CHAPTER 12:
PROBABILITY
AND
STATISTICS
Slide 1
Section 12-1
THE COUNTING PRINCIPLE
Thursday 5-11-17
Definitions:
1)Outcome-the result of a single trial.
Ex. Flipping a coin has two outcomes (head or tail).
2)Sample Space-the set of all possible outcomes.
3)Event-consists of one or more outcomes of a trial.
4)Independent Events-two events are independent if the
occurrence (or non-occurrence) of one of the events does
not affect the probability of the occurrence of the other
event.
Ex. The choices of letters and digits to be put on a
license plate, because each letter or digit chosen
does not affect the choices for the others.
12-1 Example
Probability experiment:
Sample space:
Roll a die
{1 2 3 4 5 6}
Event: { Die is even } = { 2 4 6 }
Outcome:
{3}
Section 12-1
THE COUNTING PRINCIPLE
5) Fundamental Counting Principle-If there are “a” ways for
one activity to occur and “b” ways for a second activity to
occur, then there are (a × b) ways for both to occur.
Ex. There are 6 ways to roll a die and 2 ways to flip
a coin… 6 × 2 = 12 ways for both to occur.
6) Dependent events-the outcome of one event DOES
affect the out come of another event.
12-1: Example: When two coins are tossed, what is the
probability that both are tails?
Tree Diagram
First Coin
H
/
T
\
/ \
Second Coin H
T
H
T
|
|
|
|
Possible
HH
HT
TH TT
Outcomes
1
2
3
4
Therefore, 4 possible outcomes.
Fundamental Counting Principle:
2 possible outcomes first coin X 2 possible outcomes second coin
= 4 possible outcomes
P(both tails)= 1
4
12-1: Independent and Dependent
Events
• Independent events: if one event occurs, it does
not affect the probability of the other event
– Drawing cards from two decks
• Dependent events: if one event affects the
outcome of the second event, changing the
probability
– Drawing two cards in succession from same deck
without replacement
12-1: Examples
Independent Events
A = Being female
B = Having type O blood
A = 1st child is a boy
B = 2nd child is a boy
Dependent Events
A = taking an aspirin each day
B = having a heart attack
A = being a female
B = being under 64” tall
12-1: Examples
Determine if the following events are independent or
dependent.
1. 12 cars are on a production line where 5 are defective
and 2 cars are selected at random.
A = first car is defective
B = second car is defective.
2.
Two dice are rolled.
A = first is a 4 and B = second is a 4
Skills Practice Workbook
pg. 79 (#1-11)
SG&I workbook
pg. 157 (#1-8)
12-3: PROBABILITY
Friday 5-12-17
Definition:
-Probability is the measure of how likely
something will occur.
-It is the ratio of desired outcomes to total
outcomes.
PROBABILITY= #desired
#total
-Probabilities MUST be simplified if possible.
12-3: EXAMPLE
Flip a coin…
-What is the probability you get heads?
-What is the probability you get tails?
(Remember think of all the possible outcomes)
12-3: Tree Diagrams
Two six-sided dice are rolled.
Describe the sample space.
Start
1st roll
1
2
1 2 3 4 5 6
1 2 3 4 5 6
3
4
12 3 4 5 6 1 2 3 4 5 6
5
6
1 2 3 4 5 6
1 2 3 4 5 6
2nd roll
6 possible outcomes on die #1
X 6 possible outcomes on die #2
= 36outcomes
12-3: Sample Space and Probabilities
Two dice are rolled and the sum is noted.
1,1
1,2
1,3
1,4
1,5
1,6
2,1
2,2
2,3
2,4
2,5
2,6
3,1
3,2
3,3
3,4
3,5
3,6
4,1
4,2
4,3
4,4
4,5
4,6
5,1
5,2
5,3
5,4
5,5
5,6
6,1
6,2
6,3
6,4
6,5
6,6
Find the probability the sum is 4.
3/36 = 1/12 = 0.083
Find the probability the sum is 11.
2/36 = 1/18 = 0.056
Find the probability the sum is 4 or 11.
5/36 = 0.139
12-3: Probability of Success &
Failure
Definitions:
1-Success-desired outcome
2-Failure-any other outcome
3-Random-when all outcomes have an
equally likely chance of occurring
4-Odds-another way to measure the chance
of an event occurring, it is the ratio of the
number of successes to the number of
failures.
