#### Transcript Probability

```Probability
Foldable #1
Do a hotdog bun fold and make 10
cuts
Combination
Conditional probability
Dependent events
Events
Experimental probability
Independent events
Mutually exclusive events
Permutation
Probability
Sample space
Theoretical probability
Vocabulary Words
• Combination - a grouping of items in which
order does not matter. There are generally
fewer ways to select items when order does
not matter.
• Conditional probability – the probability of
event B given that event A has occurred.
• Dependent events – the occurrence of one
event affects the probability of the other.
Vocabulary Words: These will be on
the test
• Event – an outcome or set of outcomes.
• Experimental probability - the likelihood that the
event occurs based on the actual results of an
experiment
experimental probability=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑒𝑣𝑒𝑛𝑡 𝑜𝑐𝑐𝑢𝑟𝑠
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠
• Independent Events – when the occurrence of
one event does not affect the probability of the
other
• Mutually Exclusive Events – events that cannot
both occur in the same trial of an experiment.
Vocabulary Words
• Permutation - a selection of a group of objects in
which order is important.
• Probability – the measure of how likely an event
is to occur.
• Sample space – the set of all possible outcomes.
• Theoretical probability – the likelihood of an
event based on mathematical reasoning
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
P(event)=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒
Experimental and Theoretical
Probability
• Outcome – each possible result of a probability
experiment or situation.
• Event – an outcome or set of outcomes.
• Sample space – the set of all possible outcomes
• Probability – the measure of how likely an event
is to occur
• Theoretical probability – the ratio of the number
of favorable outcomes to the total number of
outcomes.
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
P(event)=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒
Experimental Probability
Experimental probability of an event – the
number of times that the event occurs.
experimental
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑒𝑣𝑒𝑛𝑡 𝑜𝑐𝑐𝑢𝑟𝑠
probability=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡 𝑖𝑠 𝑑𝑜𝑛𝑒
Finding Experimental Probability
The table shows the results of a
spinner experiment. Find each
experimental probability.
Number
Occurrences
1
6
2
11
3
19
4
14
A. Spinning a 4
14
50
=
7
=28%
25
There were 50 occurrences
B. Spinning a number greater
than 2
33
= 66%
50
Theoretical Probability
• Theoretical probability – the ratio of the
number of favorable outcomes to the total
number of outcomes.
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
P(event)=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒
Finding Theoretical Probability
A CD has 5 upbeat dance songs and 7 slow
ballads. What is the probability that a randomly
selected song is an upbeat dance song?
P(upbeat dance
5
song)=
12
≈ 41.7%
Probability Distributions and
Frequency Tables
• Frequency table – a data display that show
how often an item appears in a category.
• Relative frequency – the ratio of the frequency
of the category to the total frequency.
relative frequency
𝑡ℎ𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑎𝑡𝑒𝑔𝑜𝑟𝑦
=
𝑡𝑜𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
• Probability distribution – shows the probability of
each possible outcome.
Finding Relative Frequencies
The results of a survey of students’ music
preferences are organized in this frequency
table. What is the relative frequency for each
type of music?
a. classical
1/8
b. Hip hop
7/40
c. Country
1/5
Type of Music
Preferred
Frequency
Rock
10
Hip Hop
7
Country
8
Classical
5
Alternative
6
Other
4
The results of a survey of students’ music
preferences are organized in this frequency
table. What is the relative frequency of
preference for rock music?
10
=
10+7+8+5+6+4
=
10
40
=
1
4
Type of Music
Preferred
Frequency
Rock
10
Hip Hop
7
Country
8
Classical
5
Alternative
6
Other
4
Calculating Probability by Using
Relative Frequencies
A student conducts a probability experiment by
tossing 3 coins one after the other. Using the
results below, what is the probability that
exactly two heads will occur in the next three
tosses?
Coin Toss Result
HHH
HHT
HTT
HTH
THH
THT
TTT
TTH
Frequency
5
7
9
6
2
9
10
2
Calculating Probability by Using
Relative Frequencies
Step1: Find the number of times a trial results in exactly two heads.
