Transcript Section 6-2 Basics of Probability

6.2 Basics of Probability
LEARNING GOAL
Know how to find probabilities using theoretical and
relative frequency methods and understand how to
construct basic probability distributions.
Copyright © 2009 Pearson Education, Inc.
Let’s begin by considering a toss of two coins. Figure 6.1
shows that there are four different ways the coins could fall.
We say that each of these four ways is a different outcome of
the coin toss.
Figure 6.1
But suppose we are interested only in the number of heads.
Because the two middle outcomes in Figure 6.1 each have 1
head, we say that these two outcomes represent the same event.
Figure 6.1 shows that there are four possible outcomes for the
two-coin toss, but only three possible events: 0 heads, 1 head,
and 2 heads.
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Slide 6.2- 2
Definitions
Outcomes are the most basic possible results of
observations or experiments.
An event is a collection of one or more outcomes that
share a property of interest.
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Slide 6.2- 3
Expressing Probability
The probability of an event, expressed as
P(event), is always between 0 and 1 inclusive.
A probability of 0 means that the event is
impossible, and a probability of 1 means that
the event is certain.
Figure 6.2 The scale shows various degrees of certainty
as expressed by probabilities.
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Slide 6.2- 4
Theoretical Probabilities
Theoretical Method for Equally Likely Outcomes
Step 1. Count the total number of possible outcomes.
Step 2. Among all the possible outcomes, count the
number of ways the event of interest, A, can occur.
Step 3. Determine the probability, P(A), from
number of ways A can occur
P(A) = total number of outcomes
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Slide 6.2- 5
EXAMPLE 1 Guessing Birthdays
Suppose you select a person at random from a large
group at a conference. What is the probability that the
person selected has a birthday in July? Assume 365
days in a year.
Solution: If we assume that all birthdays are equally
likely, we can use the three-step theoretical method.
Step 1. Each possible birthday represents an outcome,
so there are 365 possible outcomes.
Step 2. July has 31 days, so 31 of the 365 possible
outcomes represent the event of a July
birthday.
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Slide 6.2- 6
EXAMPLE 1 Guessing Birthdays
Solution: (cont.)
Step 3. The probability that a randomly selected
person has a birthday in July is
31
P(July birthday) =
≈ 0.0849
365
which is slightly more than 1 in 12.
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Slide 6.2- 7
Theoretical Probabilities
Counting Outcomes
Suppose we toss two coins and want to count the total number of
outcomes. The toss of the first coin has two possible outcomes:
heads (H) or tails (T). The toss of the second coin also has two
possible outcomes.
The two outcomes for the first coin can occur with either of the
two outcomes for the second coin.
So the total number of outcomes for two tosses is 2 × 2 = 4;
they are HH, HT, TH, and TT, as shown in the tree diagram of
Figure 6.4a (next slide).
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Slide 6.2- 8
Figure 6.4 a Tree diagram showing
the outcomes of tossing two coins.
TIME OUT TO THINK
Explain why the outcomes for tossing one coin twice in
a row are the same as those for tossing two coins at the
same time.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2- 9
We can now extend this thinking.
If we toss three coins, we
have a total of 2 × 2 × 2 = 8
possible outcomes:
HHH, HHT, HTH, HTT,
THH, THT, TTH, and TTT,
as shown in Figure 6.4 b.
Figure 6.4 b Tree diagram showing the
outcomes of tossing three coins.
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Slide 6.2- 10
Counting Outcomes
Suppose process A has a possible outcomes and
process B has b possible outcomes. Assuming the
outcomes of the processes do not affect each other, the
number of different outcomes for the two processes
combined is a × b.
This idea extends to any number of processes.
For example, if a third process C has c possible
outcomes, the number of possible outcomes for the
three processes combined is a × b × c.
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Slide 6.2- 11
EXAMPLE 2 Some Counting
a. How many outcomes are there if you roll a fair die
and toss a fair coin?
Solution:
a. The first process, rolling a fair die, has six
outcomes (1, 2, 3, 4, 5, 6). The second process,
tossing a fair coin, has two outcomes (H, T).
Therefore, there are 6 × 2 = 12 outcomes for the
two processes together (1H, 1T, 2H, 2T, . . . , 6H,
6T).
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Slide 6.2- 12
EXAMPLE 2 Some Counting
b. What is the probability of rolling two 1’s (snake
eyes) when two fair dice are rolled?