Ex. Success:failure
12-3: PROBABILITY VS. ODDS
Ex. What is the probability of rolling a die and
getting the number 5?
• Probability = #desired
total #
Probability=1
6
Now what are the odds?
• ODDS = success:failure
ODDS= 1:5
(means one chance of getting a number 5 : five chances of
not getting a number 5.)
12-3: Odds
Odds can be written as a fraction or
ratio (but must be simplified just like
probabilities):
Success
Failures
Success:Failures
Note: In sports we often look at wins
over losses.
12-3: Converting from Probability to ODDS
EXAMPLE: Flipped a coin four times…
P(two tails)= ¼ = .25 or 25% chance
What are the odds?
ODDS=
1 success
3 failures
1:3
12-3: PROBABILITY DISTRIBUTIONS
Monday 5-15-17
Definitions:
1-random variable-a variable whose value is
the numerical outcome of a random event.
Ex. When rolling a die we can let the random
variable “D” represent the number showing on
the die. So, “D” can equal 1, 2, 3, 4, 5, or 6.
12-3: PROBABILITY DISTRIBUTIONS
2-probability distribution-for a particular
random variable it is a function that maps the
sample space to the probabilities of the
outcomes in the sample space.
Ex.
# on Die
1
2
3
4
5
6
Probability
1/6
1/6
1/6
1/6
1/6
1/6
12-3: PROBABILITY DISTRIBUTIONS
3-uniform distribution-a distribution where all
of the probabilities are the same.
4-relative-frequency histogram-a table of
probabilities or graph that helps to visualize a
probability distribution.
12-3: Probability
Probability is based on observations or experiments.
THERE ARE TWO TYPES OF PROBABILITY:
1- Experimental AND
2-Theoretical
Definitions:
An experimental probability is one that happens as the
result of an experiment.
# outcomes
# trials
***The probability we have done so far has been
“theoretical probabilities,” because there was no
experiment. **Except when you do a project!
12-3: THEORETICAL AND EXPERIMENTAL PROBABILITY
The probability of an event is a number between
0 and 1 that indicates the likelihood the event will occur.
• If P(E) = 0, then event E is impossible.
• If P(E) = 1, then event E is certain.
0  P(E)  1
Impossible
0
Even
.5
Certain
1
12-3: THEORETICAL AND EXPERIMENTAL PROBABILITY
THE THEORETICAL PROBABILITY OF AN EVENT
The theoretical
probability
of anlikely,
eventthe
is
When
all outcomes
are equally
often simplyprobability
called the probability
of theA event.
theoretical
that an event
will occur is:
P (A) =
number of outcomes in A
total number of outcomes
all possible
outcomes
P (A) = 4
9
outcomes
in event A
You can express a probability as a fraction, a decimal, or a percent.
For example:1 , 0.5, or 50%.
2
12-3: EXAMPLE Theoretical
P(A) =
number if ways A can occur
total number of outcomes
In a bag you have 3 red marbles, 2 blue
marbles and 7 yellow marbles. If you
select one marble at random,
P(red) =
12-3: EXAMPLE Theoretical
P(A) =
number if ways A can occur
total number of outcomes
In a bag you have 3 red marbles, 2 blue
marbles and 7 yellow marbles. If you
select one marble at random,
P(red) = 3 / (3+2+7) = 3/12 = 1/4
12-3: Experimental examples
• Prentice went fishing at a pond that contains
three types of fish: blue gills, red gills and
crappies. He caught 40 fish and recorded the
type. The following frequency distribution
shows his results.
Fish Type Number of times caught
Blue gill
13
Red gill
17
Crappy
10
If you catch a fish, what is the probability that it
is a blue gill? A red gill? A crappy?
12-3: Subjective Probability
• Subjective probability results from
educated guesses, intuition and estimates.
Examples…
• A doctor’s prediction that a patient has a 90%
chance of full recovery
• A business analyst predicting an employee strike
being 0.25
12-3: Summary
• Classical (Theoretical)
• The number of outcomes in a sample space is
known and each outcome is equally likely to
occur.
• Empirical (Statistical)
A.K.A. Experimental
• The frequency of outcomes in the sample space is
estimated from experimentation.
• Subjective (Intuition)
• Probabilities result from intuition, educated
guesses, and estimates.
