Download Probability of two dependent events

12-1: The Counting Principle
Learning Targets:
 I can distinguish between independent
and dependent events.
 I can solve problems involving
independent and dependent events.
The Counting Principle
Definitions
trial: an experiment (like flipping a coin)
outcome: the result of a single trial
sample space: a list of all possible outcomes
event: one or more outcomes of a trial
The Fundamental
Counting Principle
If event M can occur in m ways and event N can
occur in n ways, then event M followed by event
N can occur in m•n ways.
-works with dependent events
-works with independent events
Dependent and Independent
Events
independent: the outcome of one event does not
impact the outcome of another event
(rolling a die or tossing a coin)
dependent: the outcome of one event does
impact the outcome of another event
(taking a sock out of a drawer and then taking
another sock out of the same drawer without
replacement of the first one)
Independent Events
A sandwich menu offers customers a choice of white,
rye, or cheese bread with one spread chosen from
butter, mustard, or mayonnaise. How many different
combinations of bread and spread are there?
Make a tree diagram to see your sample space:
White
Butter
Mustard
Rye
Mayo
Butter
Mustard
Cheese Bread
Mayo
Butter
Mustard
Mayo
Nine possible combinations.
You could also do:
Number of Breads ● Number of Spreads =
3●3=9
Independent Events
A pizza place offers customers a choice of American,
mozzarella, Swiss, feta, or provolone cheese with one
topping chosen from pepperoni, mushrooms or
sausage. How many different combinations of cheese
and toppings are there?
Make a tree diagram to see your sample space:
Give a few minutes to complete.
Independent Events – Tree Diagram
American
Pep. Mush. Sau.
Mozzarella
Swiss
Pep. Mush. Sau. Pep. Mush. Sau.
Feta
Provolone
Pep. Mush. Sau. Pep. Mush. Sau.
15 possible combinations.
You could also do:
Number of Cheeses ● Number of Toppings =
5 ● 3 = 15
More Than Two Independent
Events
Communication How many codes are possible if an
answering machine requires a 2-digit code to
retrieve messages?
Two Digit Code:
10
10
____ ____
How many digits can you choose from for each spot?
More Than Two Independent
Events
Communication How many codes are possible if an
answering machine requires a 2-digit code to
retrieve messages?
Two Digit Code: _10_ ● _10_
There are 100 different codes to choose from.
Possibilities
Digits
0-9 = 10
Letters
A-Z = 26
Cards
52 total
4 suits 13 cards per suit
Dependent Events
How many different schedules could a student have
who is planning to take 4 different classes? Assume
each class is offered each period.
First Period Choices ● Second Period Choices ● Third Period Choices ● Fourth Period Choices
If a class is choosen for first hour, it can not been choosen again.
Dependent Events
How many different schedules could a student have
who is planning to take 4 different classes? Assume
each class is offered each period.
4 ● 3 ● 2 ● 1 = 24
Assignment
Work on the 21 problems that follow in the note
packet.
Algebra 2A - Chapter 12
Section 2
Permutations and
Combinations
12-2: Permutations and Combinations
Learning Targets:
 I can solve problems with
permutations.
 I can solve problems with
combinations.
Permutations
permutation: when a group of objects or people
are arranged in a certain order
- order of objects very important
The number of permutations of n distinct
objects taken r at a time is given by
n!
P(n,r ) 
(n  r )!
Also written n Pr
Permutations
n!
P(n,r ) 
(n  r )!
Also written n Pr
Eight people enter the Best Pic contest. How many
ways can blue, red, 
and green ribbons be awarded?
Order Matters!!!
Permutations
n!
P(n,r ) 
(n  r )!
Also written n Pr
How many permutations of the letters MATH are
possible?

Order Matters!!!
Permutations
n!
P(n,r ) 
(n  r )!
Also written n Pr
How many different four-letter code words can be
 EQUATIONS ?
formed from the word
Order Matters!!!
Also known as factorial:
Permutations
with Repetition
The number of permutations of n objects of which
p are alike and q are alike is:
n!
p!q!
or
n!
p!q! r!

How many different ways can the
letters of the word
BANANA be arranged?
You will notice some repetition
here. The letter A appears thrice
and the letter N appears twice.
Combinations
combination: an arrangement or selection of
objects in which order is not important
The number of combinations of n distinct
objects taken r at a time is given by
n!
C(n,r ) 
(n  r )! r!
Also written n Cr
Combinations
n!
C(n,r ) 
(n  r )! r!
Also written n Cr
Five cousins at a family reunion decide that three of
them will go to pick up a pizza. How many ways can
 to go?
they choose three people
Order Does Not Matters!!!
Combinations
n!
C(n,r ) 
(n  r )! r!
Also written n Cr
There are 60 players on a football team. Seven of
them will be chosen for a random drug test. How
many ways can they
be chosen?
Order Does Not Matters!!!
Multiple Events
n!
C(n,r ) 
(n  r )! r!
Also written n Cr
Six cards are drawn from a standard deck of cards.
How many hands consist
of two hearts and