The possible results that show exactly two heads are HHT, HTH, and
THH. The frequency of these results is 7+6+2=15
Step 2: Find the total of all the frequencies.
5+7+9+6+2+9+10+2=50
Step 3: Find the relative frequency of ta trial with exactly two heads.
relative frequency=
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑡𝑤𝑜 ℎ𝑒𝑎𝑑𝑠
𝑡𝑜𝑡𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠
=
15
50
=
3
10
A student conducts a probability experiment by
spinning the spinner shown. Using the results in
the frequency table, what is the probability of
the spinner pointing at 4 on the next spin?
Spinner
Result
1
2
3
4
Frequency
29
32
21
18
18/100=9/50
Finding a Probability Distribution
In a recent competition, 50 archers shot 6
arrows each at a target. Three archers hit no
bull’s eyes; 5 hit one bull’s eye; 7 hit two bull’s
eyes; 7 hit three bull’s eye; 11 hit four bull’s eye;
10 hit five bull’s eye; and 7 hit six bull’s eye.
What is the probability distribution for the
number of bull’s eyes each archer hit?
Finding a Probability Distribution
Plan
Know
Need
The possible outcome and
the frequency of each
outcome
Make a frequency table
and use relative
frequencies to complete
the probability distribution.
The probabilities of each
outcome
First, create a frequency table showing all the possible outcomes: 0, 1, 2,
3, 4, 5, or 6 bull’s eyes and the frequencies for each.
Next, use the table to find the relative frequencies for each number of
bull’s eyes. The relative frequencies are the probability distribution.
Probability Distribution of Bull’s Eye’s Hits
Number of bull’s eyes Hit
0
1
2
3
4
5
6
Frequency
3
5
7
7
11
10
7
Probability
3/50
1/10
7/50
7/50
11/50 1/5
7/50
On a math test, there were 10 scores between
90 and 100, 12 scores between 80 and 89, 15
scores between 70 and 79, 8 scores between 60
and 69, and 2 scores below 60. What is the
probability distribution for the test scores?
Probability Distribution of Bull’s Eyes Hits
Scores Ranges
90-100
80-89
70-79
60-69
Below 60
Frequency
10
12
15
8
2
Probability
10/47
12/47
15/47
8/47
2/47
Permutations and Combinations
• Fundamental Counting Principle says that if event
M occurs in m ways and event N occurs in n
ways, then event M followed by event N can
occur in m x n ways.
• Permutation – an arrangement of items in which
the order of the objects is important.
• n factorial – The factorial of a number is the
product of the natural numbers less than or equal
to the number.
• Combination – an arrangement of items in which
the order is not important.
Fundamental Counting Principle
You have a red shirt, a blue shirt, a pair of black pants and
a pair of khaki pants. How many different outfits can you
make.
Red shirt and black pants
Red shirt and khaki pants
Blue shirt and black pants
Blue shirt and khaki pants
Total of 4 different outfits
2 shirts times 2 pairs of pants = 4 different outfits.
Fundamental Counting Principle
The Greasy Spoon Restaurant offers 6 appetizers
and 14 main courses. In how many ways can a
person order a two-course meal?
Solution: Choosing from one of 6 appetizers and
one of 14 main courses, the total number of
two-course meals is 6× 14 = 84
A restaurant offers 10 appetizers and 15 main
courses. In how many ways can you order a twocourse meal?
Solution: 10× 15 = 150 two-course meals
The Fundamental Counting Principle
with More Than Two Groups of Items
Next semester you are planning to take three
courses – math, English and humanities. Based on
time blocks and highly recommended professors,
there are 8 sections of math, 5 of English, and 4 of
humanities that you find suitable. Assuming no
scheduling conflicts, how many different threecourse schedules are possible?
Solution: math English
humanities
8
×5
× 4 = 160
A pizza can be ordered with two choices of
size(medium or large), three choices of crust
(thin, thick, or regular), and five choices of
toppings ( ground beef, sausage, pepperoni,
bacon, or mushrooms.) How many different
one-topping pizzas can be ordered?
