Transcript B - Cengage

A Mathematical View
of Our World
1st ed.
Parks, Musser, Trimpe,
Maurer, and Maurer
Chapter 10
Probability
Section 10.1
Simple Experiments
• Goals
• Study probability
• Experimental probability
• Theoretical probability
• Study probability properties
• Mutually exclusive events
• Unions and intersections of events
• Complements of events
10.1 Initial Problem
• Three cards were removed from
wedding presents and then randomly
replaced.
• What are the chances that at least one
of the gifts was paired with the correct
card?
• The solution will be given at the end of the section.
Interpreting Probability
• Probability is the mathematics of
chance.
• For example, the statement “The
chances of winning the lottery game are
1 in 150,000” means that only 1 of
every 150,000 lottery tickets printed is a
winning ticket.
Probability Terminology
• Making an observation or taking a
measurement is called an experiment.
• An outcome is one of the possible results of
an experiment.
• The set of all possible outcomes is called the
sample space.
• An event is any collection of possible
outcomes.
Example 1
• The experiment consists of rolling a
standard six-sided die and recording
the number of dots showing on the top
face.
• List the sample space.
• List one possible event.
Example 1, cont’d
• Solution: The sample space contains 6
possible outcomes and can be written
{1, 2, 3, 4, 5, 6}.
• One possible event is {2, 4, 6}, which is
the event of getting an even number of
dots.
Example 2
• The experiment consists of tossing a
coin 3 times and recording the results in
order.
• List the sample space.
• List one possible event.
Example 2, cont’d
• Solution: The sample space contains 8
possible outcomes and can be written
{HHH, HHT, HTH, THH, HTT, THT,
TTH, TTT}.
• One possible event is {HTH, HTT, TTH,
TTT}, which is the event of getting a tail
on the second coin toss.
Example 3
• The experiment
consists of spinning
a spinner twice and
recording the colors
it lands on.
• List the sample
space.
• List one possible
event.
Example 3, cont’d
• Solution: The sample space contains
16 possible outcomes and can be
written {RR, RY, RG, RB, YR, YY, YG,
YB, GR, GY, GG, GB, BR, BY, BG, BB}.
• One possible event is {RR, YY, GG,
BB}, which is the event of getting the
same color on both spins.
Example 4
• The experiment consists of rolling 2
standard dice and recording the
number appearing on each die.
• List the sample space.
• List one possible event.
Example 4, cont’d
• Solution: The sample space contains 36
possible outcomes and can be written {(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)}.
Example 4, cont’d
• Solution, cont’d: One possible event is
{(6,1), (5,2), (4,3), (3,4), (2,5), (1,6)},
which is the event of getting a total of 7
dots on the two dice.
Question:
1. An experiment consists of tossing a
coin and then rolling a 4-sided die. List
the outcomes in the sample space
a. { H1, H2, H3, H4, T1, T2, T3, T4 }
b. { H, T, 1, 2, 3, 4 }
c. { H1, H2, H3, H4, H5, H6, T1, T2, T3,
T4, T5, T6 }
d. { H, T, 1, 2, 3, 4, 5, 6 }
Probability, cont’d
• The probability of an event is a number
from 0 to 1, and can be written as a
fraction, decimal, or percent.
• The greater the probability, the more likely
the event is to occur.
• An impossible event has probability 0.
• A certain event has probability 1.
Experimental Probability
• One way to find the probability of an
event is to conduct a series of
experiments.
• The experimental probability is the relative
frequency with which an event occurs in a
particular sequence of trials.
Example 5
• An experiment
consisted of tossing
2 coins 500 times
and recording the
results
• Let E be the event of
getting a head on the
first coin and find the
experimental
probability of E.
Example 5, cont’d
• Solution: The event
E is {HH, HT}.
• Event E occurred a
total of 137 + 115 =
252 times out of
500.
• The experimental
probability of E is
252
 0.504
500
Question:
A total of 200 people are given a
taste test of 2 kinds of crackers. The
results are that 148 of them prefer
Cracker A, 41 of them prefer Cracker
B, and 11 have no preference.
Find the experimental probability of a
randomly selected person preferring
Cracker B.
a. 74.0%
c. 20.5%
b. 5.5%
d. 27.7%
Theoretical Probability
• Another way to find the probability of an
event is to use the theory of what
“should” happen rather than conducting
experiments.
• The theoretical probability is the chance
an event will occur based on the situation,
such as tossing a fair coin and knowing
each side should come up half of the time.
Theoretical Probability, cont’d
• If all the outcomes in a sample space
are equally likely to occur, then the
probability of event E is equal to the
number of outcomes in E divided by the
number of outcomes in the sample
space S.
• The probability of event E is written P(E).
Example 6
• An experiment consists of tossing 2
fair coins.
• Find the theoretical probability of:
a) Each outcome in the sample space.
b) The event E of getting a head on the first
coin.
c) The event of getting at least one head.
Example 6, cont’d
•
Solution:
a) There are 4 outcomes in the sample space:
{HH, HT, TH, TT}. Each outcome is equally
likely to occur.
Example 6, cont’d
• Solution, cont’d:
b) The event E is {HH, HT} and the
theoretical probability of E is the number
of outcomes in E divided by the number
of outcomes in the sample space.
•
2 1
P  E     0.5
4 2
Example 6, cont’d
• Solution, cont’d:
c) The event of getting at least one head is
E = {HH, HT, TH}.
•
3
P  E    0.75
4
Example 7
• An experiment consists of rolling 2 fair
dice.
• Find the theoretical probability of:
a) Event A: getting 7 dots.
b) Event B: getting 8 dots.
c) Event C: getting at least 4 dots.
Example 7, cont’d
•
Solution: There are 36 outcomes in the
sample space.
a) The event A contains 6 outcomes: {(1,6), (2,5),
(3,4), (4,3), (5,2), (6,1)}. Each outcome is
equally likely to occur.
•
6 1
P  A 