four spades?
Order Does Not Matters!!!
There are 13 cards per suit.
Hearts Spades
Multiple Events
n!
C(n,r ) 
(n  r )! r!
Also written n Cr
Thirteen cards are drawn from a standard deck of
cards. How many hands
 consist of six hearts and
seven diamonds?
Order Does Not Matters!!!
There are 13 cards per suit.
Hearts Diamonds
How many of you have parents
that play the Lottery?
Let’s calculate the number of different
combinations there possibly are.
Mega Millions Total Combinations
Since the total number of combinations for Mega Millions numbers is
used in all the calculations, we will calculate it first. The number of ways 5
numbers can be randomly selected from a field of 56 is: COMBIN(56,5) =
3,819,816.
For each of these 3,819,816 combinations there are COMBIN(46,1) = 46
different ways to pick the sixth number (the “Mega” number). The total
number of ways to pick the 6 numbers is the product of these. Thus, the
total number of equally likely Mega Millions combinations is 3,819,816 x
46 = 175,711,536.

for Understanding
How many different ways can the letters of the word
ALGEBRA be arranged?
Six friends at a party decide that three of them will go
to pick up a movie. How many ways can they choose
three people to go?
Ten people are competing in a swim race where 4
ribbons will be given. How many ways can blue,
red, green, and yellow ribbons be awarded?
Assignment
p. 641: 4-32
Reflect
A class of 250 students wants to
elect a committee of 4 to buy
supplies for the homecoming float.
How many different committees are
possible?
Algebra 2A - Chapter 12
Section 3
Probability
12-3: Probability
Learning Targets:
 I can find probability and odds of
events.
 I can create and use graphs of
probability distributions.
Probability
success: desired outcome
failure: any outcome that is not a success
If an event can succeed in s ways and fail in f
ways, then the probabilities of success P(S), and
of failure, P(F), are as follows:
s
P(S) 
sf
f
P(F) 
sf
Probability
If an event can succeed in s ways and fail in f
ways, then the probabilities of success P(S), and
of failure, P(F), are as follows:
s
P(S) 
sf
f
P(F) 
sf
Probability is between 0 and 1, inclusive.

The closer to 1, the more likely the event is to
occur. The closer to 0, the less likely the event is
to occur.
Probability
When two coins are tossed, what is the
probability that both are tails?
Use a sample space, tree diagram:
Toss #1:
Toss #2:
H
H
T
T
H
T
Probability with Combinations
Monica has a collection of 32 CDs, of which 18
are R&B and 14 are rap. As she’s leaving for a
trip, she grabs 6 CDs. What is the probability
that she selects 3 R&B and 3 rap?
Probability with Combinations
Roman has a collection of 26 books–16 are fiction and
10 are nonfiction. He randomly chooses 8 books to
take with him on vacation. What is the probability that
he chooses 4 fiction and 4 nonfiction?
Odds
The odds that an event will occur can be
expressed as the ratio of the number of ways it
can succeed to the number of ways it can fail. If
an event can succeed in s ways and fail in f
ways, then the odds of success and of failure
are as follows:
Odds of success = s : f
Odds of failure = f : s
Notice: s + f = Total Possibilities
Odds
According to the CDC, the chances of a male
born in 1990 living to age 65 are about 3 in 4. For
females the chances are about 17 in 20.
What are the odds of a male living to be at least
65?
3:1
What are the odds of a female living to be at
least 65?
17:3
Probability Distributions
Which outcomes are least likely? most likely?
Suppose two dice
are rolled. The
table and the
relative-frequency
histogram show
the distribution of
the sum of the
numbers rolled.
S = Sum
Probability
2
3
4
5
6
7
8
9
10
11
12