Solutions: 2× 3 × 5 = 30
Permutations
Order Matter
One item can be arranged one way:
1 permutation
A
Two items can be arranged two ways:
2 x 1 permutations
AB and BA
Three items can be arranged six ways:
ABC, ACB, BAC, BCA, CAB, CBA
3 x 2 x1 permutations
n Factorial
The factorial of a number is the product of the
natural numbers less than or equal to the
number.
0! Is defined as 1
6!=6 x 5 x4 x3 x2 x 1=720
n!=n x (n-1) x (n-2) x (n-3) x … x 1
Finding the Number of Permutations
you play the songs using the random shuffle
option, how many different ways can the
sequence of songs be played?
8!=8x7x6x5x4x3x2x1=40,320
In how many ways can you arrange 12 books on
a shelf?
12!=12x11x10x9x8x7x6x5x4x3x2x1=479,001,600
Sometimes you may not want to order an entire
set of items. Suppose you want to select and
order 3 people from a group of 7. One way to
find possible permutations is to use the
Fundamental Counting Principle.
First person
Second person Third person
7 choices
x
6 choices x 5 choices
=210 permutations.
Another way to find the possible permutations is to use
factorials. You can divide the total number of
arrangements by the number of arrangements that are
not used.
𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 7 𝑝𝑒𝑜𝑝𝑙𝑒 7! 7𝑥6𝑥5𝑥4𝑥3𝑥2𝑥1
= =
𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 4 𝑝𝑒𝑜𝑝𝑙𝑒 4!
4𝑥3𝑥2𝑥1
= 210
Notice: 4x3x2x1 will cancel so 7x6x5=210
Permutation Formula
The number of permutations of n items taken r
at a time.
nPr=
𝑛!
𝑛−𝑟 !
The number of permutation of 7 items taken 3
at a time.
7P3=
7!
7!
=
7−3 ! 4!
Finding a Permutation
How many ways can a club select a president, a
vice president, and a secretary from a group of 5
people?
n=5 since there are 5 people to choose from.
r=3 since there are 3 officer positions.
5P3=
5!
5−3 !
=
5!
2!
=
5𝑥4𝑥3𝑥2𝑥1
2𝑥1
= 5x4x3 = 60
An art gallery has 9 fine-art photographs from
an artist and will display 4 from left to right
along a wall. In how many ways can the gallery
select and display the 4 photographs?
9P4=
9!
9−4 !
= 3024
Combinations Formula
To find the number of combinations, the formula for
permutations can be modified.
Number of
𝑤𝑎𝑦𝑠 𝑡𝑜 𝑎𝑟𝑟𝑎𝑛𝑔𝑒 𝑎𝑙𝑙 𝑖𝑡𝑒𝑚𝑠
permutations=
𝑤𝑎𝑦𝑠 𝑡𝑜 𝑎𝑟𝑟𝑎𝑛𝑡 𝑖𝑡𝑒𝑚𝑠 𝑛𝑜𝑡 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑
Because order does not matter, divide the number of
permutations by the number of ways to arrange the selected
items.
Number of
𝑤𝑎𝑦𝑠 𝑡𝑜 𝑎𝑟𝑟𝑎𝑛𝑔𝑒 𝑎𝑙𝑙 𝑖𝑡𝑒𝑚𝑠
combinations=(𝑤𝑎𝑦𝑠 𝑡𝑜 𝑎𝑟𝑟𝑎𝑛𝑔𝑒 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑
𝑖𝑡𝑒𝑚𝑠)(𝑤𝑎𝑦𝑠 𝑡𝑜 𝑎𝑟𝑟𝑎𝑛𝑔𝑒 𝑖𝑡𝑒𝑚𝑠 𝑛𝑜𝑡 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑)
Combinations Formula
The number of combinations of n items taken r at a
time is
𝑛!
nCr=𝑟! 𝑛−𝑟 !
The number of combinations of 7 items taken 3 at a
time is
7!
7C3=3! 7−3 !
Using the Combination Formula
Katie is going to adopt kittens from a litter of 11. How
many ways can she choose a group of 3 kittens?