COMPLETE PROBABILITY AND ODDS WORKSHEET
12-4: Independent Events
Tuesday 5-16-17
Whatever happens in one event has absolutely nothing
to do with what will happen next because:
1. The two events are unrelated
OR
2. You repeat an event with an item whose
numbers will not change (ex. spinners or
dice)
OR
3. You repeat the same activity, but you
REPLACE the item that was removed.
The probability of two independent events, A and B, is equal to the
probability of event A times the probability of event B.
P(A∩B) = P(A) ● P(B)
Slide 28
12-4: Multiplication Rule for
Independent Events
• To get probability of both events
occurring, multiply probabilities of
individual events
• Ace from first deck and spade from
second
– Probability of ace is 4/52 = 1/13
– Probability of spade is 13/52 = 1/4
– Probability of both is 1/13 x 1/4 = 1/52
12-4: INDEPENDENT EVENTS
Practice:
Roll a die and flip a coin:
1-P(heads and 6) =
2-P(tails and a 5) =
12-4: INDEPENDENT EVENTS
Practice:
Roll a die and flip a coin:
1-P(heads and 6) = ½ x 1/6 =
2-P(tails and a 5) =
12-4: INDEPENDENT EVENTS
Practice:
Roll a die and flip a coin:
1-P(heads and 6) = ½ x 1/6 = 1/12
2-P(tails and a 5) =
12-4: INDEPENDENT EVENTS
Practice:
Roll a die and flip a coin:
1-P(heads and 6) = ½ x 1/6 = 1/12
2-P(tails and a 5) = ½ x 1/6 =
12-4: INDEPENDENT EVENTS
Practice:
Roll a die and flip a coin:
1-P(heads and 6) = ½ x 1/6 = 1/12
2-P(tails and a 5) = ½ x 1/6 = 1/12
12-4: Independent Events
Example: Suppose you spin each of these two spinners. What
is the probability of spinning an even number and a vowel?
P(even) =
P(vowel) =
P(even and vowel) =
1
6
P
S
5
2
O
T
3
4
R
Slide 35
12-4: Independent Events
Example: Suppose you spin each of these two spinners. What
is the probability of spinning an even number and a vowel?
1
P(even) =
(3 evens out of 6 outcomes)
2
1
(1 vowel out of 5 outcomes)
P(vowel) =
5
P(even and vowel) =
1
6
P
S
5
2
O
T
3
4
R
Slide 36
12-4: Independent Events
Example: Suppose you spin each of these two spinners. What
is the probability of spinning an even number and a vowel?
1
P(even) =
(3 evens out of 6 outcomes)
2
1
(1 vowel out of 5 outcomes)
P(vowel) =
5
1 1 1
P(even and vowel) =  
2 5 10
1
6
P
S
5
2
O
T
3
4
R
Slide 37
12-4: Independent Events
Find the probability
P(jack and factor of 12)
x
=
Slide 38
12-4: Independent Events
Find the probability
P(jack and factor of 12) 1
5
5
x
8
=
Slide 39
12-4: Independent Events
Find the probability
P(jack and factor of 12) 1
5
5
x
8
5
=
40
1
8
Slide 40
12-4: Independent Events
Find the probability
• P(6 and not 5)
x
=
Slide 41
12-4: Independent Events
Find the probability
• P(6 and not 5)
1
6
5
x
6
=
Slide 42
12-4: Independent Events
Find the probability
• P(6 and not 5)
1
6
5
x
6
5
=
36
Slide 43
12-4: Dependent Event
• What happens during the second event depends upon
what happened before.
• In other words, the result of the second event will change
because of what happened first.
• Determining the probability of a dependent event is
usually more complicated than finding the probability of a
independent event.
The probability of two dependent events, A and B, is equal to the
probability of event A times the probability of event B. However,
the probability of event B now depends on event A.
P(A∩B) = P(A) ● P(B)
Slide 44
12-4:
So for both INDEPENDENT & DEPENDENT EVENTS…
To find the probability that two events, A and B will occur
in sequence, multiply the probability A occurs by the
conditional probability B occurs, given A has occurred.
P(A and B) = P(A) x P(B )
***EXCEPTION: For dependent,
P(B given A)
DEPENDENT EXAMPLE:
Two cars are selected from a production line of 12 where 5
are defective. Find the probability both cars are defective.
A = first car is defective B = second car is defective.
