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3.1 Events, Sample Spaces, and Probability

Let’s begin our treatment of probability with straightforward examples that are easily described. With the aid of these simple examples, we can introduce important definitions that will help us develop the notion of probability more easily.

Suppose a coin is tossed once and the up face is recorded. The result we see is called an observation, or measurement, and the process of making an observation is called an experiment. Notice that our definition of experiment is broader than the one used in the physical sciences, which brings to mind test tubes, microscopes, and other laboratory equipment. Statistical experiments may include, in addition to these things, recording an Internet user’s preference for a Web browser, recording a voter’s opinion on an important political issue, measuring the amount of dissolved oxygen in a polluted river, observing the level of anxiety of a test taker, counting the number of errors in an inventory, and observing the fraction of insects killed by a new insecticide. The point is that a statistical experiment can be almost any act of observation, as long as the outcome is uncertain.

An experiment is an act or process of observation that leads to a single outcome that cannot be predicted with certainty.

Consider another simple experiment consisting of tossing a die and observing the number on the up face. The six possible outcomes of this experiment are as follows:

Observe a 1.
Observe a 2.
Observe a 3.
Observe a 4.
Observe a 5.
Observe a 6.

Note that if this experiment is conducted once, you can observe one and only one of these six basic outcomes, and the outcome cannot be predicted with certainty. Also, these outcomes cannot be decomposed into more basic ones. Because observing the outcome of an experiment is similar to selecting a sample from a population, the basic possible outcomes of an experiment are called sample points.*

A sample point is the most basic outcome of an experiment.

Example 3.1 Listing Sample Points for a Coin-Tossing Experiment

Problem

Two coins are tossed, and their up faces are recorded. List all the sample points for this experiment.

Solution

Even for a seemingly trivial experiment, we must be careful when listing the sample points. At first glance, we might expect one of three basic outcomes: Observe two heads; Observe two tails; or Observe one head and one tail. However, further reflection reveals that the last of these, Observe one head and one tail, can be decomposed into two outcomes: Head on coin 1, Tail on coin 2; and Tail on coin 1, Head on coin 2.

A useful tool for illustrating this notion is a tree diagram. Figure 3.1 shows a tree diagram for this experiment. At the top of the “tree” there are two branches, representing the two outcomes (H or T) for the first tossed coin. Each of these outcomes results in two more branches, representing the two outcomes (H or T) for the second tossed coin. Consequently, after tossing both coins, you can see that we have four sample points:

Figure 3.1
Tree diagram for the coin-tossing experiment
1. Observe HH
2. Observe HT
3. Observe TH
4. Observe TT
where H in the first position means “Head on coin 1,” H in the second position means “Head on coin 2,” and so on.

Look Back

Even if the coins are identical in appearance, they are, in fact, two distinct coins. Thus, the sample points must account for this distinction.

Now Work Exercise 3.15a

We often wish to refer to the collection of all the sample points of an experiment. This collection is called the sample space of the experiment. For example, there are six sample points in the sample space associated with the die-toss experiment. The sample spaces for the experiments discussed thus far are shown in Table 3.1.

The sample space of an experiment is the collection of all its sample points.

Just as graphs are useful in describing sets of data, a pictorial method for presenting the sample space will often be useful. Figure 3.2 shows such a representation for each of the experiments in Table 3.1. In each case, the sample space is shown as a closed figure, labeled S, containing all possible sample points. Each sample point is represented by a solid dot (i.e., a “point”) and labeled accordingly. Such graphical representations are called Venn diagrams.