Step 1: Is this a permutation or a combination? This is a
combination since order doesn’t matter. Kitty, Smoky and
Tiger is the same as Tiger, Kitty and Smoky.
Step 2: Use the formula
11!
11C3=3! 11−3 !
=
11𝑥10𝑥9𝑥8𝑥7𝑥65𝑥4𝑥3𝑥2𝑥1
3𝑥2𝑥1𝑥(8𝑥7𝑥6𝑥5𝑥4𝑥3𝑥2𝑥1)
=
11𝑥10𝑥9
3𝑥2𝑥1
= 165
The swim team has 8 swimmers. Two swimmers
will be selected to swim in the first heat. How
many ways can the swimmers be selected?
8!
8C2=2! 8−2 !
= 28
Identifying combinations and
Permutations
To determine whether to use the permutation
formula or the combination formula, you must
decide whether order is important.
A. A college student is choosing 3 classes to take
during first, second and third semester from the
5 elective classes offered in his major. How many
possible ways can the student schedule the
three classes.
Since the order does matter this is a permutation.
Identifying Combinations and
Permutations
B. A jury of 12 people is chosen from a pool of
35 potential jurors. How many different juries
can be chosen?
Since order doesn’t matter this is a combination.
A yogurt shop allows you to choose any 3 of the
10 possible mix-ins for a Just Right Smoothie.
How many different Just Right Smoothies are
possible?
Since order doesn’t matter this is a combination.
Compound Probability
• Compound event – an event that is made up of
two or more events
• Independent events – an event that does not
affect how another event occurs.
• Dependent events – an event that does affect
how another event occurs.
• Mutually exclusive events – events that cannot
happen at the same time
• Overlapping events – have common outcomes
Identifying Independent and
Dependent Events
Are the outcomes of each trial independent or dependent events?
A.
Choose a number tile from 12 tiles. Then spin a spinner.
The choice of number tile does not affect the spinner result. The
events are independent.
B. Pick on card from a set of 15 sequentially numbered cards. Then,
without replacing the card, pick another card.
The first card chosen affects the possible outcomes of the second pick,
so the events are dependent.
You roll a standard number cube. Then you flip a
coin. Are the outcomes independent or
dependent events. Explain.
They are independent because the roll of a die
does not affect the flip of a coin.
Independent Events
If A and B are independent event,
then P(A and B)=P(A) x P(B)
Finding the Probability of Independent
Events
Find each probability.
A. Spinning 4 then 4 again on
the spinner.
P(4 and then 4)=P(4) x
1 1
1
P(4)=( )
=
6
6
36
B. Spinning 5 then 3.
P(5 and then 3)=P(5) x
1 1
1
P(3)=( )
=
3
4
12
Find each probability.
A. Rolling a 6 on one number cube and a 6 on another.
P(6 on one cube and 6 on another)=P(6) x P(6)=
1 1
1
( )
=
6
6
36
3 times.
1 1
1
1
=
2
2
2
8
Dependent Events
To find the probability of dependent events you
can use conditional probability P( 𝐵 𝐴
If A and B are dependent events, then P(A and
B)=P(A) x P( 𝐵 𝐴 , where P( 𝐵 𝐴 is the
probability of B given that A has occurred.
Finding the Probability of Dependent
Events
Two number cubes are rolled – one red and one blue. Explain why the
events are dependent. Then find the indicated probability.
A. The red cube shows a 1 and the sum is less than 4.
Why the events are dependent. The events “the red cube shows a 1”
and “the sum is less than 4” are dependent because P(sum<4) is
different when it is known that a red 1 has occurred.
1
6
P(red 1)=
1
3
P( 𝑠𝑢𝑚 < 4 𝑟𝑒𝑑 1 = only 1 and 2 will meet the condition.
P(red1 and sum<4)=P(red 1) x P( 𝑠𝑢𝑚 < 4 𝑟𝑒𝑑 1 =
1 1
( )
6 3
=
1
18
Finding the Probability of Dependent
Events
Two number cubes are rolled – one red and one blue. Explain why the
events are dependent. Then find the indicated probability.
