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Probability
What’s the odds?
Warm Up
Write each fraction in simplest form.
16
20
4
5
2. 12
36
1
3
8
3. 64
1
8
39
4. 195
1
5
1.
Vocabulary
experiment
trial
outcome
sample space
event
probability
An experiment is an activity in which results are
observed. Each observation is called a trial, and
each result is called an outcome. The sample
space is the set of all possible outcomes of an
experiment.
Experiment
Space

flipping a coin
rolling a number cube
Sample
heads, tails
1, 2, 3, 4, 5, 6
An event is any set of one or more outcomes. The
probability of an event is a number from 0 (or
0%) to 1 (or 100%) that tells you how likely the
event is to happen. You can write probability as a
fraction, a decimal, or a percent.
•
A probability of 0 means the event is impossible,
or can never happen.
•
A probability of 1 means the event is certain, or
will always happen.
•
The probabilities of all the outcomes in the
sample space add up to 1.
Never
happens
Happens about
half the time
Always
happens
0
1
4
0.25
1
2
0.5
3
4
0.75
1
0%
25%
50%
75%
100%
0
1
Additional Example 1A: Finding Probabilities of
Outcomes in a Sample Space
Give the probability for each outcome.
The basketball team has a
70% chance of winning.
P(win) = 70% = 0.7.
P(lose) = 1 – 0.7 = 0.3, or 30%
Additional Example 1B: Finding Probabilities of
Outcomes in a Sample Space
Give the probability for each outcome.
P(1) = 3
8
Three of the eight sections of the
spinner are labeled 1, so 3 is a
8
reasonable estimate.
Additional Example 1B Continued
3
P(2) =
8
P(3) =
2
8
Three of the eight sections of the
spinner are labeled 2, so 3 is a
8
reasonable estimate.
Two of the eight sections of the
spinner are labeled 3, so 2 is a
8
reasonable estimate.
Check The probabilities of all the outcomes
must add to 1.
3 + 3+ 2
= 1
8 8
8
Check It Out! Example 1A
Give the probability for each outcome.
The polo team has a
50% chance of winning.
P(win) = 50% = 0.5.
P(lose) = 1 – 0.5 = 0.5, or 50%.
Check It Out! Example 1B
Give the probability for each outcome.
Outcome
Teal
Red
Orange
Probability
P(teal) =
3
8
Three of the eight sections of the
spinner are teal, so 3 is a
8
reasonable estimate.
Check It Out! Example 1B Continued
Three of the eight sections of the
3
spinner are red, so 3 is a
P(red) =
8
8
reasonable estimate.
P(orange) =
2
8
Two of the eight sections of the
spinner are orange, so 2 is a
8
reasonable estimate.
Check The probabilities of all the outcomes
must add to 1.
3 + 3+ 2
= 1
8 8
8
To find the probability of an event, add
the probabilities of all the outcomes
included in the event.
Additional Example 2A: Finding Probabilities of Events
A quiz contains 5 true-false questions. Suppose
you guess randomly on every question. The table
below gives the probability of each score.
What is the probability of guessing 3 or more
correct?
The event “three or more correct” consists of the
outcomes 3, 4, and 5.
P(3 or more correct) = 0.313 + 0.156 + 0.031
= 0.5, or 50%.
Additional Example 2B: Finding Probabilities of Events
A quiz contains 5 true-false questions. Suppose
you guess randomly on every question. The table
below gives the probability of each score.
What is the probability of guessing fewer
than 2 correct?
The event “fewer than 2 correct” consists of the
outcomes 0 and 1.
P(fewer than 2 correct) = 0.031 + 0.156
= 0.187, or 18.7%.
Check It Out! Example 2A
A quiz contains 5 true-false questions. Suppose
you guess randomly on every question. The table
below gives the probability of each score.
What is the probability of guessing 2 or more
correct?
The event “two or more correct” consists of the
outcomes 2, 3, 4, and 5.
P(2 or more) = 0.313 + 0.313 + 0.156 + 0.031
= 0.813, or 81.3%.
Check It Out! Example 2B
A quiz contains 5 true-false questions. Suppose
you guess randomly on every question. The table
below gives the probability of each score.
What is the probability of guessing fewer
than 3 correct?
The event “fewer than 3” consists of the outcomes
0, 1, and 2.
P(fewer than 3) = 0.031 + 0.156 + 0.313 =
0.5, or 50%.
Let’s Solve some problems…
Use the table to find the probability of each
event.
1. 1 or 2 occurring 0.351
2. 3 not occurring 0.874
3. 2, 3, or 4 occurring
0.794
Warm Up
Use the table to find the probability of
each event.