Solution:
b. Rolling a single die has six equally likely outcomes.
Therefore, when two fair dice are rolled, there are
6 × 6 = 36 different outcomes.
Of these 36 outcomes, only one is the event of
interest (two 1’s). So the probability of rolling two
1’s is
number of ways two 1’s can occur
1
P(two 1’s) =
=
= 0.0278
total number of outcomes
36
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Slide 6.2- 13
EXAMPLE 3 Counting Children
What is the probability that, in a randomly selected
family with three children, the oldest child is a boy, the
second child is a girl, and the youngest child is a girl?
Assume boys and girls are equally likely.
Solution: There are two possible outcomes for each
birth: boy or girl. For a family with three children, the
total number of possible outcomes (birth orders) is 2 ×
2 × 2 = 8 (BBB, BBG, BGB, BGG, GBB, GBG, GGB,
GGG).
The question asks about one particular birth order (boygirl-girl), so this is 1 of 8 possible outcomes.
Therefore, this birth order has a probability of 1 in 8, or
1/8 = 0.125.
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Slide 6.2- 14
By the Way ...
Births of boys and girls are not equally likely.
Naturally, there are approximately 105 male births for
every 100 female births. However, male death rates
are higher than female death rates, so female adults
outnumber male adults.
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Slide 6.2- 15
TIME OUT TO THINK
How many different four-child families are possible if
birth order is taken into account? What is the probability
of a couple having a four-child family with four girls?
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Slide 6.2- 16
Theoretical Probabilities
Relative Frequency Probabilities
The second way to determine probabilities is to
approximate the probability of an event A by making
many observations and counting the number of times
event A occurs.
This approach is called the relative frequency (or
empirical) method.
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Slide 6.2- 17
Relative Frequency Method
Step 1. Repeat or observe a process many times and
count the number of times the event of interest,
A, occurs.
Step 2. Estimate P(A) by
number of times A occurred
P(A) =
total number of observations
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Slide 6.2- 18
EXAMPLE 4 500-Year Flood
Geological records indicate that a river has crested above a
particular high flood level four times in the past 2,000 years.
What is the relative frequency probability that the river will crest
above the high flood level next year?
Solution: Based on the data, the probability of the river cresting
above this flood level in any single year is
number of years with flood
4
1
= 2000 =
total number of years
500
Because a flood of this magnitude occurs on average once every
500 years, it is called a “500-year flood.”
The probability of having a flood of this magnitude in any given
year is 1/500, or 0.002.
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Slide 6.2- 19
Theoretical Probabilities
Relative Frequency Probabilities
Subjective Probabilities
The third method for determining probabilities is to estimate a
subjective probability using experience or intuition.
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Slide 6.2- 20
Three Approaches to Finding Probability
A theoretical probability is based on assuming that all
outcomes are equally likely. It is determined by dividing
the number of ways an event can occur by the total
number of possible outcomes.
A relative frequency probability is based on
observations or experiments. It is the relative frequency
of the event of interest.
A subjective probability is an estimate based on
experience or intuition.
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Slide 6.2- 21
EXAMPLE 5 Which Method?
Identify the method that resulted in the following statements.
a. The chance that you’ll get married in the next year is zero.
b. Based on government data, the chance of dying in an
automobile accident is 1 in 7,000 (per year).
c. The chance of rolling a 7 with a twelve-sided die is 1/12.
Solution:
a. This is a subjective probability because it is based on a feeling
at the current moment.
b. This is a relative frequency probability because it is based on
observed data on past automobile accidents.
c. This is a theoretical probability because it is based on assuming
that a fair twelve-sided die is equally likely to land on any of its
twelve sides.
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Slide 6.2- 22
Probability of an Event Not Occurring
Probability of an Event Not Occurring
If the probability of an event A is P(A), then the
probability that event A does not occur is P(not A).
Because the event must either occur or not occur, we
can write
P(A) + P(not A) = 1
or
P(not A) = 1 – P(A)
Note: The event not A is called the complement of the
event A; the “not” is often designated by a bar, so Ā
means not A.
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Slide 6.2- 23
Probability Distributions
Suppose you toss two coins simultaneously. The outcomes are
the various combinations of a head and a tail on the two coins.
Because each coin can land in two possible ways (heads or
tails), the two coins can land in 2 × 2 = 4 different ways.