36 6
Example 7, cont’d
• Solution, cont’d:
b) The event B contains 5 equally likely
outcomes: {(2,6), (3,5), (4,4), (5,3),
(6,2)}.
•
5
P  B 
36
Example 7, cont’d
• Solution, cont’d:
c) The event C contains 33 equally likely
outcomes: {(1,3), (1,4), (1,5), (1,6), (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)}.
33 11

• P C  
36 12
Example 8
• A jar contains four
marbles: 1 red, 1
green, 1 yellow, and 1
white.
Example 8, cont’d
•
If we draw 2 marbles in a row, without
replacing the first one, find the probability
of:
a) Event A: One of the marbles is red.
b) Event B: The first marble is red or yellow.
c) Event C: The marbles are the same color.
d) Event D: The first marble is not white.
e) Event E: Neither marble is blue.
Example 8, cont’d
• Solution: The sample space contains
12 outcomes: {RG, RY, RW, GR, GY,
GW, YR, YG, YW, WR, WG, WY}.
a) Event A: One of the marbles is red.
• A = {RG, RY, RW, GR, YR, WR}.
• P  A  6  1
12 2
Example 8, cont’d
•
Solution, cont’d:
b) Event B: The first marble is red or yellow.
• B = {RG, RY, RW, YR, YG, YW}.
6 1

• P  B 
12 2
c) Event C: The marbles are the same color.
• C = { }.
0
P C  
0
•
12
Example 8, cont’d
•
Solution, cont’d:
d) Event D: The first marble is not white.
• D = {RG, RY, RW, GR, GY, GW, YR, YG,
YW}.
•
9 3
P  D 

