Download 5.5 ROBABILITY AS A THEORETICAL CONCEPT Equally Likely

that each engine will fire is 0.75. The engines
fire independently.
a. Conduct an experiment to estimate the
probability of a liftoff.
b. Verify the law of experimental independence.
5. Assume that a two-engine airplane can make
a safe flight if one of its engines stops in operation. We want to determine the probability of
not having a safe flight (that is, of both engines
failing) assuming that the engines operate independently of each other. Each engine has a
0.10 chance of failing in flight.
a. Since the engines are assumed to be independent of each other, which formula tells
us what P(Both engines fail) approximately
equals?
b. Using the formula for P(Both engines fail)
you found in part (a), estimate the chance
that a flight will not be safe.
6. Suppose automobile accidents are equally
likely to occur on each day of the week. What
is the probability that if two accidents occur,
they will both be on the weekend (that is, on
Saturday or Sunday)? Assume that the two
accidents are independent of each other.
For additional exercises, see page 723.
5.5 PROBABILITY AS A THEORETICAL CONCEPT
So far, we have solved probability problems by estimating the required
probability after conducting some sort of experiment and collecting data.
But probability may be approached from another point of view: in many
cases we can calculate the theoretical probability in advance by determining
the number of ways a successful outcome can occur and dividing by the
total number of possible outcomes in the trial (no data collection needed for
this). We have already used this viewpoint in simple cases—for example,
when we decided which model to use in a given experiment.
We first need to discuss the idea of equally likely outcomes. When the
outcomes are equally likely, the method for computing the theoretical
probability just proposed in the preceding paragraph is justified. First, two
definitions. The sample space is the set of all possible outcomes in a trial.
An event is merely a collection of outcomes. For example, the event that a
tossed die gives an even number consists of the subset of outcomes 兵2, 4, 6其
from the sample space of all possible outcomes 兵1, 2, 3, 4, 5, 6其.
Equally Likely Outcomes
Many real-world trials or experiments have a finite set of outcomes possible
on each trial. In many situations involving probability, all the outcomes
are equally likely. That is, if the situation were to be repeated a large
number of times, each of the outcomes would occur approximately the
same proportion of times. A coin toss, for example, has two equally likely
outcomes: heads and tails (see Figure 5.1a). A roll of an ordinary die has six
equally likely outcomes: 1, 2, 3, 4, 5, and 6 (see Figure 5.1b).
Suppose we seek the theoretical probability of an event, such as the
event that a rolled die produces an even number of spots. Then of the
possible outcomes (1, 2, 3, 4, 5, 6) we see that there are three ways (2, 4, 6) of
H
T
(a) Two outcomes of
tossing a coin
1
2
3
4
5
6
(b) Six outcomes of
rolling a die
Possible outcomes of tossFigure 5.1
ing a coin and rolling a die.
obtaining a “successful” outcome. By successful we simply mean an outcome
that causes the event of interest—namely, the die being even—to occur. For
such probability problems in which we are willing to assume equally likely
outcomes, the following rule is extremely useful.
Theoretical probability of an event when outcomes
are equally likely
Suppose we have a set of N equally likely
outcomes, of which S of them are successes
in the sense that each “successful” outcome
causes the event to occur. Then the theoretical probability of the event is
number of ways of obtaining
a successful outcome
p(event) ⳱
total number of outcomes
⳱
S
N
For the toss of a coin, we can say
p(heads) ⳱
1
⳱ 0.5
2
For the probability of getting a 1 or a 2 on the roll of a die, we can say
p(1 or 2) ⳱
2
1
⳱
6
3
because 2 of the possible outcomes (1, 2) cause the event to happen and
there are 6 possible outcomes. We did not do an experiment to find this
probability. Instead, we used some basic ideas about equally likely outcomes
and successful outcomes.
Observe that we use a lowercase p, as in
p(heads) ⳱
1
⳱ 0.5
2
to indicate a theoretical probability. Earlier in this chapter we referred to the
coin-tossing experiment of Kerrich and found, based on 10,000 simulated
tosses of the coin, that
P(heads) ⳱ 0.507
The capital P shows that we are talking about an experimental probability
based on data obtained from a certain number of trials. So P(heads) is an
estimate of p(heads). That is, the experimental probability of getting heads
is an estimate of the theoretical probability of getting heads on the toss of
a fair coin. Remember this p, P convention; it is used throughout the
book!