1. A or B occurring 0.494
2. C not occurring 0.742
3. A, D, or E occurring
0.588
Vocabulary
experimental probability
In experimental probability, the likelihood of
an event is estimated by repeating an
experiment many times and observing the
number of times the event happens. That
number is divided by the total number of trials.
The more times the experiment is repeated, the
more accurate the estimate is likely to be.
Additional Example 1A: Estimating the Probability
of an Event
A marble is randomly drawn out of a bag and
then replaced. The table shows the results
after fifty draws.
Estimate the probability of drawing a red marble.
Outcome
Draw
probability 
Green
Red
Yellow
12
15
23
number of red marbles drawn = 15
total number of marbles drawn 50
The probability of drawing a red marble is about 0.3,
or 30%.
Additional Example 1B: Estimating the Probability
of an Event
A marble is randomly drawn out of a bag and
then replaced. The table shows the results
after fifty draws.
Estimate the probability of drawing a green
marble.
Outcome
Draw
probability 
Green
Red
Yellow
12
15
23
number of green marbles drawn = 12
50
total number of marbles drawn
The probability of drawing a green marble is about
0.24, or 24%.
Additional Example 1C: Estimating the Probability
of an Event
A marble is randomly drawn out of a bag and
then replaced. The table shows the results
after fifty draws.
Estimate the probability of drawing a yellow
marble.
Outcome
Draw
probability 
Green
Red
Yellow
12
15
23
number of yellow marbles drawn = 23
50
total number of marbles drawn
The probability of drawing a yellow marble is about
0.46, or 46%.
Check It Out! Example 1A
A ticket is randomly drawn out of a bag and
then replaced. The table shows the results
after 100 draws.
Estimate the probability of drawing a purple
ticket.
Outcome
Purple
Orange
Brown
Draw
55
22
23
probability 
number of purple tickets drawn = 55
100
total number of tickets drawn
The probability of drawing a purple ticket is about
0.55, or 55%.
Check It Out! Example 1B
A ticket is randomly drawn out of a bag and
then replaced. The table shows the results
after 100 draws.
Estimate the probability of drawing a brown
ticket.
Outcome
Purple
Orange
Brown
Draw
55
22
23
probability 
number of brown tickets drawn = 23
100
total number of tickets drawn
The probability of drawing a brown ticket is about
0.23, or 23%.
Check It Out! Example 1C
A ticket is randomly drawn out of a bag and
then replaced. The table shows the results
after 1000 draws.
Estimate the probability of drawing a blue
ticket.
Outcome
Red
Blue
Pink
Draw
285
112
603
probability 
number of blue tickets drawn = 112
1000
total number of tickets drawn
The probability of drawing a blue ticket is about
0.112, or 11.2%.
Additional Example 2: Sports Application
Use the table to compare the probability
that the Huskies will win their next game
with the probability that the Knights will
win their next game.
Additional Example 2 Continued
probability 
number of wins
total number of games
probability for a Huskies win  79  0.572
138
probability for a Knights win  90  0.616
146
The Knights are more likely to win their next
game than the Huskies.
Check It Out! Example 2
Use the table to compare the probability
that the Huskies will win their next game
with the probability that the Cougars will
win their next game.
Check It Out! Example 2 Continued
probability 
number of wins
total number of games
probability for a Huskies win  79  0.572
138
probability for a Cougars win  85  0.567
150
The Huskies are more likely to win their next
game than the Cougars.
Lesson Quiz: Part I
1. Of 425 students,234 seniors were enrolled in
a math course. Estimate the probability that
a randomly selected senior is enrolled in a
math course. 0.55, or 55%
2. Mason made a hit 34 out of his last 125
times at bat. Estimate the probability that he
will make a hit his next time at bat.
0.27, or 27%
Lesson Quiz: Part II
3. Christina polled 176 students about their
favorite yogurt flavor. 63 students’ favorite
flavor is vanilla and 40 students’ favorite
flavor is strawberry. Compare the probability
of a student’s liking vanilla to a student’s
liking strawberry.
about 36% to about 23%
Warm Up
1. If you roll a number cube, what are the
possible outcomes?
1, 2, 3, 4, 5, or 6
2. Add 1 + 1 =
6
12
1
4
1
2
3. Add 2 +36
5
9
=
Vocabulary
theoretical probability
equally likely
fair
mutually exclusive
disjoint events
Theoretical probability is used to estimate
probabilities by making certain assumptions
about an experiment. Suppose a sample space
has 5 outcomes that are equally likely, that is,
they all have the same probability, x. The
probabilities must add to 1.
x+x+x+x+x=1
5x = 1
x=1
5
A coin, die, or other object is called fair if all
outcomes are equally likely.