12 4
e) Event E: Neither marble is blue.
• E=S
• P E  12  1
 
12
Question:
A jar contains 1 red marble, 1 green
marble, and 1 blue marble. You
draw 2 marbles in a row, without
replacing the first one. What is the
probability of the event E of the first
marble being red or yellow?
a. P(E) = 0
c. P(E) = ½
b. P(E) = 1/3
d. P(E) = 2/3
Union and Intersection
• The union of two events, A U B,
refers to all outcomes that are in one,
the other, or both events.
• The intersection of two events, A ∩ B,
refers to outcomes that are in both
events.
Mutually Exclusive Events
• Events that have no outcomes in
common are said to be mutually
exclusive.
•
A B  
• If A and B are mutually exclusive
events, then P  A  B   P  A  P  B 
Example 9
• A card is drawn from a standard deck of
cards.
• Let A be the event the card is a face card.
• Let B be the event the card is a black 5.
• Find and interpret P(A U B).
Example 9, cont’d
• Solution: The sample space contains
52 equally likely outcomes.
Example 9, cont’d
• Solution, cont’d: Event A has 12
outcomes, one for each of the 3 face
cards in each of the 4 suits.
• P(A) = 12/52.
• Event B has 2 outcomes, because
there are 2 black fives.
• P(B) = 2/52.
Example 9, cont’d
• Solution, cont’d: Events A and B are
mutually exclusive because it is
impossible for a 5 to be a face card.
• P(A U B) = 12/52 + 2/52 = 14/52 = 7/26.
• This is the probability of drawing either a
face card or a black 5.
Complement of an Event
• The set of outcomes in a sample space S,
but not in an event E, is called the
complement of the event E.
• The complement of E is written Ē.
Complement of an Event, cont’d
• The relationship between the probability
of an event E and the probability of its
complement Ē is given by:
 
• P  E   1 P E
 
• P E  1 P  E 
Example 10
•
In a number matching game,
•
First Carolan chooses a whole number from 1
to 4.
•
Then Mary guesses a number from 1 to 4.
a) What is the probability the numbers are
equal?
b) What is the probability the numbers are
unequal?
Example 10, cont’d
•
Solution: The sample space contains 16
outcomes: { (1,1), (1,2), (1,3), (1,4), (2,1),
(2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4),
(4,1), (4,2), (4,3), (4,4) }.
a) Let E be the event the numbers are equal.
•
P(E) = ¼
b) Then Ē is the event the numbers are
unequal.
•
P(Ē) = 1 – ¼ = ¾
Example 11
• A diagram of a sample space S for an
experiment with equally likely outcomes is
shown.
Example 11, cont’d
• Find the probability of each of the events:
• S
•
• A
• B
• C
• A B
• A B
• AC
• C
Example 11, cont’d
• Solution:
Properties of Probability
• For a sample space S and events A
and B:
1) For any event A,
2) P     0
3) P(S) = 1
0  P  A  1
Properties of Probability, cont’d
•
For a sample space S and events A and B:
4) If events A and B are mutually exclusive,
then P  A  B   P  A  P  B 
5) If A and B are any events, then
P  A  B   P  A  P  B   P  A  B 
6) For any event A and its complement:
 
P A  1  P  A
Example 12
• An experiment
consists of spinning
the spinner once
and recording the
number on which it
lands.
Example 12, cont’d
•
Define 4 events:
•
A: an even number
•
B: a number greater than 5
•
C: a number less than 3
•
D: a number other than 2
a) Find P(A), P(B), P(C), and P(D).
b) Find and interpret P(A U B) and P(A ∩ B).
c) Find and interpret P(B U C) and P(B ∩ C).
Example 12, cont’d
• Solution: The sample space has 8
equally likely outcomes: {1, 2, 3, 4, 5,
6, 7, 8}.
a) Find P(A),P(B), P(C), and P(D).
• A = {2, 4, 6, 8}, so P(A) = 4/8 = 1/2
• B = {6, 7, 8}, so P(B) = 3/8
• C = {1, 2}, so P(C) = 2/8 = 1/4
• D = {1, 3, 4, 5, 6, 7, 8}, so P(D) = 7/8.
Example12, cont’d
• Solution, cont’d:
b) Find and interpret P(A U B) and
P(A ∩ B).
• A and B are not mutually exclusive, so
P  A  B   P  A  P  B   P  A  B 
• A ∩ B = {6, 8}, so P(A ∩ B) = 2/8
4 3 2 5
P
A