Table 6.2 has a row for each of these four outcomes.
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Slide 6.2- 24
Notice that the four outcomes represent only three
different events:
2 heads (HH), 2 tails (TT), and 1 head and 1 tail (HT or
TH).
The probability of two heads is P(HH) = 1/4 = 0.25.
The probability of two tails is P(TT) = 1/4 = 0.25.
The probability of one head and one tail is P(H and T) =
2/4 =0.50.
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Slide 6.2- 25
These probabilities result in a probability distribution that
can be displayed as a table (Table 6.3) or a histogram
(Figure 6.6).
Figure 6.6 Histogram showing the
probability distribution for the results
of tossing two coins.
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Slide 6.2- 26
By the Way ...
Another common way to express likelihood is to use
odds. The odds against an event are the ratio of
the probability that the event does not occur to the
probability that it does occur. For example, the odds
against rolling a 6 with a fair die are (5/6)/(1/6), or 5 to
1. The odds used in gambling are called payoff
odds; they express your net gain on a winning bet.
For example, suppose that the payoff odds on a
particular horse at a horse race are 3 to 1. This
means that for each $1 you bet on this horse, you will
gain $3 if the horse wins (and get your original $1
back).
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Slide 6.2- 27
Making a Probability Distribution
A probability distribution represents the probabilities of
all possible events. Do the following to make a display
of a probability distribution:
Step 1. List all possible outcomes. Use a table or figure
if it is helpful.
Step 2. Identify outcomes that represent the same
event. Find the probability of each event.
Step 3. Make a table in which one column lists each
event and another column lists each probability. The
sum of all the probabilities must be 1.
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Slide 6.2- 28
EXAMPLE 8 Tossing Three Coins
Make a probability distribution for the number of heads that
occurs when three coins are tossed simultaneously.
Solution: We apply the three-step process.
Step 1. The number of different outcomes when three coins are
tossed is 2 × 2 × 2 = 8. All eight possible outcome are
HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT.
Step 2. There are four possible events: 0 heads, 1 head, 2 heads,
and 3 heads. By looking at the eight possible outcomes, we
find the following:
• Only one of the eight outcomes represents the event of 0
heads, so its probability is 1/8.
• Three of the eight outcomes represent the event of 1 head
(and 2 tails): HTT, THT, and TTH. This event therefore
has a probability of 3/8.
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Slide 6.2- 29
EXAMPLE 8 Tossing Three Coins
Solution: (cont.)
• Three of the eight outcomes represent the event of 2
heads (and 1 tail): HHT, HTH, and THH. This event also
has a probability of 3/8.
• Only one of the eight outcomes represents the event of 3
heads, so its probability is 1/8.
Step 3. We make a table
with the four events listed
in the left column and their
probabilities in the right
column. Table 6.4 shows
the result.
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Slide 6.2- 30
TIME OUT TO THINK
When you toss four coins, how many different outcomes
are possible? If you record the number of heads, how
many different events are possible?
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Slide 6.2- 31
EXAMPLE 9 Two Dice Distribution
Make a probability distribution for the sum of the dice when two
dice are rolled. Express the distribution as a table and as a
histogram.
Solution: Because there are six ways for each die to land, there
are 6 × 6 = 36 outcomes of rolling two dice. We enumerate all 36
outcomes in Table 6.5 by listing one die along the rows and the
other along the columns. In each cell, we show the sum of the
numbers on the two dice.
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Slide 6.2- 32
EXAMPLE 9 Two Dice Distribution
Solution: (cont.)
The possible events are the sums from 2 to 12. These are the
events of interest in this problem. We find the probability of each
event by counting all the outcomes for each sum and then dividing
the number of outcomes by 36.
For example, the five highlighted outcomes in Table 6.5 (previous
slide) have a sum of 8, so the probability of a sum of 8 is 5/36.
Table 6.6 shows the complete probability distribution, and Figure
6.7 (next slide) shows the distribution as a histogram.
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Slide 6.2- 33
0.18
0.16
Probability
0.14
0.12
0.10
0.08
0.06
0.04
0.02
2
3
4
5
6
7
8
Sum of two dice
9
10
11
12
Figure 6.7 Histogram showing the probability distribution for the sum of two dice.
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Slide 6.2- 34
The End
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Slide 4.- 35