Additional Example 1: Calculating Theoretical
Probability
An experiment consists of spinning this spinner
once. Find the probability of each event.
A. P(4)
The spinner is fair, so all 5 outcomes
are equally likely: 1, 2, 3, 4, and 5.
number of outcomes for 4 1
P(4) =
=
5
5
Additional Example 1: Calculating Theoretical
Probability
An experiment consists of spinning this spinner
once. Find the probability of each event.
B. P(even number)
There are 2 outcomes in the
event of spinning an even
number: 2 and 4.
P(even number) = number of possible even numbers
5
2
=
5
Check It Out! Example 1
An experiment consists of spinning this spinner
once. Find the probability of each event.
A. P(1)
The spinner is fair, so all 5 outcomes
are equally likely: 1, 2, 3, 4, and 5.
number of outcomes for 1 1
P(1) =
=
5
5
Check It Out! Example 1
An experiment consists of spinning this spinner
once. Find the probability of each event.
B. P(odd number)
There are 3 outcomes in the
event of spinning an odd
number: 1, 3, and 5.
P(odd number) = number of possible odd numbers
5
3
=
5
Additional Example 2: Calculating Probability for a
Fair Number Cube and a Fair Coin
An experiment consists of rolling one fair
number cube and flipping a coin. Find the
probability of the event.
A. Show a sample space that has all outcomes
equally likely.
The outcome of rolling a 5 and flipping heads can
be written as the ordered pair (5, H). There are
12 possible outcomes in the sample space.
1H
1T
2H
2T
3H
3T
4H
4T
5H
5T
6H
6T
Additional Example 2: Calculating Theoretical
Probability for a Fair Coin
An experiment consists of rolling one fair
number cube and flipping a coin. Find the
probability of the event.
B. P(tails)
There are 6 outcomes in the event “flipping tails”:
(1, T), (2, T), (3, T), (4, T), (5, T), and (6, T).
P(tails) =
6
1
=
12
2
Check It Out! Example 2
An experiment consists of flipping two coins.
Find the probability of each event.
A. P(one head & one tail)
There are 2 outcomes in the event “getting one
head and getting one tail”: (H, T) and (T, H).
P(head and tail) =
2
1
=
4
2
Check It Out! Example 2
An experiment consists of flipping two coins.
Find the probability of each event.
B. P(both tails)
There is 1 outcome in the event “both tails”:
(T, T).
P(both tails) =
1
4
Additional Example 3: Altering Probability
Stephany has 2 dimes and 3 nickels. How
many pennies should be added so that the
probability of drawing a nickel is 3 ?
7
Adding pennies to the bag will increase the number
of possible outcomes.
3
3
=
5+x
7
3(5 + x) = 3(7)
Set up a proportion. Let x equal
the number of pennies
Find the cross products.
Additional Example 3 Continued
15 + 3x = 21
–15
– 15
3x = 6
3
3
Multiply.
Subtract 15 from both sides.
Divide both sides by 3.
x= 2
2 pennies should be added to the bag.
Check It Out! Example 3
Carl has 3 green buttons and 4 purple buttons.
How many white buttons should be added so
that the probability of drawing a purple button
is 2 ?
9
Adding buttons to the bag will increase the number
of possible outcomes. Let x equal the number of
white buttons.
4
2
=
7+x
9
2(7 + x) = 9(4)
Set up a proportion. Let x equal
the number of white buttons.
Find the cross products.
Check It Out! Example 3 Continued
14 + 2x = 36
–14
– 14
2x = 22
2
2
Multiply.
Subtract 14 from both sides.
Divide both sides by 2.
x = 11
11 white buttons should be added to the bag.
Two events are mutually exclusive, or
disjoint events, if they cannot both occur in
the same trial of an experiment. For example,
rolling a 5 and an even number on a number
cube are mutually exclusive events because
they cannot both happen at the same time.
Additional Example 4: Finding the Probability of
Mutually Exclusive Events
Suppose you are playing a game in which you
roll two fair number cubes. If you roll a total of
five you will win. If you roll a total of two, you
will lose. If you roll anything else, the game
continues. What is the probability that the
game will end on your next roll?
It is impossible to roll both a total of 5 and a total of
2 at the same time, so the events are mutually
exclusive. Add the probabilities to find the
probability of the game ending on your next flip.
Additional Example 4 Continued
The event “total = 2” consists of 1 outcome, (1, 1), so
P(total = 2) = 1 .
36
The event “total = 5” consists of 4 outcomes, (1, 4),
(2, 3), (4, 1), (3, 2), so P(total = 5) =
.4
36
P(game ends) = P(total = 2) + P(total = 5)
= 1 +4 = 5
36 36
36
The probability the game will end is 5 , or about 14%.