B






•
8 8 8 8
Example 12, cont’d
• Solution, cont’d:
c) Find and interpret P(B U C) and
P(B ∩ C).
• B and C are mutually exclusive, so
P  B  C   0 and P  B  C   P  B   P C 
3 2 5
• PB C   
8 8 8
10.1 Initial Problem Solution
• Three cards were removed from
wedding presents and then randomly
replaced. What are the chances that at
least one of the gifts was paired with the
correct card?
Initial Problem Solution, cont’d
• Solution: Let E be the
event that at least
one gift is paired with
the right card.
• Label the gifts A, B,
and C and the cards
a, b, and c.
• The sample space
contains 6 outcomes.
Initial Problem Solution, cont’d
• Solution: Four of the 6 outcomes correspond
to at least one gift being paired with the right
card.
• The probability of this event occurring is
4 2
PE  
6 3
Section 10.2
Multistage Experiments
• Goals
• Study tree diagrams
• Study the fundamental counting principle
• Study probability tree diagrams
• Study the additive property
• Study the multiplicative property
10.2 Initial Problem
• A friend who likes to gamble wagers
that if you toss a coin repeatedly you
will get 2 tails before you get 3 heads.
• Should you take the bet?
• The solution will be given at the end of the
section.
Tree Diagrams
• A tree diagram is a visual aid that can
be used to represent the outcomes of
an experiment.
• You can construct a one-stage tree
diagram by starting from a single point:
1)Draw one branch for each outcome.
2)Place a label at the end of the branch to
represent each outcome.
Example 1
• A tree diagram is
drawn for the
experiment of
drawing 1 ball from
a box containing 1
red ball, 1 white ball,
and 1 blue ball.
Tree Diagrams, cont’d
• You can construct a two-stage tree diagram
for experiments that consist of a sequence
of 2 actions:
1) Draw a one-stage tree diagram, made of
primary branches, for the outcomes of the first
action.
2) Starting at the end of each branch of the tree
from Step 1, draw a one-stage tree diagram,
made of secondary branches, for each
outcome of the second action.
Example 2
• There are
12
possible
outcomes.
Question:
How many outcomes in
the tree diagram
correspond to getting
either a red or green
marble first and either a
yellow or green marble
second?
a. 3
c. 4
b. 6
d. 2
Fundamental Counting Principle
• The number of outcomes in an experiment
can also be determined using the
fundamental counting principle:
• If an event or action A can occur in r ways, and ,
for each of these r ways, an event or action B
can occur in s ways, then the number of ways
events or actions A and B can occur, in
succession, is rs.
• The principle can be extended to more than two
events or actions.
Example 3
• The options on a pizza are:
• Small, medium, or large
• White or wheat crust
• Sausage, pepperoni, bacon, onion,
mushrooms
• How many different one-topping pizzas
are possible?
Example 3, cont’d
• Solution:
• The first action is choosing 1 of 3 sizes.
• The second action is choosing 1 of 2
crusts.
• The third action is choosing 1 of 5
toppings.
• There are 3(2)(5) = 30 different onetopping pizzas possible.
Question:
In the previous example, what is the
probability of someone randomly
selecting a large, whole wheat pizza
with any one of the 5 toppings?
a. P(E) = 1
c. P(E) =1/3
b. P(E) = 1/30
d. P(E) = 1/6
Example 4
• Find the probability of getting a sum of
11 when tossing a pair of fair dice.
Example 4, cont’d
• Solution: There are 36 equally likely
outcomes in the sample space.
• 6 possible outcomes on the first roll
• 6 possible outcomes on the second roll
• 6(6) = 36
• There are 2 ways of rolling a sum of 11:
(5,6) and (6,5).
• The probability is 2/36 = 1/18.
Example 5
• Suppose 2 cards are drawn from a
standard deck.
• Use a two-stage tree diagram to find
the probability of getting a pair.
Example 5, cont’d
• Solution: A partial
tree diagram is
shown at right.
• There are 52 primary
branches.
• There are 51
secondary branches.
• So there are a total
of 52(51) = 2652
possible outcomes.
Example 5, cont’d
• Solution, cont’d:
Consider only the
outcomes that result
in a pair.
• There are 52 primary
branches.
• There are 3
secondary branches.
• So there are a total
of 52(3) = 156 pairs.
Example 5, cont’d
• Solution, cont’d:
• The probability of drawing a pair is:
156
1