B. The blue cube sows a multiple of 3 and the sum is 8.
The events are dependent because P(sum is 8) is different when the
blue cube shows a multiple of 3.
P(blue multiple of 3)=
1
3
2
1
P 𝑠𝑢𝑚 𝑖𝑠 8 𝑏𝑙𝑢𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3 = = (3+5 , 6+2)
12
6
P(blue multiple of 3 and sum is 8)= P(blue multiple of 3) x
1
1
1
P 𝑠𝑢𝑚 𝑖𝑠 8 𝑏𝑙𝑢𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3 =
=
3
6
18
Two number cubes are rolled – one red and one black. Explain why the
events are dependent, and then find the indicated probability.
The red cube shows a number greater than 4 and the sum is greater
than 9.
P(sum of 9) changes when it is know that the red cube is >4.
2
6
P(>4)= =
1
3
5
x P 𝑠𝑢𝑚 > 9 > 4 = five of the possible 12 outcomes are >9. (5+5,
12
5+6, 6+4, 6+5, 6+6)
1
3
P(>4 and sum >9)=P(>4) x P 𝑠𝑢𝑚 > 9 > 4 =( )(
5
12)
=
5
36
Mutually Exclusive Events
If A and B are mutually exclusive events, then
P(A and B)=0, and P(A or B)=P(A) +P(B).
Finding the Probability of Mutually
Exclusive Events
Student athletes at a local high school may participate in only
one sport each season. During the fall season, 28% of student
athletes play basketball and 24% are on the swim team. What
is the probability that a randomly selected student athlete
plays basketball or is on the swim team?
Because athletes participate in only one sport each season,
the events are mutually exclusive. Use the formula P(A or B)=
P(A) + P(B)
team)=28%+24%=52%
In the Spring season, 15% of the athletes play
baseball and 23% are on the track team. What is
the probability of an athlete either playing
baseball or being on the track team?
P(baseball or track team)=P(baseball)+P(track
team)=15%+23%=38%
Probability of Overlapping Events
If A and B are overlapping events, the P(A or
B)=P(A) + P(B) – P(A and B)
Suppose you have 7 index cards, each having one of
the following letters written on it: A B C D E F G
P(FACE), the probability of selecting a letter from
the word FACE, is 4/7.
P(CAB), the probability of selecting a letter from the
word CAB, is 3/7
D
F
E
G
A
C
B
Consider P(FACE or CAB), the probability of choosing a letter from either the word
FACE or the word CAB. These events overlap since the words have two letters in
common. If you simply add P(FACE) and P(CAB), you get 4/7+3/7=7/7. The value of
the numerator should be the number of favorable outcomes, but there are only 5
distinct letters in the words FACE and CAB. The problem is that when you simply
add, the letters A and C are counted twice, once in the favorable outcomes for the
word FACE, and once for the favorable out comes for the word CAB. You must
subtract the number of letters that the two words have in common so they are only
counted once.
P(FACE or CAB)=(4+5-2)/7=4/7+5/7-2/7=P(FACE)+P(CAB)-P(AC)
Find Probabilities of Overlapping
Events
What is the probability of rolling either an even
number or a multiple of 3 when rolling a
standard number cube?
Know
Need
You are rolling a
standard number
cube. The events
are overlapping
events because 6 is
both even and a
multiple of 3.
You need the
probability of
rolling an even
number and the
probability of
rolling a multiple
of 3
Plan
Find the
probabilities and
use the formula for
probabilities of
overlapping events.
P(even or multiple of 3)=P(even) +P(multiple of
3)-P(even and multiple of 3)=3/6+2/61/6=4/6=2/3
What is the probability of rolling either an odd
number or a number less than 4 when rolling a
standard number cube?
P(odd or <4)=P(odd)+P(<4)-P(odd and
<4)=3/6+3/6-2/6=4/6=2/3
Probability Models
• Two-way frequency table – displays the
frequencies of data in two different
categories.
• Conditional probability – the probability that
an event will occur, given that another event
Using a Two-Way Frequency Table
The table shows data about student
involvement in extracurricular activities at a
local high school. What is the probability that a
randomly chosen student is a female who is not
involved in extracurricular activities?