For a roll of a die, the probability of getting a 3 is
p(3) ⳱
1
6
As a slightly more complicated example, suppose we want to find the
probability of getting an even number when rolling a die, as was briefly
mentioned above. According to Figure 5.1b, three of the outcomes give even
numbers. So
p(even number when rolling a die) ⳱
3
1
⳱ ⳱ 0.5
6
2
Now let’s see how to find the theoretical probability of getting two
heads in the toss of two coins. Figure 5.2, called a tree diagram, shows the
possible outcomes. The tree diagram helps us keep track of the possible
outcomes when there are multiple subtrials, as in the tossing of two coins.
Along the top branch of the tree, we find the successful outcome HH. Along
the bottom branch we have one of the other outcomes, TT, and so on.
From this diagram we count a total of four equally likely outcomes; only
one of these is two heads. So the theoretical probability is
p(two heads) ⳱
First coin
H
T
Second coin
1
⳱ 0.25
4
Possible outcomes
H
T
H H
H T
H
T
T H
T T
2 heads
1 head
0 heads
Figure 5.2
Possible outcomes of tossing two
coins, showing outcome of two heads.
First coin
Second coin
H
T
H
T
H
T
Outcomes
H
H
T
T
H
T
H
T
One
or more
heads
Figure 5.3
more heads.
Possible outcomes yielding one or
First child
Second child
G
G
B
G G
G B
G
B
B
B
B
Figure 5.4
Possible families
G
B
2 girls
1 girl
One
or more
girls
0 girls
Possible outcomes yielding one or more girls.
If we want the theoretical probability of getting one or more heads
when tossing two coins, we can count four possible outcomes from the tree
diagram (Figure 5.3). We find that three of the outcomes give one or more
heads. So
p(one or more heads) ⳱
3
⳱ 0.75
4
If we want to find the theoretical probability of one or more girls in a
two-child family, for example, we can draw a tree diagram that is just like
the diagram for the two coins (see Figure 5.4). Again, counting the number
of families with one or more girls, we find three. So
p(one or more girls) ⳱
3
⳱ 0.75
4
This process can be extended. For example, for a three-child family, as
we considered in Section 5.2, the number of possible family combinations is
found as shown in Figure 5.5. You can see that a tree diagram is not required
to carry out the counting process, but it is a very helpful tool.
We can use a rule to find the total number of outcomes (families, in
this case) possible in a trial. In our example, the trial—an observation of a
three-child family—can be divided into three subtrials, one for each child
in the family. To find the total number of trials, we find the product of
the numbers of possible outcomes of all individual subtrials. Since in our
First child
Second child
Third child
G
B
G
B
G
B
G
B
G
B
G
B
G
B
Figure 5.5
Possible families
G
G
G
G
B
B
B
B
G
G
B
B
G
G
B
B
G
B
G
B
G
B
G
B
3 girls
2 girls
1 girl
2 girls
1 girl
0 girls
Possible family combinations for three children.
example each child can be either a boy or a girl and there are three children,
we have
2⫻2⫻2⳱ 8
possible family combinations. Drawing the tree diagram also makes this
fact clear.
Suppose one of two subtrials can occur in four ways and the second
subtrial can occur in three ways. Then drawing a tree diagram would tell
us that the total number of possible ways the trial can occur is 4 ⫻ 3, as
we see in Figure 5.6. This rule is sometimes called the multiplication rule
for counting the total number of ways a trial made up of subtrials can
occur.
Subtrial 1
Subtrial 2
a
1
2
3
b
1
2
3
c
1
2
3
d
1
2
3
Figure 5.6
Possible outcomes
for subtrials 1 and 2.
Returning to the example of the three-child family, we find that the case
of three girls happens in one of the eight possible family combinations. So
p(three girls) ⳱
1
⳱ 0.125
8
Some problems involving equally likely outcomes require complex reasoning involving permutations, combinations, and other counting formulas.
For example, what is the theoretical probability of being dealt five hearts in a
five-card hand from an ordinary deck of cards (a nice hand if one is playing
poker)? We would need to study permutations and combinations before
being able to answer a problem like this. One important theoretical probability model involving equally likely outcomes—the binomial probability
model—will be discussed in Chapter 8.