2652 17
Probability Tree Diagrams
• Tree diagrams can also be used to
determine probabilities in multistage
experiments.
• Tree diagrams that are labeled with the
probabilities of events are called
probability tree diagrams.
Probability Tree Diagrams, cont’d
Example 6
• Draw a probability tree diagram to
represent the experiment of drawing
one ball from a container holding 2 red
balls and 3 white balls.
Example 6, cont’d
• Solution: The
first tree has one
branch for each
ball.
• The second tree
was simplified
by combining
branches.
Probability Tree Diagrams, cont’d
• If an event E is the union of events E1,
E2, …, En, where each pair of events is
mutually exclusive, then:
P  E   P  E1  E2 
 P  E1   P  E2  
En 
P  En 
• This is called the additive property of
probability tree diagrams.
Example 7
• Draw a probability
tree diagram to
represent the
experiment of
spinning the spinner
once.
• Find the probability
of landing on white
or on green.
Example 7, cont’d
• Solution: There
are 4 outcomes in
the sample space.
• They are not all
equally likely.
• They are all
mutually
exclusive.
Example 7, cont’d
• Solution, cont’d: Use
the central angles to
find the probability of
each outcome and
draw the probability
tree diagram.
Example 7, cont’d
• Solution, cont’d:
The probability of
white or green is:
P W  G 
 P W   P  G 
1 1 7
  
3 4 12
Example 8
• A jar contains 3 marbles, 2
black and 1 red.
• A marble is draw and
replaced, and then a
second marble is drawn.
What is the probability
both marbles are black?
Example 8, cont’d
• Solution: Draw a probability tree diagram to
represent the experiment.
Example 8, cont’d
• Solution, cont’d: Assign a probability to the end of
each secondary branch.
Example 8, cont’d
• Solution, cont’d: Either tree diagram can be
used to find that P(BB) = 4/9.
Probability Tree Diagrams, cont’d
• The previous example illustrates another
property of probability tree diagrams.
• If an experiment consists of a sequence of
actions represented by branches of a tree
diagram, the probability of the sequence of
actions is the product of all the probabilities
on those branches.
• This is called the multiplicative property of
probability tree diagrams.
Example 9
• A jar contains 3 red balls and 2 green balls.
Example 9, cont’d
• First a coin is tossed.
• If the coin lands heads, a red ball is added to
the jar.
• If the coin lands tails, a green ball is added to
the jar.
• Second a ball is selected from the jar.
• What is the probability a red ball is chosen?
Example 9, cont’d
• Solution: A probability tree diagram is
created.
Example 9, cont’d
• Solution, cont’d: The
probability of
choosing a red ball
is found by adding
the probabilities at
the end of the
branches labeled R.
4 3
7
P  R   
•
12 12 12
Example 10
• A jar contains 3 marbles, 2
black and 1 red.
• A marble is drawn and not
replaced before a second
marble is drawn.
• What is the probability that
both marbles were black?
Example 10, cont’d
• Solution: Create a probability tree diagram
to represent the experiment.
Example 10, cont’d
• Solution, cont’d: The probability of choosing
2 black marbles is:
2 1 1
P  BB    
3 2 3
Question:
What is the probability of drawing a
black marble second?
a. P(E) = ½
b. P(E) = 2/3
c. P(E) = 1/6
d. P(E) = 1/3
Example 11
• A jar contains 2 red gumballs and 2
green gumballs.
• An experiment consists of drawing
gumballs one at a time from a jar
until a red one is chosen.
• Find the probability of:
• A: only 1 draw is needed
• B: exactly 2 draws are needed
• C: exactly 3 draws are needed
Example 11, cont’d
• Solution: Create a
probability tree diagram.
• The probabilities are:
3
• P  A 
5
2 3 3
• P  B   
5 4 10
• P  C   2  1 1  1
5 4
10
Example 12
• Both spinners are spun.
• Find the probability both spinners stop
on the same color.
Example 12, cont’d
• Solution: Create a probability tree diagram.
• The desired event is {WW, RR, GG}.
Example 12, cont’d
• Solution, cont’d: The 3 outcomes are
mutually exclusive.
• P WW   1  1  1
3 2
•
•
6
1 1 1
P  RR    
4 6 24
1 1 1
• P  GG   4  3  12
1 1 1
7
P WW  RR  GG   
 