Involved in
Activities
Not Involved in
Activities
Totals
Male
112
145
257
Female
139
120
259
Total
251
265
516
To find the probability, calculate the relative frequency.
Relative
0.233
𝑓𝑒𝑚𝑎𝑙𝑒𝑠 𝑛𝑜𝑡 𝑖𝑛𝑣𝑜𝑙𝑣𝑒𝑑
frequency=
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑒𝑟 𝑜𝑓 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠
=
120
516
≈
The two-way frequency table at the right shows
the number of male and female students by
grade level on the prom committee. What is the
probability that a member of the prom
committee is a male who is a junior?
Male
Female
Totals
Juniors
3
4
7
Seniors
3
2
5
Totals
6
6
12
𝑗𝑢𝑛𝑖𝑜𝑟 𝑚𝑎𝑙𝑒
3
Relative frequency=𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑟𝑜𝑚 𝑐𝑜𝑚𝑚𝑖𝑡𝑡𝑒𝑒 = 12 = .25
Finding Probability
Respondents of a poll were
the state legislature that
would increase the
minimum wage. What is the
probability that a randomly
selected person is over 60
years old, given that the
the state bill?
Age
For
Group
Against
No opinion
Totals
18-29
310
50
20
380
30-45
200
30
10
240
46-60
120
20
30
170
Over
60
150
20
40
210
Totals
780
120
100
1000
The condition that the person selected has no
opinion on the minimum-wage bill limits the
total outcomes to the 100 people who had no
opinion. Of those 100 people, 40 respondents
were over 60 years old.
40
P 𝑜𝑣𝑒𝑟 60 𝑛𝑜 𝑜𝑝𝑖𝑛𝑖𝑜𝑛 = 100=.4
Respondents of a poll were
the state legislature that
would increase the
minimum wage. What is the
probability that a randomly
selected person is 30-45
years old, given that the
person is in favor of the
minimum-wage bill?
Age
For
Group
Against
No opinion
Totals
18-29
310
50
20
380
30-45
200
30
10
240
46-60
120
20
30
170
Over
60
150
20
40
210
Totals
780
120
100
1000
200
P 30 − 45 𝐹𝑜𝑟 = 780 ≈ .256
Using Relative Frequencies
A company has 150 sales
representatives. Two months after a
sales seminar, the company vicepresident made the table based on
sales results. What is the probability
that someone who attended the
seminar had an increase in sales?
Attended Did not Totals
Seminar Attend
Seminar
Increased
Sales
.48
.02
.5
No
Increase
in Sales
.32
.18
.5
Totals
.8
.2
1
P 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑 𝑠𝑎𝑙𝑒𝑠 𝑠𝑎𝑙𝑒𝑠 𝑠𝑒𝑚𝑖𝑛𝑎𝑟 =
𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑎𝑡𝑡𝑒𝑛𝑑 𝑠𝑒𝑚𝑖𝑛𝑎𝑟 𝑎𝑛𝑑 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑 𝑠𝑎𝑙𝑒𝑠
𝑡𝑜𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑎𝑡𝑡𝑒𝑛𝑑𝑒𝑑 𝑠𝑒𝑚𝑖𝑛𝑎𝑟
P 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑 𝑠𝑎𝑙𝑒𝑠 𝑠𝑎𝑙𝑒𝑠 𝑠𝑒𝑚𝑖𝑛𝑎𝑟 =
.48
.8
= .6
A company has 150 sales
representatives. Two months after a
sales seminar, the company vicepresident made the table based on
sales results. What is the probability
that a randomly selected sales
representative, who did not attend
the seminar, did not see an increase
in sales?
