Download Introduction to Probability - Finite Mathematics Section 6.2

Section 6.2 – Introduction to Probability
In probability, a sample space includes all possible outcomes of an experiment. Each element in
the sample space is called a simple event. If S is the sample space and S = {s1 , s2 , … , sn } , then
the simple events are {s1} , {s2 } , … , {sn } .
If an experiment involves rolling a die, the sample space is S = {1, 2, 3, 4, 5, 6} and the simple
events are {1} , {2} , {3} , {4} , {5} , and {6} .
If an experiment involves tossing a coin, the sample space is S = { H , T } , where H denotes heads
and T denotes tails. The simple events are { H } and {T } .
A compound event is a combination of two or more simple events. The word “or” indicates a
union of the events.
If an experiment involves rolling a die one time and observing the number on the uppermost
face, some compound events are rolling a(n):
2 or a 4:
odd number:
2 or an odd number:
number greater than 3:
number that is not a 2:
even or odd number:
{2, 4}
{1, 3, 5}
{1, 2, 3, 5}
{4, 5, 6}
{1, 3, 4, 5, 6}
{1, 2, 3, 4, 5, 6}
Probability is a measure of the likelihood that an event occurs. The probability of an event can
be written as a decimal or fraction between 0 and 1. An event that cannot occur has probability 0,
and an event that is certain to occur has probability 1. If there is a 20% chance of rain tomorrow,
we can say that the probability that it will rain is 0.20, or we can alternatively write the
20
probability as the fraction 100
= 15 .
We use the notation P ( E ) to indicate the probability of a given event, E. An event can be a
simple event, or it can be the union, intersection and/or complement of 2 or more simple events.
Some important facts about probability are written below.
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Section 6.2
Important Facts about Probability
The probability that an event will occur is a number between 0 and 1, inclusive.
An event that cannot occur has probability 0.
An event that is certain to occur has probability 1.
The more likely an event is to occur, the closer the probability is to 1.
The less likely an event is to occur, the closer the probability is to 0.
The probabilities of all of the outcomes in the sample space add up to 1.
Simple events are mutually exclusive (i.e. they cannot occur at the same time).
If an event includes 2 or more members in the sample space, we find the
probability of the event by adding the probabilities of the simple events together.
Example 1: Suppose that an experiment consists of selecting a number from the set
S = {1, 2, 3, … , 10} , E is the event of selecting a number greater than 20 and F is the event of
selecting a number less than 15. Find P ( E ) and P ( F ) .
Solution: Since there are no numbers in the sample space that are greater than 20, E is an
impossible event, and P ( E ) = 0. Since all of the numbers in the sample space are less than 15, F
is a certain event and P ( F ) = 1 .
***
Example 2: Suppose that there are 3 contestants remaining on a reality television show, Anna,
Ben and Carl. The probability that Ben will win the competition is estimated to be 0.42. The
probability that Carl will win is estimated to be 0.26. What is the probability that Anna will win?
Solution: In this case, the sample space contains 3 elements, the possible eventual winners. The
probabilities that each of the 3 remaining contestants will win must add up to 1. Define E to be
the event that Anna wins. Then P ( E ) = 1 − ( 0.42 + 0.26 ) = 1 − 0.68 = 0.32 .
Therefore, the probability that Anna will win is 0.32.
***
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Section 6.2
We will discuss two different types of probability in this section: theoretical probability and
empirical probability.
Theoretical Probability
The theoretical probability for an event in a uniform sample space is shown below.
Theoretical Probability Formula
Let E represent an event in a uniform sample space S. The probability of E,
denoted P ( E ) , can be found as follows:
P(E) =
The number of ways event E can occur
The number of outcomes in the sample space
If S represents the sample space, the above formula can be written as follows:
P(E) =
n(E)
n(S )
With some experiments, the elements in the sample space are equally likely. This means that
each element is as likely to occur as any other element. In this case, the sample space is called a
uniform sample space. If m is the number of elements in a uniform sample space, then the
probability of each simple event is m1 . Examples of uniform sample spaces include experiments
that involve fair coins, fair dice and cards.
Example 3: If a coin is flipped once, what is the probability that it lands up tails?
Solution: This is a straightforward problem with an answer of
1
2
, but we will use the above
definition to illustrate its meaning. The uniform sample space is S = { H , T } , where H represents
heads and T represents tails. Unless stated otherwise, we assume that the coin is a fair coin,
which means that we have an equal chance of flipping heads or tails. The desired event is tails,
so E = {T } .