36
Check It Out! Example 4
Suppose you are playing a game in which you
flip two coins. If you flip both heads you win
and if you flip both tails you lose. If you flip
anything else, the game continues. What is the
probability that the game will end on your next
flip?
It is impossible to flip both heads and tails at the
same time, so the events are mutually exclusive.
Add the probabilities to find the probability of the
game ending on your next flip.
Check It Out! Example 4 Continued
The event “both heads” consists of 1 outcome, (H, H),
1
so P(both heads) = 4 . The event “both tails” consists of
1
1 outcome, (T, T), so P(both tails) = 4 .
P(game ends) = P(both tails) + P(both heads)
=1 + 1
4 4
1
=
2
1
The probability that the game will end is , or 50%.
2
Let’s work some problems…
An experiment consists of rolling a fair
number cube. Find each probability.
1
1. P(rolling an odd number)
21
2. P(rolling a prime number)
2
An experiment consists of rolling two fair
number cubes. Find each probability.
3. P(rolling two 3’s) 1
36
4. P(total shown > 10) 1
12
Warm Up
Multiply. Write each fraction in simplest form.
1.
2
5

3
5
6
25
2.
1
6

3
4
1
8
Write each fraction as a decimal.
2
3.
5
0.4
32
4. 125
0.256
Vocabulary
compound events
independent events
dependent events
A compound event is made up of two or more separate
events. To find the probability of a compound event, you
need to know if the events are independent or dependent.
Events are independent events if the occurrence of one
event does not affect the probability of the other. Events
are dependent events if the occurrence of one does
affect the probability of the other.
Let’s work some problems…
Determine if the events are dependent or independent.
A. getting tails on a coin toss and rolling a 6 on a number
cube
Tossing a coin does not affect rolling a number cube, so the two
events are independent.
B. getting 2 red gumballs out of a gumball machine
After getting one red gumball out of a gumball machine, the
chances for getting the second red gumball have changed, so the
two events are dependent.
Let’s work some problems…
Determine if the events are dependent or independent.
A. rolling a 6 two times in a row with the same number cube
The first roll of the number cube does not affect the second
roll, so the events are independent.
B. a computer randomly generating two of the same numbers
in a row
The first randomly generated number does not affect the
second randomly generated number, so the two events are
independent.
Let’s work some problems…
Three separate boxes each have one blue marble and one
green marble. One marble is chosen from each box.
A. What is the probability of choosing a blue marble from
each box?
The outcome of each choice does not affect the outcome of the
other choices, so the choices are independent.
1
In each box, P(blue) =
.
2
P(blue, blue, blue) =
1
2
·
1
2
·
1 = 1 = 0.125
8
2
Multiply.
Let’s work some problems…
Three separate boxes each have one blue marble and one green
marble. One marble is chosen from each box.
B. What is the probability of choosing a blue marble, then a
green marble, and then a blue marble?
1
In each box, P(blue) =
.
2
1
In each box, P(green) =
.
2
P(blue, green, blue) =
1
2
·
1
2
·
1 = 1 = 0.125 Multiply
2
8
Let’s work some problems…
Three separate boxes each have one blue marble and one green
marble. One marble is chosen from each box.
C. What is the probability of choosing at least one blue
marble?
Think: P(at least one blue) + P(not blue, not blue, not
blue) = 1.
In each box, P(not blue) =
1
.
2
P(not blue, not blue, not blue) =
1 · 1 · 1 = 1 = 0.125
Multiply.
2
2
2
8
Subtract from 1 to find the probability of choosing at least
one blue marble.
1 – 0.125 = 0.875
Let’s work some problems…
Two boxes each contain 4 marbles: red, blue, green, and
black. One marble is chosen from each box.
A. What is the probability of choosing a blue marble from
each box?
The outcome of each choice does not affect the outcome of the
other choices, so the choices are independent.
In each box, P(blue) =
1
.
4
P(blue, blue) =
1
4
·
1 = 1 = 0.0625
16
4
Multiply.
Let’s work some problems…
Two boxes each contain 4 marbles: red, blue, green, and black.
One marble is chosen from each box.
B. What is the probability of choosing a blue marble and then a
red marble?
1
In each box, P(blue) =
4
1
In each box, P(red) =
.
.
4
P(blue, red) =
1
4
·
1 = 1 = 0.0625
16
4
Multiply.
Let’s work some problems…
Two boxes each contain 4 marbles: red, blue, green, and black.
One marble is chosen from each box.
C. What is the probability of choosing at least one blue marble?
Think: P(at least one blue) + P(not blue, not blue) = 1.
1
In each box, P(blue) =
P(not blue, not blue) =
3
4
·
.
4
3 = 9 = 0.5625
16
4
Multiply.