Attended Did not Totals
Seminar Attend
Seminar
Increased
Sales
.48
.02
.5
No
Increase
in Sales
.32
.18
.5
Totals
.8
.2
1
P 𝑛𝑜 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑 𝑠𝑎𝑙𝑒𝑠 𝑛𝑜 𝑠𝑎𝑙𝑒𝑠 𝑠𝑒𝑚𝑖𝑛𝑎𝑟 =
𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑛𝑜𝑡 𝑎𝑡𝑡𝑒𝑛𝑑 𝑠𝑒𝑚𝑖𝑛𝑎𝑟 𝑎𝑛𝑑 𝑛𝑜 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑 𝑠𝑎𝑙𝑒𝑠
𝑡𝑜𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑑𝑖𝑑 𝑛𝑜𝑡 𝑎𝑡𝑡𝑒𝑛𝑑 𝑠𝑒𝑚𝑖𝑛𝑎𝑟
P 𝑛𝑜 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑 𝑠𝑎𝑙𝑒𝑠 𝑛𝑜 𝑠𝑎𝑙𝑒𝑠 𝑠𝑒𝑚𝑖𝑛𝑎𝑟 =
.18
.5
= .36
Conditional Probability Formulas
For any two events A and B, the probability of B
occurring, given that event A has occurred, is
P 𝐵𝐴 =
𝑃(𝐴 𝑎𝑛𝑑 𝐵)
, 𝑤ℎ𝑒𝑟𝑒
𝑃(𝐴)
𝑃(𝐴) ≠ 0
Using Conditional Probabilities
In a study designed to test the effectiveness of a new drug, half of the
a placebo, a tablet or pill containing no medication. The probability of
a volunteer receiving the drug and getting well was 45%. What is the
probability of someone getting well, given that he receives the drug?
Step 1: Identify the probabilities
P 𝐵 𝐴 = 𝑃(𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑤𝑒𝑒, 𝑔𝑖𝑣𝑒𝑛 𝑡𝑎𝑘𝑖𝑛𝑔 𝑡ℎ𝑒 𝑛𝑒𝑤 𝑑𝑟𝑢𝑔)
P(A)=P(taking the new drug)=1/2 =.5
P(A and B)=P(taking the new drug and getting well)=45% or .45
Step 2: Find P 𝐵 𝐴
P 𝐵𝐴 =
𝑃(𝐴 𝑎𝑛𝑑 𝐵)
𝑃(𝐴)
=
.45
.5
= .9
The probability of a volunteer receiving the placebo
and having his or her health improve was 20%.
What is the conditional probability of a volunteer’s
health improving, given that they received the
placebo?
P 𝐵 𝐴 =P(getting well, given the placebo)
P(A)= P(taking the placebo)=.5
P(A and B)=(taking the placebo and getting well)=.2
P 𝐵 𝐴 =.2/.5=.4
Comparing Conditional Probabilities
In a survey of pet owners, 45% own a dog, 27%
own a cat and 12% own both a dog and a cat.
What is the conditional probability that a dog
owner also owns a cat? What is the conditional
probability that a cat owner also owns a dog?
P 𝑐𝑎𝑡 𝑑𝑜𝑔 =
.12
.45
= .267
P 𝑑𝑜𝑔 𝑐𝑎𝑡 =
.12
.27
= .444
The same survey showed that 5% of the pet owners own a dog, a
cat, and at least one other type of pet.
a. What is the conditional probability that a pet owner owns a
cat and some other type of pet, given that they own a dog?
.05/.45=.111
b. What is the conditional probability that a pet owner owns a
dog and some other type of pet, given that they own a cat?
.05/.27=.185
Modeling Randomness
You can use probability to make choices and to help
make decisions based on prior experience.
A random event has no predetermined pattern or
bias toward one outcome or another. You can use
random number tables or randomly generated
numbers using graphing calculators or computer
using a random number table.
Making Random Selections
There are 28 students in a homeroom. Four students are chosen at random to
represent the homeroom on a student committee. How can a random number table
be used to fairly choose the students?
Step 1: Select a line from a random number table.
18823 18160 93593 67294 09632 62617 86779
Step 2: Group the line from the table into two digit numbers.
18 82 31 81 60 93 59 36 72 94 09 63 26 26 17 86 77 9
Step 3: Match the first four numbers less than 28 with the position of the students’
names on a list. Duplicates and numbers greater than 28 are discarded because they
don’t correspond to any student on the list.
18 09 26 17
The students listed 18th , 9th, 26th, and 17th on the list are chosen fairly.