P(A) = 5/12
P(B given A) = 4/11
P(A and B) = 5/12 x 4/11 = 5/33 = 0.1515
12-4: Probability Examples
Ex.1) Independent Events:
Spinner #1 is partitioned into three equal sections, colored
black, white, and grey. Spinner #2 is partitioned into four
equal sections, colored red, blue, green, and yellow. If
both spinners are spun, what is the probability of getting
black and red?
12-4: Probability Example
Since we expect to get black one-third of the time, and we
expect to get red one-quarter of the time, then we expect
to get black one-third and red one-quarter of the time. . .
P (black and red )

Imagine a tree diagram where the first column shows the
three outcomes for Spinner #1, each of which is followed by
the four outcomes for Spinner #2 in the second column.
Three groups of four branches creates 12 possible outcomes.
12-4: Probability Example
P (black and red )
1

of
3
1


3
1

12
1
4
1
4
Imagine a tree diagram where the first column shows the
three outcomes for Spinner #1, each of which is followed by
the four outcomes for Spinner #2 in the second column.
Three groups of four branches creates 12 possible outcomes.
12-4: Probability Example
Ex.2) Dependent Events:
A bag contains 10 marbles; 5 red, 3 blue, and 2 silver. If you
draw one marble at random and hold it in your left hand, and
then draw a second marble at random and hold it in your
right hand, what is the probability that you are holding two
silver marbles?
It’s easy to determine the probability of the first marble being
silver. However, notice that if you start by getting a silver
marble and then try for the second, the bag will be different.
How? Now, there is only one silver marble in a bag
containing a total of 9 marbles. . .
12-4: Probability Example
P( silver and silver ) 
12-4: Probability Example
2 1
P ( silver and silver ) 

10 9
1
1


5
9

12-4: Probability Example
2 1
P( silver and silver) 

10 9
1
1


5
9
1

45
12-4: Dependent Event
Example: There are 6 black pens and 8 blue pens in a jar. If you
take a pen without looking and then take another pen without
replacing the first, what is the probability that you will get 2
black pens?
P(black first) =
P(black second) =
Slide 53
12-4: Dependent Event
Example: There are 6 black pens and 8 blue pens in a jar. If you
take a pen without looking and then take another pen without
replacing the first, what is the probability that you will get 2
black pens?
P(black first) =
6
3
or
14
7
5
P(black second) =
(There are 13 pens left and 5 are black)
13
THEREFORE………………………………………………
P(black and black) =??
Slide 54
12-4: Dependent Event
Example: There are 6 black pens and 8 blue pens in a jar. If you
take a pen without looking and then take another pen without
replacing the first, what is the probability that you will get 2
black pens?
P(black first) =
6
3
or
14
7
5
P(black second) =
(There are 13 pens left and 5 are black)
13
THEREFORE………………………………………………
3 5
15
P(black and black) = 
or
7 13
91
Slide 55
12-4: Dependent Events
Find the probability
• P(Q∩Q)
• All the letters of the
alphabet are in the
bag 1 time.
• Do not replace the
letter.
• ∩ means “and”.
• “AND” MEANS
MULTIPLY!!!
x
=
Slide 56
12-4: Dependent Events
Find the probability
• P(Q∩Q)
• All the letters of the
alphabet are in the
bag 1 time.
• Do not replace the
letter.
• ∩ means “and”.
• “AND” MEANS
MULTIPLY!!!
1
26
0
x
25
=
Slide 57
12-4: Dependent Events
Find the probability
• P(Q∩Q)
• All the letters of the
alphabet are in the
bag 1 time.
• Do not replace the
letter.
• ∩ means “and”.
• “AND” MEANS
MULTIPLY!!!
1
26
0
x
25
0
=
650
0
Slide 58
12-4: Examples
Finding Probabilities of Events
You roll a six-sided die whose sides are numbered from
1 through 6.
Find the probability of rolling a
4.
SOLUTION
Only one outcome corresponds to rolling a 4.
number of ways to roll a 4
=
P (rolling a 4) =
number of ways to roll the die
Finding Probabilities of Events
You roll a six-sided die whose sides are numbered from
1 through 6.
Find the probability of rolling a
4.
SOLUTION
Only one outcome corresponds to rolling a 4.