Subtract from 1 to find the probability of choosing at least one blue
marble.
1 – 0.5625 = 0.4375
To calculate the probability of two dependent events
occurring, do the following:
1. Calculate the probability of the first event.
2. Calculate the probability that the second event would
occur if the first event had already occurred.
3. Multiply the probabilities.
Let’s work some problems…
The letters in the word dependent are placed in a box.
A. If two letters are chosen at random, without replacing the
first letter, what is the probability that they will both be
consonants?
Because the first letter is not replaced, the sample space is
different for the second letter, so the events are dependent. Find
the probability that the first letter chosen is a consonant.
P(first consonant) =
6 = 2
9
3
If the first letter chosen was a consonant, now there would be 5
consonants and a total of 8 letters left in the box. Find the
probability that the second letter chosen is a consonant.
P(second consonant) =
5
8
2 · 5 = 5
12
3
8
Multiply.
The probability of choosing two letters that are both consonants
5
is
.
12
Let’s work some problems…
B. If two letters are chosen at random, what is the probability
that they will both be consonants or both be vowels?
There are two possibilities: 2 consonants or 2 vowels. The
probability of 2 consonants was calculated in Example 3A. Now
find the probability of getting 2 vowels.
P(first vowel) =
3 = 1
9
3
Find the probability that
the first letter chosen is a
vowel.
If the first letter chosen was a vowel, there are now only 2
vowels and 8 total letters left in the box.
P(second vowel) =
2 = 1
8
4
Find the probability that the
second letter chosen is a vowel.
1
· 1 =
Multiply.
12
3
4
The events of both consonants and both vowels are mutually
exclusive, so you can add their probabilities.
1
5
12
+
1 =
12
6 = 1
2
12
P(consonant) + P(vowel)
The probability of getting two letters that are either both
consonants or both vowels is 1 .
2
Remember!
Two mutually exclusive events cannot both
happen at the same time.
Let’s work some problems…
The letters in the phrase I Love Math are placed in a box.
A. If two letters are chosen at random, without replacing the
first letter, what is the probability that they will both be
consonants?
Because the first letter is not replaced, the sample space is
different for the second letter, so the events are dependant. Find
the probability that the first letter chosen is a consonant.
P(first consonant) =
5
9
If the first letter chosen was a consonant, now there would
be 4 consonants and a total of 8 letters left in the box. Find
the probability that the second letter chosen is a consonant.
P(second consonant) =
5 · 1 = 5
18
9
2
4 = 1
8
2
Multiply.
The probability of choosing two letters that are both consonants
5
is
.
18
Let’s work some problems…
B. If two letters are chosen at random, what is the probability
that they will both be consonants or both be vowels?
There are two possibilities: 2 consonants or 2 vowels. The
probability of 2 consonants was calculated in Check It Out 3A.
Now find the probability of getting 2 vowels.
P(first vowel) =
4
Find the probability that the first
letter chosen is a vowel.
9
If the first letter chosen was a vowel, there are now only 3 vowels
and 8 total letters left in the box.
P(second vowel) =
3
8
Find the probability that the
second letter chosen is a vowel.
4 · 3 = 12 = 1
Multiply.
72
9
8
6
The events of both consonants and both vowels are mutually
exclusive, so you can add their probabilities.
5
18
+
1 = 8 =
18
6
4
P(consonant) + P(vowel)
9
The probability of getting two letters that are either both
consonants or both vowels is 4 .
9
Let’s work some problems…
Determine if each event is dependent or independent.
1. drawing a red ball from a bucket and then drawing a green
ball without replacing the first
dependent
2. spinning a 7 on a spinner three times in a row
independent
3. A bucket contains 5 yellow and 7 red balls. If 2 balls are
selected randomly without replacement, what is the probability
that they will both be yellow?
5
33
Probability and Counting
Continued
Grade 7 Pre-Algebra
Drawing a Tree Diagram
• You can use a tree diagram to display and count
possible choices.
– Example: A school team sells caps in two colors (blue or
white), two sizes (child or adult), and two fabrics (cotton or
polyester)
cotton
child
blue
polyester
Each branch of the “tree” represents
one choice – for example, bluechild-cotton
cotton
adult
polyester
cotton
child
white
polyester
cotton
adult
polyester
There are 8 possible cap choices
Let’s work some
problems…
Suppose the caps in the previous example also
came in black. Draw a tree diagram. How
many cap choices are there?
Answer:
Suppose the caps in the previous example also
came in black. Draw a tree diagram. How
many cap choices are there?
There are 12 possible cap choices
Counting Principle
• Another way to count choices is to use the Counting
Principle.