P(E) =
n ({T } )
n(E)
The number of ways event E can occur
1
=
=
=
The number of outcomes in the sample space n ( S ) n ({ H , T } ) 2
***
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Section 6.2
Example 4: If a fair die is tossed once, find the probability that the uppermost face displays a(n):
A. 3
B. 2 or 4
C. odd number
D. 2 or an odd number:
E. number greater than 3
F.
G. even or an odd number
H. number greater than 8
number that is not a 2
Solution: The sample space is S = {1, 2, 3, 4, 5, 6} , so n ( S ) = 6 . This is an example of a uniform
sample space, since each of the simple events in the sample space are equally likely. Notice that
the compound events in parts B through G were mentioned on the first page of this section.
n( E)
P(E) =
n( E)
n( S)
Event E, in words
Event E, as a set
A.
3
{3}
1
B.
2 or 4
{2, 4}
2
2
6
= 13
C.
odd number
{1, 3, 5}
3
3
6
= 12
D.
2 or an odd number
{1, 2, 3, 5}
4
4
6
=
E.
number greater than 3
{4, 5, 6}
3
3
6
= 12
F.
number that is not a 2
{1, 3, 4, 5, 6}
5
G.
even or an odd number
{1, 2, 3, 4, 5, 6}
6
6
6
=1
H.
number greater than 8
{}
0
0
6
=0
1
6
2
3
5
6
***
Example 5: Suppose that an experiment consists of flipping a coin 3 times and observing the
sequence of heads and tails that results. Find the probability that at least 2 heads are observed.
Solution: In section 5.3, we created the following tree diagram to represent all of the outcomes
for this situation:
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Section 6.2
The sample space for this experiment is:
S = { HHH , HHT , HTH , HTT , THH , THT , TTH , TTT }
The sample space is a uniform sample space, since each of the above outcomes is equally likely.
Event E is defined to be the outcomes with at least 2 heads, so E = {HHH , HHT , HTH , THH } .
Therefore P ( E ) =
n(E)
n(S )
=
4 1
= .
8 2
***
Example 6: Suppose that an experiment consist of pulling a marble at random from a bag that
contains 4 red marbles, 8 blue marbles and 3 yellow marbles.
A.
What is the probability that a marble pulled from the bag is red?
B.
What is the probability that a marble pulled from the bag is yellow?
Solution: The computation for this problem is quite simple, but we will first explain how we can
use our current probability formula in this situation. The outcomes for this experiment are
S = {red, blue, yellow} but it is not a uniform sample space, since each color is not equally
likely to be drawn. However, each individual marble is just as likely to be pulled out of the bag
as any other marble.
Let us consider the sample space in a different way. There are 15 total marbles. Suppose that
they are labeled as follows:
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Section 6.2
4 red marbles:
r1 , r2 , r3 , r4
8 blue marbles:
b1 , b2 , b3 , b4 , b5 , b6 , b7 , b8
3 yellow marbles:
y1 , y2 , y3
Let T represent the sample space where each individual marble is listed. Then
T = {r1 , r2 , r3 , r4 , b1 , b2 , b3 , b4 , b5 , b6 , b7 , b8 , y1 , y2 , y3 }
We can now use sample space T to compute the indicated probabilities.
A.
Let E represent the event where a red marble is pulled out of the bag. Then
E = {r1 , r2 , r3 , r4 } .
There are 15 marbles in the bag, of which 4 are red. If we define the event E to be the
n(E) 4
event that a red marble is drawn, P ( E ) =
= ≈ 0.2667.
n ( T ) 15
A simpler way to compute the probability of a red marble is:
# of red marbles
P ( red ) =
total # of marbles
B.
Let F represent the event where a yellow marble is pulled out of the bag.
P ( yellow ) =
# of yellow marbles 3 1
= = = 0.2
total # of marbles
15 5
***
Empirical Probability
Suppose that we conduct an experiment where we flip a coin 400 times. Based on theoretical
probability, we would expect the coin to come up heads 12 of the time (200 heads) and to come
up tails
1
2
of the time (200 tails). In reality, however, we know that might not occur.
Empirical probability, also known as experimental probability or relative frequency, is the
observed probability of an event based on repeated trials of an experiment. Some examples of
empirical probability experiments are:
Testing 1000 light bulbs to see how many are defective
Surveying 250 people to find out their favorite movie
Tossing a coin 80 times to see how many times it turns up “heads”
Notice that all of the above situations are based on an actual experiment where data is gathered.