Figure 3.2
Venn diagrams for the three experiments from Table 3.1

Table 3.1 Experiments and Their Sample Spaces

Experiment:	Observe the up face on a coin.
Sample Space:	1. Observe a head. 2. Observe a tail.
This sample space can be represented in set notation as a set containing two sample points:
[&S: \|cbo\|H, T\|cbc\|&] $S : {H, T}$ $S : {H, T}$
Here, `H` represents the sample point Observe a head and `T` represents the sample point Observe a tail.
Experiment:	Observe the up face on a die.
Sample Space:	1. Observe a 1. 2. Observe a 2. 3. Observe a 3. 4. Observe a 4. 5. Observe a 5. 6. Observe a 6.
This sample space can be represented in set notation as a set of six sample points:
[&S: \|cbo\|1, 2, 3, 4, 5, 6\|cbc\|&] $S : {1, 2, 3, 4, 5, 6}$ $S : {1, 2, 3, 4, 5, 6}$
Experiment:	Observe the up faces on two coins.
Sample Space:	1. Observe HH. 2. Observe HT. 3. Observe TH. 4. Observe TT.
This sample space can be represented in set notation as a set of four sample points:
[&S: \|cbo\|HH, HT, TH, TT\|cbc\|&] $S : {H H, H T, T H, T T}$ $S : {H H, H T, T H, T T}$

Now that we know that an experiment will result in only one basic outcome—called a sample point—and that the sample space is the collection of all possible sample points, we’re ready to discuss the probabilities of the sample points. You have undoubtedly used the term probability and have some intuitive idea about its meaning. Probability is generally used synonymously with “chance,” “odds,” and similar concepts. For example, if a fair coin is tossed, we might reason that both the sample points Observe a head and Observe a tail have the same chance of occurring. Thus, we might state, “The probability of observing a head is 50%” or “The odds of seeing a head are 50:50.” Both of these statements are based on an informal knowledge of probability. We’ll begin our treatment of probability by using such informal concepts and then solidify what we mean later.

The probability of a sample point is a number between 0 and 1 that measures the likelihood that the outcome will occur when the experiment is performed. This number is usually taken to be the relative frequency of the occurrence of a sample point in a very long series of repetitions of an experiment.* For example, if we are assigning probabilities to the two sample points in the coin-toss experiment (Observe a head and Observe a tail), we might reason that if we toss a balanced coin a very large number of times, the sample points Observe a head and Observe a tail will occur with the same relative frequency of .5.

Biography John Venn (1834–1923)

The English Logician

Born in Hull, England, John Venn is probably best known for his pictorial representation of unions and intersections (the Venn diagram). While lecturing in moral science at Cambridge University, Venn wrote The Logic of Chance and two other treatises on logic. In these works, Venn probably had his greatest contribution to the field of probability and statistics: the notion that the probability of an event is simply the long-run proportion of times the event occurs. Besides being a well-known mathematician, Venn was a historian, philosopher, priest, and skilled machine builder. (His machine for bowling cricket balls once beat a top star on the Australian cricket team.)

Our reasoning is supported by Figure 3.3. The figure plots the relative frequency of the number of times that a head occurs in simulations (by computer) of the toss of a coin N times, where N ranges from as few as 25 tosses to as many as 1,500 tosses. You can see that when N is large (e.g., $N = 1, 500$ $N = 1, 500$ ), the relative frequency is converging to .5. Thus, the probability of each sample point in the coin-tossing experiment is .5.

For some experiments, we may have little or no information on the relative frequency of occurrence of the sample points; consequently, we must assign probabilities to the sample points on the basis of general information about the experiment. For example, if the experiment is about investing in a business venture and observing whether it succeeds or fails, the sample space would appear as in Figure 3.4.

Figure 3.3
Proportion of heads in `N` tosses of a coin

Figure 3.4
Experiment: invest in a business venture and observe whether it succeeds (`S`) or fails (`F` )

We are unlikely to be able to assign probabilities to the sample points of this experiment based on a long series of repetitions, since unique factors govern each performance of this kind of experiment. Instead, we may consider factors such as the personnel managing the venture, the general state of the economy at the time, the rate of success of similar ventures, and any other information deemed pertinent. If we finally decide that the venture has an 80% chance of succeeding, we assign a probability of .8 to the sample point Success. This probability can be interpreted as a measure of our degree of belief in the outcome of the business venture; that is, it is a subjective probability. Notice, however, that such probabilities should be based on expert information that is carefully assessed. If not, we may be misled on any decisions based on these probabilities or based on any calculations in which they appear. [Note: For a text that deals in detail with the subjective evaluation of probabilities, see Winkler (1972) or Lindley (1985).]