– If there are m ways of making one choice, and n ways of
making a second choice, then there are m · n ways of the
first choice followed by the second.
• The Counting principle is particularly useful when a
tree diagram would be too large to draw.
• The Counting Principle is sometimes called the
“Multiplication Counting Principle”.
Using the Counting Principle
• Example: How many two-letter monograms
are possible?
first letter
second letter
monograms
Possible choices
possible choices
possible choices
26
x
=
26
676
There are 676 possible two-letter monograms
Let’s work some
problems…
How many three-letter monograms are possible?
Answer:
• How many three-letter monograms are
possible?
26 x 26 x 26 = 17,576
Using a Tree Diagram
Example: Use a tree diagram to find the sample
space for tossing two coins. Then find the
probability of tossing two tails.
heads
heads
tails
The tree diagram shows there are
four possible outcomes, one of
which is tossing two tails.
heads
tails
tails
P(event) = number of favorable outcomes
number of possible outcomes
P(two tails) = number of two-tail outcomes
number of possible outcomes
=¼
The probability of tossing two tails is ¼ .
Let’s work some
problems…
You toss two coins. Find P(one head and one tail).
Answer:
You toss two coins. Find P(one head and one tail).
2/4 or ½
Using the Counting Principle
Example: In lotteries the winning number is made up of four digits chosen at
random. Suppose a player buys two tickets with different numbers. What
is the probability that the player has a winning ticket?
First find the number of possible outcomes. For each digit there are 10
possible outcomes, 0 through 9.
1st digit
2nd digit
3rd digit
4th digit
total
Outcomes
Outcomes
Outcomes
Outcomes
Outcomes
10
x
10
x
10
x
10
=
10,000
Then find the probability when there are two favorable outcomes.
P(winning ticket) = number of favorable outcomes =
2
number of possible outcomes
10,000
The probability is 2/10,000 or 1/5,000
Let’s work some
problems…
A lottery uses five digits chosen at random. Find
the probability of buying a winning ticket.
Answer:
A lottery uses five digits chosen at random. Find the probability of
buying a winning ticket.
10 x 10 x 10 x 10 x 10 = 100,000 (outcomes)
Then the probability when there is one favorable outcomes:
P(winning ticket) = # favorable outcomes
# possible outcomes
=
1
100,000
The probability is 1/100,000
Independent Events
• Independent Events are events for which the occurrence of
one event does not effect the probability of the occurrence of
the other.
– Suppose there are 10 cards with one number from 1 to 10 on them.
You are interested to draw an even number and then again a second
card with even number.
– If you replace your first card, the probability of getting an even
number on the second card is unaffected.
• Probability of Independent Events
– For two independent events A and B, the probability of both events
occurring is the product of the probabilities of each event occurring.
– P(A, then B) = P(A) x P(B)
Finding Probability of Independent Events
Example: You roll a number cube once. Then you roll it
again. What is the probability that you get 2 on the
first roll and a number greater than 4 on the second
roll?
There is one 2 among 6 numbers on a number cube.
P(2) = 1/6
There are two numbers greater than 4 on a number cube.
P(greater than 4) = 2/6
P(2, then greater than 4) = P(2) x P(greater than 4)
= 1/6 x 2/6
= 2/36, or 1/18
The probability is 1/18.
Let’s work some problems…
You toss a coin twice. Find the probability of
getting two heads.
Answer:
You toss a coin twice. Find the probability of
getting two heads.
½x½=¼
Let’s work some problems…
Under the best conditions, a wild bluebonnet seed has
20% probability of growing. If you select two seeds at
random, what is the probability that both will grow,
under best conditions?
Answer:
Under the best conditions, a wild bluebonnet seed has
20% probability of growing. If you select two seeds at
random, what is the probability that both will grow,
under best conditions?
P(a seed grows) = 20% or 0.20
P(two seeds grow) = P(a seed grows) x P(a seed grows)
= 0.20 x 0.20
= 0.04
= 4%
The probability that two seeds grow is 4%
Let’s work some problems…
Chemically treated bluebonnet seeds have a
30% probability of growing. You select two
such seeds at random. What is the probability
that both will grow?
Answer:
Chemically treated bluebonnet seeds have a
30% probability of growing. You select two
such seeds at random. What is the probability
that both will grow?
0.3 x 0.3 = 0.09 = 9%
Dependent Events
Dependent Events are events for which the occurrence of one
event affects the probability of the occurrence of the other.
– Suppose you want to draw two even-numbered cards from cards
having numbers from 1 to 10. You draw one card. Then , without
replacing the first card, you draw a second card. The probability of
drawing an even number on the second card is affected.
Probability of Dependent Events
– For two dependent events A and B, the probability of both events
occurring is the product of the probability of the first event and the
probability that, after the first event, the second event occurs.
