51 ∙ 50 = 132600 = 10200
13 12 11
1716
11
 𝑃 3 𝑐𝑙𝑢𝑏𝑠 = 52 ∙ 51 ∙ 50 = 132600 = 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 𝐵𝐴 =
𝑃(𝐴 𝑎𝑛𝑑 𝐵)
𝑃(𝐴)
Example
 A box contains black chips and white chips. A person
selects 2 chips without replacement. If the probability
15
of selecting a black chip and a white chip is 56 , and
the probability of selecting a black chip on the first
3
draw is 8 , 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 =
𝑃(𝐵 𝑎𝑛𝑑 𝑊)
15
=
𝑃(𝐵)
15
3
÷ =
56
8
56
3 8
15 8
5
∙ =
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 theGender
survey areYes
shown. 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 𝑌 𝐹 =
𝑃(𝐹 𝑎𝑛𝑑 𝑌)
.
𝑃(𝐹)
8
P(F and Y) is 100.
P(F) is
50
.
100
8 100
100
Then P(Y F)=50
4
= 25
b. Respondent was male, given
the respondent answered no.
The problem is to find P(M N).
𝑃 𝑀𝑁 =
𝑃(𝑁 𝑎𝑛𝑑 𝑀) 18 100
3
=
=
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
1
𝑃 𝑎𝑙𝑙 ℎ𝑒𝑎𝑑𝑠 = ( )5 =
2
32
𝑃 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 1 𝑡𝑎𝑖𝑙 = 1 −
1
31
=
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 different
Latex,
oil of paint can be made if you can select
 How many
kinds
one color,
one type,
texture,
andgloss
one use?
Texture
Flat,one
semigloss,
high
Use
 Solution:
7 × 2 × 3Outdoor,
× 2 = 84 indoor
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 written as 𝑛𝑃𝑟 and the formula is
𝑛!
𝑃
=
𝑛 𝑟
(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?
8!
 Solution: 𝑛𝑃𝑟 = 8𝑃3 = (8−3)! = 336
 Note: You can use your calculator to do the
calculations. Type in number for n, then Math
nPr, then number for r.
PRB
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?
9!
9!
 Solution: 9𝑃2 = (9−2)! = 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 is given by the formula
𝑛!
𝐶
=
𝑛 𝑟
(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?
8!
8!
 Solution: 𝑛𝐶𝑟 = 8𝐶3 = (8−3)!3! = 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 are drawn.
48
1
(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 = 4
𝐶2∙20𝐶2
𝐶
24 4
1140
190
= 10626 = 1771
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 = 20𝐶4 = 10626 = 3542
𝐶
24 4
4845
1615
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 − 𝑃 𝑛𝑜 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 = 1 −
1615
1927
=
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:
𝑃 1 𝑇𝑉 𝑎𝑛𝑑 1 𝑁𝑒𝑤𝑠𝑡𝑖𝑚𝑒 =
6𝐶1
∙ 8𝐶1 48
=
𝐶
91
14 2
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:
𝑃 𝑚𝑎𝑟𝑟𝑖𝑒𝑑 𝑡𝑜 𝑒𝑎𝑐ℎ 𝑜𝑡ℎ𝑒𝑟 =
8
1
=
82 8
Exercises 4-5
1-17 odd
Wednesday