1
number of ways to roll a 4
=
P (rolling a 4) =
number of ways to roll the die 6
Finding Probabilities of Events
You roll a six-sided die whose sides are numbered from
1 through 6.
Find the probability of rolling an odd number.
SOLUTION
Three outcomes correspond to rolling an odd number:
rolling a 1, 3, or a 5.
P (rolling odd number) =
number of ways to roll an odd number
number of ways to roll the die
=??
Finding Probabilities of Events
You roll a six-sided die whose sides are numbered from
1 through 6.
Find the probability of rolling an odd number.
SOLUTION
Three outcomes correspond to rolling an odd number:
rolling a 1, 3, or a 5.
3
number of ways to roll an odd number
P (rolling odd number) =
= 6
number of ways to roll the die
=
1
2
Finding Probabilities of Events
You roll a six-sided die whose sides are numbered from
1 through 6.
Find the probability of rolling a number less than 7.
SOLUTION:
All six outcomes correspond to rolling a number less than 7.
P (rolling less than 7 ) =
number of ways to roll less than 7
number of ways to roll the die
=??
Finding Probabilities of Events
You roll a six-sided die whose sides are numbered from
1 through 6.
Find the probability of rolling a number less than 7.
SOLUTION:
All six outcomes correspond to rolling a number less than 7.
P (rolling less than 7 ) =
number of ways to roll less than 7 6
=
number of ways to roll the die
6
= 1
12-4:SUMMARY: Independent &
Dependent
1-Independent & Dependent Events
P(A∩B) = P(A) ● P(B)
P(A∩B) means probability of event A
and event B occurring. Multiply!
Symbol ∩ means the same as the
word “and”….both means multiply.
12-4: Independent
Events
Find the probability
P(jack and factor of 12) 1
5
5
x
8
5
=
40
1
8
Slide 67
12-4: Dependent
Events
Find the probability
• P(Q∩Q)
• All the letters of the
alphabet are in the
bag 1 time.
• Do not replace the
letter.
• ∩ means “and”.
• “AND” MEANS
MULTIPLY!!!
1
26
0
x
25
0
=
650
0
Skills Practice Workbook pg. 82 (1-15)
Slide 68
TWO OR MORE EVENTS
12-5: Adding Probabilities
Wednesday 5-17-17
Definitions:
1) Simple event-consists of only one event
2) Compound event-consists of two or more
simple events
3) Mutually exclusive events-cannot occur at
the same time
4) Inclusive events (not mutually exclusive)can occur at the same time
Compare “A and B” to “A or B”
Multiplication vs. Adding
The compound event “A and B” means that A and
B both occur in the same trial. Use the
multiplication rule to find P(A and B).
The compound event “A or B” means either A
can occur without B, B can occur without A or
both A and B can occur. Use the addition rule to
find P(A or B).
B
A
B
A
A or B
A and B
Mutually Exclusive Events
Mutually Exclusive events-two events that cannot occur at the
same time.
P(A U B)=P(A) + P(B)
U means “or”. “OR” MEANS ADD!!!
Ex. Probability of drawing a 2 or an ace is found by adding
their individual probabilities.
P(2 or ace)= P(2) + P(ace) *Add probabilities.
4
52
+
4 *There are 4 two’s and 4
52 aces in a deck.
8 =
2 *Simplify.
52
13
The probability of drawing a 2 or an ace is 2/13.
Inclusive Events
Inclusive events (not mutually exclusive)-can
occur at the same time.
P(A U B)=P(A) + P(B) – P(A and B)
- Redundancy
Redundancy-meaning subtract what both have in
common!!
Ex. Drawing a spade or drawing an ace
– Probability of drawing a spade: 13 outcomes, so 13/52 =
1/4
– Probability of drawing an ace: 4 outcomes, so 4/52 = 1/13
– Ace of spades is common to both events, probability is
1/52
– So probability of drawing a spade or an ace is 1/4 + 1/13 –
1/52 = 16/52 = 4/13
Mutually Exclusive Events
Two events, A and B, are mutually exclusive if
they cannot occur in the same trial.
EXAMPLES:
A = A person is under 21 years old
B = A person is running for the U.S. Senate
A = A person was born in Philadelphia
B = A person was born in Houston
Mutually exclusive…
A
B
P(A and B) = 0
*No Redundancy
No common ground
When event A occurs it excludes event B in the same trial.