– P(A, then B) = P(A) x P(B after A)
Finding Probability for Dependent Events
Example: Three girls and two boys volunteer to represent their
class at a school assembly. The teacher selects one name and
then another from a bag containing the five students’ names.
What is the probability that both representatives will be girls?
• P(girl) = 3/5
• P(girl after girl) = 2/4
Three of five students are girls.
If a girl’s name is drawn, two of the four remaining
students are girls.
• P(girl, then girl) = P(girl) x P(girl after girl)
= 3/5 x 2/4
= 6/20 or 3/10
The probability that both representatives will be girls is 3/10.
Let’s work some problems…
Three girls and two boys volunteer to represent their
class at a school assembly. The teacher selects one
name and then another from a bag containing the
five students’ names. Find:
• P(boy, then girl)
• P(girl, then boy)
Answer:
Three girls and two boys volunteer to represent
their class at a school assembly. The teacher
selects one name and then another from a
bag containing the five students’ names. Find:
• P(boy, then girl) 2/5 x ¾ = 6/20 = 3/10
• P(girl, then boy) 3/5 x 2/4 = 6/20 = 3/10
More Problems…
1.
You can choose a burrito having one filling wrapped in one
tortilla. Draw a tree diagram to count the number of burrito
choices.
Tortillas: flour or corn
Fillings: beef, chicken, bean, cheese, or vegetable
Answer:
1.
You can choose a burrito having one filling wrapped in one tortilla. Draw
a tree diagram to count the number of burrito choices.
Tortillas: flour or corn
Fillings: beef, chicken, bean, cheese, or vegetable
10 choices
flour
beef
beef
chicken
chicken
bean
cheese
vegetable
corn
bean
cheese
vegetable
More Problems…
There are four roads from Marsh to Taft and
seven roads from Taft to Polk. Use the
Counting Principle to find the number of
routes from Marsh to Polk through Taft.
Answer:
There are four roads from Marsh to Taft and
seven roads from Taft to Polk. Use the
Counting Principle to find the number of
routes from Marsh to Polk through Taft.
28 routes
More Problems…
Use a tree diagram to find the sample space for tossing
three coins. Then find the probability: P(three
heads).
Answer:
Use a tree diagram to find the sample space for tossing
three coins. Then find the probability: P(three
heads).
1/8
More Problems…
Use the Counting Principle to find the probability of
choosing the three winning lottery numbers when
the numbers are chosen at random from 1 to 50.
Numbers can repeat.
Answer:
Use the Counting Principle to find the probability of
choosing the three winning lottery numbers when
the numbers are chosen at random from 1 to 50.
Numbers can repeat.
1/125,000
More Problems…
Find the probability. You roll two odd numbers and pick
a vowel (when you roll two number cubes and pick
a letter of the alphabet at random).
Answer:
Find the probability. You roll two odd numbers and pick
a vowel (when you roll two number cubes and pick
a letter of the alphabet at random).
½ · ½ · 5/26 =
5/104
More Problems…
Weather forecasters are accurate 91% of the time when
predicting precipitation for the day. What is the
probability that a forecaster will make correct
precipitation predictions two days in a row?
Answer:
Weather forecasters are accurate 91% of the time when
predicting precipitation for the day. What is the
probability that a forecaster will make correct
precipitation predictions two days in a row?
About 83%
More Problems…
You pick a marble from a bag containing 1 green
marble, 4 red marbles, 2 yellow marbles, and 3
black marbles. You replace the first marble and
then select a second one. Find the probability
P(red, then yellow).
Answer:
You pick a marble from a bag containing 1 green
marble, 4 red marbles, 2 yellow marbles, and 3
black marbles. You replace the first marble and
then select a second one. Find the probability
P(red, then yellow).
8/100 or 2/25
Odds
Odds
• Another way to describe the chance of an event occurring is
with odds. The odds in favor of an event is the ratio that
compares the number of ways the event can occur to the
number of ways the event cannot occur.
• We can determine odds using the following ratios:
Odds in Favor =
Odds against =
number of successes
number of failures
number of failures
number of successes
Example
Suppose we play a game with 2 number cubes.
• If the sum of the numbers rolled is 6 or less – you win!
• If the sum of the numbers rolled is not 6 or less – you lose
In this situation we can express odds as follows:
Odds in favor =
numbers rolled is 6 or less
numbers rolled is not 6 or less
Odds against =
numbers rolled is not 6 or less
numbers rolled is 6 or less
Example
• A bag contains 5 yellow marbles, 3 white marbles, and 1 black
marble. What are the odds drawing a white marble from the
bag?