6 24 12 24
10.2 Initial Problem Solution
• A friend who likes to gamble wagers
that if you toss a coin repeatedly you
will get 2 tails before you get 3 heads.
Should you take the bet?
• You can figure out what you should do
by creating a probability tree diagram.
Initial Problem Solution, cont’d
Initial Problem Solution, cont’d
• Add the probabilities of the outcomes
that result in a win for you.
•
1 1 1 1 5
P Win      
16 16 16 8 16
• You are much more likely to lose than to
win, so you probably should not take the
bet.
Section 10.3
Conditional Probability, Expected
Value, and Odds
• Goals
• Study conditional probability
• Study independent events
• Study odds
10.3 Initial Problem
• If you bet $100 on
one number on
the roulette wheel,
what is your
expected gain or
loss?
• The solution will be given
at the end of the section.
Conditional Probability
• Sometimes an imposed condition forces
us to focus on a portion of the sample
space called the conditional sample
space.
Conditional Probability, cont’d
• For example, in the experiment of
tossing 3 fair coins suppose you know
the first coin came up heads.
• The sample space is {HHH, HHT, HTH,
THH, HTT, THT, TTH, TTT}.
• The conditional sample space is {HHH,
HHT, HTH, HTT}.
Conditional Probability, cont’d
• The probability of A given B is called a
conditional probability.
• The probability of A given B means the
probability of event A occurring within
the conditional sample space of event
B.
• The probability of A given B is written
P(A | B),
Conditional Probability, cont’d
• Suppose A and B are events in a
sample space S and that the probability
of B is not zero.
• The formula for conditional probability is
P  A B 
P  A  B
P  B
Example 1
• One jar of marbles contains 2 white marbles and 1
black marble.
• Another jar of marbles contains 1 white marble and
2 black marbles.
Example 1, cont’d
• A coin is tossed.
• Heads: a marble is selected from the first
jar.
• Tails: a marble is selected from the second
jar.
• Find the probability that the coin landed
heads up, given that a black marble was
drawn.
Example 1, cont’d
• Solution: Create a
probability tree
diagram.
• The sample space is
{HW, HB, TW, TB}.
• The probabilities are
labeled on the
diagram.
Example 1, cont’d
• Solution, cont’d:
•
1 2 3
P  B   
6 6 6
1
• P  H  B 
6
• P  H B 
P  H  B
P  B
1
1
6


3
3
6
Question:
For the experiment in the previous
example, find the probability the
coin landed tails given that a white
marble was drawn.
a. P(T|W) = ½
b. P(T|W) = 1/6
c. P(T|W) = 2/3
d. P(T|W) = 1/3
Example 2
•
Suppose a test for a viral infection is not
100% accurate.
•
Of the population, ¼ is infected and ¾ is not.
•
Of those infected, 90% test positive.
•
Of those not infected, 80% test negative.
a) What is the probability the test is correct?
b) Given that a person’s test is positive, what
is the probability the person is infected?
Example 2, cont’d
•
Solution: Create a
probability tree
diagram.
•
Of the population, ¼
is infected and ¾ is
not.
•
Of those infected,
90% test positive.
•
Of those not infected,
80% test negative.
Example 2, cont’d
•
Solution, cont’d:
a) The probability of a
correct test is the
probability of a
positive test for an
infected person or a
negative test for an
uninfected person.
1 9 3 4 33
• 
  
 82.5%
4 10 4 5 40
Example 2, cont’d
• Solution, cont’d:
b) Find the
conditional
probability:
• P(positive)
= 9  3  15
40 20 40
Example 2, cont’d
b) Solution, cont’d:
• P(infected and
positive) = 9
40
• P(infected|positive)
9
40  9  60%
= 15
15
40
Example 3
• Results from an inspection of a candy company’s 2
production lines are shown in the table.
• If a customer find a sub-standard piece of candy,
what is the probability it came from the Bay City
factory?
Example 3, cont’d
• Solution: Use the numbers in the table
to solve:
• P(Bay City and sub-standard) =
4
4

212  4  137  7 360
Example 3, cont’d
• Solution, cont’d:
• P(sub-standard) = 4  7  11
360
360
Example 3, cont’d
• Solution, cont’d:
• P(Bay City|sub-standard) =
4
360  4
11
11
360
Independent Events
• Two events are called independent if
one event does not influence the other.
• When 2 events are independent, their
probabilities follow the rule given below:
•
P  A  B   P  A  P  B 
Example 4
• Draw 2 marbles without replacement
from a jar containing 8 white marbles
and 2 red marbles.
• Find the probability that:
a)The first marble is red.
b)The second marble is red.
c) Both marbles are red.
Example 4, cont’d
• Solution: Create a probability tree
diagram.
Example 4, cont’d
• Solution, cont’d:
a)P(red first) = P(RW) + P(RR)
16 2 18 1