12-5: The Addition Rule
EXAMPLES:
A card is drawn from a deck. Find the
probability the card is a king or a 10.
A = the card is a king B = the card is a 10.
REDUNDANCY!
P(A) = 4/52 AND P(B) = 4/52 AND P(A and B)= 0/52
P(A or B) = 4/52 + 4/52 – 0/52
= 8/52
P(A) + P(B) – P(A and B)
When events are mutually exclusive,
P(A or B) = P(A) + P(B)
Non-Mutually Exclusive Events or
INCLUSIVE
If two events can occur in the same trial, they are
non-mutually exclusive (inclusive).
EXAMPLES:
A = A person is under 25 years old
B = A person is an attorney
A = A person was born in Philadelphia
B = A person watches Walking Dead on TV
A and B
Non-mutually exclusive
P(A and B) ≠ 0
A
B
12-5 : INCLUSIVE EVENTS
The probability that one or the other of two events
will occur is:
P(A) + P(B) – P(A and B)
- Redundancy
EXAMPLES:
A card is drawn from a deck. Find the
probability it is a king or it is red.
A = the card is a king
B = the card is red.
P(A) = 4/52 or P(B) = 26/52
but P(A and B) = 2/52
P(A or B) = 4/52 + 26/52 – 2/52
= 28/52
12-5: Examples
1. A die is rolled. Find the probability of
rolling a 6 or an odd number.
-are the events mutually exclusive?
-find P(A), P(B) and, if necessary,
P(A and B)
-use the addition rule to find the
probability
12-5: Examples
1. A die is rolled. Find the probability of
rolling a 6 or an odd number.
-are the events mutually exclusive (is a 6
an odd number)?
-find P(A), P(B) and, if necessary,
P(A and B)meaning if you have
redundancies…
-use the addition rule to find the
probability
12-5: Examples
1. A die is rolled. Find the probability of
rolling a 6 or an odd number.
-are the events mutually exclusive (is a 6
an odd number)? NO
-find P(A), P(B) and, if necessary,
P(A and B) Redundancies? NO
1/6 + 3/6 =
12-5: Examples
1. A die is rolled. Find the probability of
rolling a 6 or an odd number.
-are the events mutually exclusive (is a 6
an odd number)? NO
-find P(A), P(B) and, if necessary,
P(A and B) Redundancies? NO
1/6 + 3/6 = 4/6 = 2/3
12-5: Example
2. A card is selected from a standard deck.
Find the probability that the card is a face
card or a heart.
How many face cards in a deck? 12
How many heart cards in a deck? 13
-are there any redundancies (heart & face
card)?
-find P(A), P(B) and, if necessary,
P(A and B)
-use the addition rule to find the probability
12-5: Example
2. A card is selected from a standard deck.
Find the probability that the card is a face
card or a heart.
How many face cards in a deck? 12
How many heart cards in a deck? 13
-are there any redundancies (heart & face
card)? YES
SO…
-find P(A), P(B) and, if necessary,
P(A and B)
12-5: Example
2. A card is selected from a standard deck.
Find the probability that the card is a face
card or a heart.
How many face cards in a deck? 12
How many heart cards in a deck? 13
-find P(A) + P(B) - P(A and B)
-REDUNDANCIES
12/52 + 13/52 – 3/52 =
12-5: Example
2. A card is selected from a standard deck.
Find the probability that the card is a face
card or a heart.
How many face cards in a deck? 12
How many heart cards in a deck? 13
-find P(A) + P(B) - P(A and B)
-REDUNDANCIES
12/52 + 13/52 – 3/52 = 22/52
12-5: Contingency Table
Mark conducted a survey asking adults in three
different cities if they liked a new juice that just
came out. The results were as follows:
Omaha
Yes
100
No
125
Undecided 75
Total
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
3. P(Miami or Yes)
2. P(Miami and Seattle)
4. P(Miami or Seattle)
12-5: Contingency Table
Omaha
Yes
100
No
125
Undecided 75
Total
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
= 250/1000 • 150/250 = 150/1000
= 3/20
2. P(Miami and Seattle) = 0
12-5: Contingency Table
Omaha
Yes
100
No
125
Undecided 75
Total
300
Seattle
150
130
170
450
3. P(Miami or Yes)
4. P(Miami or Seattle)
Miami
150
95
5
250
Total
400
350
250
1000
=250/1000 + 400/1000 – 150/1000
= 500/1000 = 1/2
=250/1000 + 450/1000 – 0/1000
= 700/1000 = 7/10
Summary
Probability at least one of two events occur
P(A or B) = P(A) + P(B) - P(A and B)
Add the simple probabilities, BUT to prevent
double counting, don’t forget to subtract the
probability of both occurring…
THE REDUNDANCY.