Theoretical Independence
We have already learned the law of experimental independence, namely
that
P(A and B) ⬇ P(A) ⫻ P(B)
It should not surprise us that this is an approximate empirical stand-in for an
analogous law for theoretical probabilities for which exact equality holds.
The law of theoretical independence
If two events are independent, then the theoretical probability that they both occur is
exactly equal to the product of the two individual probabilities.
If we denote the two independent events
as A and B, then according to this law,
p(A and B) ⳱ p(A) ⫻ p(B)
Suppose, for example, we toss a coin and roll a six-sided die. What is
the theoretical probability of heads on the coin and a 6 on the die?
Since
1
p(heads) ⳱
2
and
1
p(6 on die) ⳱
6
and since the tossing of a coin and the rolling of a die are independent, we
have
1 1
1
⬇ 0.08
p(heads and 6) ⳱ ⫻ ⳱
2 6
12
Example 5.8
Suppose we choose a pair of random digits independently, each taken from the
digits 0 to 9. What is the theoretical probability of getting the pair 2, 3 (that is, first 2
and then 3)? (Note that this is what sampling two digits from Table B.3 amounts to.)
Solution
We have
p(2) ⳱
1
10
p(3) ⳱
1
10
and
So
p(2 and then 3) ⳱ p(2) ⫻ p(3)
⳱
1
1
1
⫻
⳱
⳱ 0.01
10 10
100
This property of independence also applies to more than two events for
both estimated and theoretical probabilities.
Example 5.9
Three Independent Events Involving Estimated Probabilities
Suppose we collect weather data on three cities—San Juan, Chicago, and Calcutta—
and find these estimated probabilities of rain on a certain day:
P(rain in San Juan) ⳱ 0.58
P(rain in Chicago) ⳱ 0.47
P(rain in Calcutta) ⳱ 0.34
We make the model assumption that rainfall in each city is independent of rainfall in
the other cities. So, using the law of experimental independence, the experimental
probability of rain simultaneously in all three cities must obey
P(rain in all three cities) ⬇ P(rain in San Juan)
⫻ P(rain in Chicago)
⫻ P(rain in Calcutta)
⬇ 0.58 ⫻ 0.47 ⫻ 0.34 ⬇ 0.093
We know that p(rain in all three cities) ⬇ P(rain in all three cities) ⬇ 0.093. Thus we
have an experimental estimate of p(rain in all three cities), obtained by assuming
experimental independence!
Let’s now consider the theoretical probability of three independent
events all occurring together.
Example 5.10
Three Independent Events Involving Theoretical Probabilities
What is the theoretical probability of three heads occurring in three tosses of a fair
coin?
Solution
For each toss,
p(heads) ⳱
1
2
Therefore,
p(three heads) ⳱ p(heads on first toss)
⫻ p(heads on second toss)
⫻ p(heads on third toss)
1 1 1
1
⳱ ⫻ ⫻ ⳱ ⳱ 0.125
2 2 2
8
Compare this answer with the one obtained earlier in this section for p(three
girls) using a tree diagram (see Figure 5.5) and treating the eight outcomes possible
in the trial as equally likely. We can thus reach the same correct answer of 0.125
in two different ways, one using the idea of eight equally likely outcomes and the
other using the idea of three independent subtrials.
What is the probability of three heads occurring if the coin is loaded with
p(heads) ⳱ 0.6 on each trial? Here the tree diagram approach fails, and you must
use the multiplication rule of theoretical independence.
Working with probability as a theoretical concept with rules and formulas for computing the probability of events is further developed in Chapter 8 and in the beginning sections of Chapter 14.
SECTION 5.5 EXERCISES
Unless otherwise noted, the probabilities required in these exercises are theoretical, not experimental.
1. Identify three probability experiments having
multiple stages (such as tossing two dice) and
equally likely outcomes. Draw a tree diagram
for each to justify your choices.
2. A four-sided die is called a tetrahedral die.
Assume that the possible outcomes of tossing
a four-sided die are 1, 2, 3, 4. Under the
assumption of equally likely outcomes, find
the following:
a. p(1)
b. p(2)
c. p(3)
d. p(4)
e. p(even number) f. p(number less than 4)
3. A ten-spinner is shown below. The arrow is
free to spin until it stops in one of the 10
sectors.
8
7
6
9 0
1
2
5
4
3