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Section 6.2
Empirical Probability Formula
If E represents an event, then the probability of E, denoted P ( E ) , can be found as
follows:
P(E) =
The number of times that event E occurs
The total number of trials in the experiment
Notice that the formulas for theoretical probability and empirical probability are similar to one
another – but one is based on what should happen in theory, and the other is based on the
observed data from an experiment. The following example illustrates the difference between
empirical and theoretical probability.
Example 7:
A.
What is the theoretical probability of obtaining heads when a coin is tossed? Express the
answer as both a fraction and a decimal.
B.
Renata conducts an experiment where she tosses a coin 400 times. She obtains 190 heads
and 210 tails. Based on the results of Renata’s experiment, what is the probability of
obtaining heads when a coin is tossed? Express the answer as both a fraction and a
decimal.
Solution:
A.
There are 2 outcomes in the sample space when tossing a coin: heads and tails. In this
problem, heads is our desired outcome, which represents 1 of those 2 outcomes.
Therefore, the theoretical probability of tossing heads is 12 .
This answer may be written as follows: P ( heads ) =
1
2
If we wish to represent this situation using sets, S = { H , T } , and
P ({ H } ) =
B.
n ({ H } )
n(S )
=
n ({ H } )
n ({ H , T } )
=
1
= 0.5 .
2
We want to find the empirical probability of obtaining heads, since we are basing our
answer on the observed results of Renata’s experiment.
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Section 6.2
P ( heads ) =
The number of times that heads occurs
190 19
=
=
= 0.475
The total number of trials in the experiment 400 40
***
In most of the problems in this text, we will not be asked for both the theoretical and the
empirical probability. We are usually just asked to find the probability of an event, and the given
information makes it clear whether we are finding theoretical or empirical probability. If we are
asked for the probability of an event where no data has been collected, we assume that we are
finding theoretical probability.
Example 8: Violet is a quality control engineer for a flashlight company. She randomly tests 350
of the flashlights and finds that 25 of them are defective. Based on Violet’s results, find the
probability that a randomly selected flashlight is defective.
Solution: Let E represent the event that a flashlight is defective. We use the empirical probability
formula, since we are basing our analysis on Violet’s experimental results.
P(E) =
# of defective flashlights
25
5
1
=
=
=
total number of flashlights tested 350 70 14
***
Frequency Tables and Probability Distribution Tables
With some experiments, we are given a frequency table which we can use to determine the
probability of various events. A frequency table shows the number of times that each simple
event occurs. A frequency table looks like this, where m represents the number of simple events
in the experiment.
Frequency Table
Simple Event Frequency
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s1
n ( s1 )
s2
n ( s2 )
s3
n ( s3 )
⋮
⋮
sm
n ( sm )
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Section 6.2
We can then write the probabilities of the simple events in an organized fashion by creating a
probability distribution table, where the right-hand column displays the probability of each
simple event.
A probability distribution table looks like this, where m represents the number of simple events
in the experiment:
Probability Distribution Table
Simple Event
Probability
(also called Relative Frequency)
s1
P ( s1 )
s2
P ( s2 )
s3
P ( s3 )
⋮
⋮
sm
P ( sm )
To find the empirical probability, or relative frequency of each simple event, we divide the
number of occurrences in each line of the frequency table by the total number of trials.
The next example illustrates the use of a frequency table as well as a probability distribution
table.
Example 9: Nathan takes a poll outside the college cafeteria and asks students to vote for their
favorite ice cream flavor, based on the five flavors which are offered in the college cafeteria. The
results are shown in the frequency table below.
Ice Cream Flavor
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Frequency
Chocolate
35
Vanilla
45
Chocolate Chip
42
Strawberry
23
Peanut Butter
14
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Section 6.2
A.
State the sample space for this experiment.
B.
How many students were surveyed?
C.
Create a probability distribution table based on Nathan’s poll data.
D.
Find the relative frequency of a student choosing vanilla ice cream.
E.
Find the probability of a randomly selected student choosing either chocolate or
strawberry ice cream.
F.
If 900 students are expected to order ice cream tomorrow, how many of them are likely to
order peanut butter ice cream?
Solution: First, let us find the total number of outcomes in the chart by adding up all of the votes
The total number of outcomes is 35 + 45 + 42 + 23 + 14 = 159 .
A.
The sample space S consists of all of the possible responses to the survey; we can find
these responses in the left-hand column of the frequency chart:
S = {Chocolate, Vanilla, Chocolate Chip, Strawberry, Peanut Butter}
Notice that this sample space is not a uniform sample space, since the elements in the
sample space are not equally likely. (There are 5 outcomes in the sample space, but the
probability of selecting any given flavor is NOT 15 .)