***Skills Practice Worksheet-Adding Probabilities (#1-20).
Thursday 5-18
12-5 Review:
TWO OR MORE EVENTS:
(mutually exclusive and not mutually exclusive)
(1) Mutually Exclusive events-two events that
cannot occur at the same time. Exclusive-excluding
or restricted
P(A U B)=P(A) + P(B)
U means “or”.
“OR” MEANS ADD!!!
TWO OR MORE EVENTS:
(mutually exclusive and inclusive events)
(2) not mutually exclusive-or (inclusive
events) can occur at the same time. InclusiveIncluding all items normally expected or required,
containing (a specified element) as part of a
whole.
P(A U B)=P(A) + P(B) – P(A and B)
U means “or”.
“OR” MEANS ADD!!!
Conditional Probability
3-P(B│A) means probability of
event B given event A occurred.
Symbol “│” tells you which
one we want… it means
given.
Conditional Probability
• Probability of second event occurring given
first event has occurred
• Drawing a spade from a deck given you have
previously drawn the ace of spade
– After drawing ace of spades have 51 cards
left.
– Remaining cards now include only 12
spades.
– Conditional probability is then 12/51.
P(A│B) means probability of event A given
event B occurred!!!
PRACTICE
Probability Practice Problems
• Suppose you have a bag of marbles
numbered 1 – 15.
• P(even)?
Probability Practice Problems
• Suppose you have a bag of marbles
numbered 1 – 15.
• P(even)?
even numbers = 2, 4, 6, 8, 10, 12, 14 = 7
total numbers
1 through 15
=15
Probability Practice Problems
• Suppose you have a bowl of cds
numbered 1 – 15.
• P(even, more than 10)?
A COMMON SENSE PROBLEM!!!
THINK ABOUT IT!!!
Probability Practice Problems
• Suppose you have a bowl of cds
numbered 1 – 15.
• P(even, more than 10)?
Even #’s & #’s greater than 10 =12, 14= 2
Total numbers
= 15 = 15
Probability Practice Problems
• Suppose you have a bowl of cards numbered 1
– 15.
• P(even or more than 10)?
The “or” indicates the card must be even or more
than 10. You must be careful not to include a
number twice
Probability Practice Problems
• Suppose you have a bowl of cards numbered 1
– 15.
• P(even or more than 10)?
even #’s = 2, 4, 6, 8, 10, 12, 14 = 7/15
#’s greater than 10 = 11, 12, 13, 14, 15 = 5/15
Since 12 and 14 are common to both sets, you
will subtract 2/15
Solution: 7/15 + 5/15 – 2/15 = 10/15 = 2/3
Probability Practice Problems
• Suppose you have a bowl of coins numbered 1
– 15. A coin is drawn, replaced, and a second
coin is drawn.
• P(even and even)?
Probability Practice Problems
• Suppose you have a bowl of coins numbered 1
– 15. A coin is drawn, replaced, and a second
coin is drawn.
• P(even and even)?
Find the probability of each independent event
and multiply
Even #’s on first draw = 2, 4, 6, 8, 10, 12, 14 =
7/15
Even #’s on second draw = 2, 4, 6, 8, 10, 12, 14 =
7/15
Solution: 7/15 x 7/15 = 49/225
Probability Practice Problems
• Suppose you have a bowl of disks numbered 1
– 15. A disk is drawn, not replaced, and a
second disk is drawn.
• P(even and even)?
Probability Practice Problems
• Suppose you have a bowl of disks numbered 1 –
15. A disk is drawn, not replaced, and a second
disk is drawn.
• P(even and even)
Find the probability of each independent event and
multiply even #’s on first draw =
2, 4, 6, 8, 10, 12, 14 = 7/15
Even #’s on second draw = one less even number
than previous set/one less disk from bowl = 6/14
Solution: 7/15 x 6/14 = 42/210 = 1/5
12-4 Enrichment Worksheet (#1-10)