Odds in favor =
Odds against =
number of white marbles
number of non-white marbles
number of non-white marbles
number of white marbles
=
3
6
=
Therefore, the odds for are 1:2
and the odds against are 2:1
6
3
Let’s work some problems…
Find the probability of randomly choosing a red or white marble
from the given bag of red and white marbles.
Number of red marbles
Total number of marbles
16
64
Answer:
1/4
Find the probability of randomly choosing a red or white marble
from the given bag of red and white marbles.
Number of red marbles
Total number of marbles
8
40
Answer:
1/5
Find the probability of randomly choosing a red or white marble
from the given bag of red and white marbles.
Number of white marbles
Total number of marbles
7
20
Answer:
13/20
Find the probability of randomly choosing a red or white marble
from the given bag of red and white marbles.
Number of white marbles
Total number of marbles
24
32
Answer:
1/4
More Problems…
Find the favorable odds of choosing the
indicated letter from a bag that contains the
letters in the name of the given state.
S; Mississippi
N; Pennsylvania
A; Nebraska
G; Virginia
Answers:
4/11
1/4
1/4
1/8
More Problems…
You toss a six-sided number cube 20 times. For
twelve of the tosses the number tossed was
3 or more. What are the favorable odds that
the number tossed was 3 or more?
Answer:
3/2
Competition problems…
Alexandria has 3 skirts that she wears to school: 1 blue, 1
yellow, and 1 red. She also has 4 blouses: 1 yellow, 1
white, 1 striped, and 1 tan. Holly is wearing a blue skirt
and white blouse today. If Alexandria chose one of her
skirt-blouse outfits at random, what is the probability she
and Holly are wearing the same outfit?
Answer:
1/12
If the probability of rain is 3/11,
what are the odds in favor of
rain?
Answer:
3/8
Or
3:8
How many different nine letter
arrangements are there for the
letters of the word
TENNESSEE?
Answer:
__9!__
4!2!2!
3780
The ratio of Auburn to Alabama fans at
Eva School is 2:5. There are 840
Alabama or Auburn fans in attendance.
How many more Auburn fans would
need to attend to make the ratio (of
Auburn to Alabama fans) 1:2?
Answer:
60
Alex and Adam are going to the
amusement park. They cannot decide in
which order to ride the 12 roller coasters
in the park. If they only have time to
ride 8 of the roller coasters, how many
ways can they do this?
Answer:
19,958,400
At a pediatrician’s office, 2 girls and 4
boys are seen in random order. What is
the probability that the 2 girls are seen
in a row?
Answer:
5 · 2/6 · 1/5
1/3
In how many ways can
the letters
MATHEMATICS be
arranged?
Answer:
4,989,600
When two fair six-sided dice are rolled,
what is the probability that the sum of the
numbers on their upper faces is five?
Answer:
1/9
When two cards are drawn without
replacement from a standard 52-card deck,
what is the probability that they are the
same suit?
Answer:
Draw any suit 1st:
♠ or ♣ or ♥ or ♦ = 4/4=1
Then, 12 cards of that suit left = 12/51
4/4 · 12/51 =
4/17
Bag B contains three red marbles and two
green marbles. Bag C contains two red
marbles and one green marble. If a marble
is drawn from each bag, what is the
probability the two marbles are the same
color?
Answer:
3/5 · 2/3 = 6/15 or
2/5 · 1/3 = 2/15
6/15 + 2/15 =
8/15
A bag contains three white marbles and
three black marbles. When two marbles are
drawn, what is the probability they are both
white?
Answer:
1/5
In how many ways can the letters in the
word “DIVISION” be arranged?
Answer:
6720
Bag J contains two blue and three white
marbles. Bag K contains one blue and four
white marbles. If a single marble is drawn
from Bag J and placed in Bag K, and then a
marble is drawn from Bag K, what is the
probability that this final marble is white?
Answer:
3/5 · 5/6 = 15/30 or(add)
2/5 · 4/6 = 8/30
15/30 + 8/30 =
23/30
When four fair coins are flipped, what is
the probability that exactly two of them
show heads?
Answer:
6(possibilities) · 1/16 =
3/8
My bookshelf has three shelves, each of
which can hold up to five books. My book
collection is four J.R.R. Tolkein books,
three E.B. White books and two Lewis
Carroll books. If books by the same author
must be touching one another, in how many
ways can I arrange my books on the
bookshelf?
Answer:
3(ways of JRR Books) · 6(ways of remainder) · 4! · 3! · 2!
5184
Challenging problems…
Two dice are rolled and the numbers on top are
added. This sum is then multiplied by 4, then
7 is added. What is the probability that the
resulting number is 23?
Answer:
23-7=16
16÷4=4
Sum=4
1/12
The End…?