B.
To find the number of students surveyed, we add the numbers in the “Frequency”
column: 35 + 45 + 42 + 23 + 14 = 159 . This gives the total number of trials for this
experiment, which is the total number of students surveyed.
C.
A probability distribution table for Nathan’s poll data is shown below.
Ice Cream Flavor
Probability /
Relative Frequency
35
159
Chocolate
Vanilla
45
159
= 15
53
Chocolate Chip
42
159
= 14
53
Strawberry
23
159
Peanut Butter
14
159
Notice that each probability is between 0 and 1 and the sum of the probabilities is 1.
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Section 6.2
D.
Relative frequency has the same meaning as empirical probability. The relative frequency
of choosing vanilla is 15
53 .
E.
To find the probability of a student choosing either chocolate or strawberry ice cream, we
can add those probabilities together.
P ( chocolate or strawberry ) =
F.
35 23
58
+
=
159 159 159
Based on the results of the poll, the probability of a student choosing peanut butter ice
14
cream is 159
. This represents the fraction of students which we expect to order peanut
butter ice cream. If 900 students are expected to order ice cream tomorrow, we would
14
14
of 900 students to order ice cream. When translating the phrase “ 159
of
then expect 159
900” to a mathematical expression, the word “of” means that we should multiply. Using a
calculator,
14
 14 
of 900 = 
 ( 900 ) ≈ 79.24
159
 159 
Therefore, if 900 students order ice cream, we expect that about 79 of them will order
peanut butter ice cream.
***
A probability distribution table can also be created for situations that are theoretical rather than
empirical, as shown in the example below.
Example 10: Suppose that you choose a card randomly from a standard deck of playing cards
and observe the suit of the selected card.
A.
What is the sample space for this experiment?
B.
Create a probability distribution table.
Solution:
A.
We are only concerned with the suit of the selected card (and are not concerned, for
example, with whether it is a 9, an ace, a jack, etc.). Therefore, the sample space of
outcomes is S = {Heart, Spade, Diamond, Club} .
B.
This is an example of a uniform sample space, since each suit is equally likely to occur.
There are 52 cards in the deck, consisting of 13 hearts, 13 spades, 13 diamonds, and 13
1
clubs. Therefore, the probability of each simple event is 13
52 = 4 .
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Section 6.2
The probability distribution table is shown below:
Event
Probability
Heart
1
4
Spade
1
4
Diamond
1
4
Club
1
4
***
The Law of Large Numbers
There is a concept in probability known as the Law of Large Numbers.
The Law of Large Numbers
In general, as the number of trials in an experiment increases, the empirical
probability of an event becomes closer to the theoretical probability of that event.
Consider the following illustration of the Law of Large Numbers:
Renata decides to repeat her coin toss experiment. She tosses the coin over the course of nine
days. After three days, she enlists the help of her sisters, and on the ninth day, she convinces her
math teacher to involve the entire class in the experiment. Renata keeps a running tally and
records the cumulative totals (for that day combined with previous days). The outcomes are
compiled in the chart below.
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Section 6.2
Day of
Experiment
# of Tosses
to Date
# of Heads
to Date
# of Tails
to Date
P(Heads)
to Date
Day 1
2
2
0
2
2
Day 2
10
7
3
7
10
Day 3
120
55
65
55
120
≈ 0.4583
Day 4
450
216
234
216
450
= 0.48
Day 5
900
428
472
428
900
≈ 0.4756
Day 6
1300
627
673
627
1300
≈ 0.4823
Day 7
1700
834
866
834
1700
≈ 0.4906
Day 8
2000
988
1012
988
2000
= 0.494
Day 9
5000
2502
2498
2502
5000
= 0.5004
=1
= 0.7
Notice that at the end of Day 1, the empirical probability of obtaining heads is 1. If Renata used
this result to predict future coin tosses, she would expect to get heads every time that the coin
was tossed (which is not a realistic prediction). At the end of Day 2, the empirical probability of
obtaining heads is 0.7. As the coin is tossed more and more times, however, notice that the
empirical probability becomes close to the theoretical probability of 0.5.
Notice that on Day 5, the probability of obtaining heads decreases slightly, and then increases
again on Day 6 to a number closer to the theoretical probability of 0.5. The Law of Large
Numbers cannot perfectly predict what happens in real-life experiments, but it describes what
generally occurs as we increase the number of trials in an experiment.
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Section 6.2