90 90 90 5
Example 4, cont’d
• Solution, cont’d:
b) P(red second) =
P(RR) + P(WR)
16 2 18 1




90 90 90 5
c) P(both red) =
P(RR)
2
1


90 45
Example 4, cont’d
• Solution, cont’d: Notice that the two
events are not independent because
1 1 1
 
5 5 45
Question:
Draw 2 marbles with replacement
from a jar containing 8 while marbles
and 2 red marbles. Are the events
“The first marble is red” and “The
second marble is red” independent?
a. yes
b. no
Example 5
•
•
A student’s name is chosen at random from the
college enrollment list and the student is
interviewed.
•
Let A be the event the student regularly eats breakfast.
•
Let B be the event the student has a 10:00 AM class.
Explain in words what is meant by:
a)
P  A  B
b)
P  A B
c)
P A
 
Example 5, cont’d
• Solution:
a) P  A  B  : the probability the student
regularly eats breakfast and has a 10:00
AM class.
b) P  A B  : the probability the student
regularly eats breakfast given that the
student has a 10:00 AM class
c) P  A :the probability the student does not
regularly eat breakfast.
Expected Value
• The average numerical outcome for many
repetitions of an experiment is called the
expected value.
• If the outcomes of an experiment are v1,
v2, …, vn and the outcomes have
probabilities of p1, p2, …, p3, respectively,
the expected value is
E = v1(p1) + v2(p2) + … + vn(pn).
• If E = 0, the game is said to be fair.
Example 6
• An experiment consists of rolling a fair
die and noting the number on top of
the die.
• Compute the expected value of one
roll of the die.
Example 6, cont’d
•
Solution: The calculations are shown below:
Example 7
•
How much should an insurance company
charge as its average premium in order to
break even?
Example 7, cont’d
• Solution: The calculations are shown
above.
• The average premium should cost $760.
Odds
• The odds in favor of an event compare
the number of favorable outcomes to
the number of unfavorable outcomes.
• If the odds in favor are a:b, then
a
PE 
ab
Odds, cont’d
• The odds against of an event compare the
number of unfavorable outcomes to the
number of favorable outcomes.
• The odds in favor of an event E are
 
PE: P E
• The odds against an event E are
 
P E : PE
Example 8
• Suppose a card is randomly drawn from
a standard deck.
• What are the odds in favor of drawing a
face card?
Example 8, cont’d
• Solution: There are 12 face cards in the
deck, and there are 40 other cards.
• The odds in favor are 12:40, which
simplifies to 3:10.
Question:
A card is randomly selected from a
standard deck. What are the odds
against drawing a diamond?
a. 4:3
b. 3:1
c. 3:4
d. 1:3
Example 9
• Find P(E) given the following odds:
a) The odds in favor of E are 3:7.
b) The odds against E are 5:13.
Example 9, cont’d
• Solution:
a) The odds in favor of E are 3:7.
3
3
PE 

3  7 10
b) The odds against E are 5:13.
13
13
PE 

5  13 18
Example 10
• Find the odds in favor of event E,
given the following probabilities:
a) P(E) = 1/4.
b) P(E) = 3/5.
Example 10, cont’d
• Solution:
a) The odds in favor
of E are 1:3.
b) The odds in favor
of E are 3:2.
1
1
3
3
4  41
3
3
1 1
4
4
5  53
2
2
1 3
5
5
10.3 Initial Problem Solution
• A roulette wheel has 38 slots numbered
00, 0 and 1 through 36. You place a bet
on a number or combination of numbers.
If you bet on the winning number, you win
your bet plus 35 times your bet. If you
lose, you lose the money you bet.
• If you bet $100 on one number, what is
your expected gain or loss?
Initial Problem Solution, cont’d
• Solution:
• The probability of winning is 1/38
and the amount won would be
$3500.
• The probability of losing is 37/38
and the amount lost would be $100.
Initial Problem Solution, cont’d
• The expected value is approximately
-$5.26. You should expect to lose this
amount, on average, for every $100
you bet.