for Understanding
When three coins are tossed, what is the probability
that all three are heads?
Life Expectancy The chances of a male born in 1980
to live to be at least 65 years of age are about 7 in 10.
For females, the chances are about 21 in 25.
Calculate the odds for each sex living at least 65
years.
Assignment
p. 647: 4-18
p. 648: 19-53
Reflect
If 7 out of 8 students prefer the
subject of math to literature, what
are the odds that students prefer
math? that students prefer
literature?
Algebra 2A - Chapter 12
Section 4
Multiplying
Probabilities
12-4: Multiplying Probabilities
Learning Targets:
 I can find the probability of two
independent events.
 I can find the probability of two
dependent events.
Probability Rules
Probability of two independent events:
P(A and B) = P(A) • P(B)
Probability of two dependent events:
P(A and B) = P(A) • P(B following A)
extends to
P(A, B, C) = P(A) • P(B following A) • P(C following A and B)
Independent Events
At a picnic Julio reaches into an ice-filled cooler
containing 8 regular and 5 diet soft drinks. He removes a
can, then decides he is not really thirsty, so he puts it
back. What is the probability that Julio and the next
person to reach into the cooler both randomly select a
regular soft drink?
This is a problem With Replacement!!
Independent Events
Gernardo has 9 dimes and 7 pennies in his pocket. He
randomly selects one coin, looks at it, and replaces it.
He then randomly selects another coin. What is the
probability that both of the coins he selects are
dimes?
This is a problem With Replacement!!
Independent Events
Extended
In a board game, three dice are rolled to determine the
number of moves for the players. what is the probability
that the first die shows a 6, the second die shows a 6, and
the third die does not?
P(6) P(6) P(not 6) =
Three Independent Events
When three dice are rolled, what is the probability
that two dice show a 5 and the third die shows an
even number?
P(5) P(5) P(even) =
Two DEPENDENT Events
In the previous Julio and the soft drink example, what is
the probability that both people select a regular soft drink
if Julio does NOT put his drink back into the cooler?
This is a problem Without Replacement!!
Two Dependent Events
The host of a game show is drawing chips from a bag to
determine prizes. Of the 10 chips in the bag, 6 show TV, 3
show VACATION, and 1 shows CAR. If the host draws the
chips at random without replacement, find the probabilities:
a. a vacation, then a car
b. two TVs
Three Dependent Events
Three cards are drawn from a standard deck without
replacement. Find the probability of drawing a
diamond, a club, and another diamond in that order.

for Understanding #1
When three dice are rolled, what is the probability
that one die is a multiple of 3, one die shows an
even number, and one die shows a 5?
P(x3) P(even) P(5) =

for Understanding #2
The host of a game show draws chips from a bag to
determine the prizes for which contestants will play.
Of the 20 chips in the bag, 11 show computer, 8 show
trip, and 1 shows truck. If the host draws the chips at
random and does not replace them, find
each probability.
a computer, then a truck

for Understanding #3
Three cards are drawn from a standard deck of cards
without replacement. Find the probability of drawing a
heart, another heart, and a spade in that order.
Assignment
p. 654: 4-12 even
p. 655: 14-34 even
p. 656: 40, 42
Reflect
When four dice are rolled, what is the probability
that two dice show a 3 and the third die shows an
even number?
Algebra 2A - Chapter 12
Section 5
Adding
Probabilities
12-5: Adding Probabilities
Learning Targets:
 I can find the probability of mutually
exclusive events.
 I can find the probability of inclusive
events.
What are mutually exclusive
events?
simple event: only one event (like rolling a 1)
compound event: two or more simple events
(like rolling an odd or a 6)
mutually exclusive events: events that cannot
occur at the same time (like when you consider
the prob of drawing a 2 or an ace---you can’t
draw a 2 and an ace at the same time, drawing a
2 and an ace are said to be mutually exclusive
events)
Probability of Mutually
Exclusive Events
If two events, A and B, are mutually exclusive,
then the probability that A or B occurs is the
sum of their probabilities.
P(A or B) = P(A) + P(B)
This can be extended to any number of mutually
exclusive events.
Two Mutually
Exclusive Events
Keisha has a stack of 8 baseball cards, 5
basketball cards and 6 soccer cards. What is the
probability that she selects a random card that
is a baseball or soccer card?
P(base or soc) = P(base) + P(soc)
P(base or soc) =
Three Mutually
Exclusive Events
There are 7 girls and 6 boys on the homecoming committee.
A subcommittee of four is being chosen at random to decide
the theme for the class float. What is the probability that the
subcommittee will have at least 2 girls?
P(at least 2 g) = P(2 girls) + P(3 girls) + P(4 girls)
What are Inclusive Events?
Inclusive events are ones whose outcomes may
be the same. They are NOT mutually exclusive
Example:
drawing a queen or a diamond
Q’s
Dia’s
Probability of Inclusive Events
If two events, A and B, are inclusive, then the
probability that A or B occurs is the sum of their
probabilities decreased by the probability of
both occurring.
P(A or B) = P(A) + P(B) - P(A and B)
Back to our
Queen of Diamonds
P(queen or diamond) =
= P(Queen) + P(Diamond) - P(Queen of Diamonds)
Inclusive Events
The enrollment at South High School is 1400.
Suppose 550 students take French, 700 take
algebra, and 400 take both French and algebra.
What is the probability that a student selected at
random takes French or algebra?
= P(French) + P(Algebra) - P(Both)

for Understanding
Sylvia has a stack of playing cards consisting of 10 hearts, 8
spades, and 7 clubs. If she selects a card at random from this
stack, what is the probability that it is a heart or a club?
There are 2400 subscribers to an Internet service provider. Of
these, 1200 own Brand A computers, 500 own Brand B, and 100
own both A and B. What is the probability that a subscriber
selected at random owns either Brand A or Brand B?
Assignment
p. 660-661: 4-16 all, 17-31 odd
Reflect
There are 200 students taking Calculus, 500 taking
Spanish, and 100 taking both. There are 1000
students in the school. What is the probability that a
student selected at random is taking Calculus or
Spanish?
Algebra 2A - Chapter 12
Review Sections 1-5
Probability.