No matter how you assign the probabilities to sample points, the probabilities assigned must obey two rules:

Probability Rules for Sample Points

Let $p_{i}$ $p_{i}$ represent the probability of sample point i. Then

All sample point probabilities must lie between 0 and 1 (i.e., $0 \leq p_{i} \leq 1$ $0 \leq p_{i} \leq 1$ ).
The probabilities of all the sample points within a sample space must sum to 1 (i.e., $Σ p_{i} = 1$ $Σ p_{i} = 1$ ).

Assigning probabilities to sample points is easy for some experiments. For example, if the experiment is to toss a fair coin and observe the face, we would probably all agree to assign a probability of $\frac{1}{2}$ $\frac{1}{2}$ to the two sample points, Observe a head and Observe a tail. However, many experiments have sample points whose probabilities are more difficult to assign.

Example 3.2 Sample Point Probabilities—Hotel Water Conservation

Problem

Going Green is the term used to describe water conservation programs at hotels and motels. Many hotels now offer their guests the option of participating in these Going Green programs by reusing towels and bed linens. Suppose you randomly select one hotel from a registry of all hotels in Orange County, California, and check whether the hotel participates in a water conservation program. Show how this problem might be formulated in the framework of an experiment with sample points and a sample space. Indicate how probabilities might be assigned to the sample points.

Solution

The experiment can be defined as the selection of an Orange County hotel and the observation of whether a water conservation program is offered to guests at the hotel. There are two sample points in the sample space for this experiment:

$\begin{array}{l} C : & {The hotel offers a water conservation program .} \\ N : & {The hotel does not offer a water conservation program .} \end{array}$ $\begin{array}{l} C : & {The hotel offers a water conservation program .} \\ N : & {The hotel does not offer a water conservation program .} \end{array}$

The difference between this and the coin-toss experiment becomes apparent when we attempt to assign probabilities to the two sample points. What probability should we assign to sample point C? If you answer .5, you are assuming that the events C and N occur with equal likelihood, just like the sample points Observe a head and Observe a tail in the coin-toss experiment. But assigning sample point probabilities for the hotel water conservation experiment is not so easy. In fact, a recent report by the Orange County Water District stated that 70% of the hotels in the county now participate in a water conservation program of some type. In that case, it might be reasonable to approximate the probability of the sample point C as .7 and that of the sample point N as .3.

Look Back

Here we see that the sample points are not always equally likely, so assigning probabilities to them can be complicated, particularly for experiments that represent real applications (as opposed to coin- and die-toss experiments).

Now Work Exercise 3.22

Although the probabilities of sample points are often of interest in their own right, it is usually probabilities of collections of sample points that are important. Example 3.3 demonstrates this point.

Example 3.3 Probability of a Collection of Sample Points— Die-Tossing Experiment

Problem

A fair die is tossed, and the up face is observed. If the face is even, you win $1. Otherwise, you lose $1. What is the probability that you win?

Solution

Recall that the sample space for this experiment contains six sample points:

[&S: |cbo|1,2,3,4,5,6|cbc| &]
$S : {1, 2, 3, 4, 5, 6}$ $S : {1, 2, 3, 4, 5, 6}$

Since the die is balanced, we assign a probability of $\frac{1}{6}$ $\frac{1}{6}$ to each of the sample points in this sample space. An even number will occur if one of the sample points Observe a 2, Observe a 4, or Observe a 6 occurs. A collection of sample points such as this is called an event, which we denote by the letter A. Since the event A contains three sample points—all with probability $\frac{1}{6}$ $\frac{1}{6}$ —and since no sample points can occur simultaneously, we reason that the probability of A is the sum of the probabilities of the sample points in A. Thus, the probability of A is $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}$ $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}$ .

Look Back

On the basis of this notion of probability, in the long run, you will win $1 half the time and lose $1 half the time.

Now Work Exercise 3.14

Figure 3.5 is a Venn diagram depicting the sample space associated with a die-toss experiment and the event A, Observe an even number. The event A is represented by the closed figure inside the sample space S. This closed figure A comprises all of the sample points that make up event A.

Figure 3.5
Die-toss experiment with event `A`, Observe an even number

To decide which sample points belong to the set associated with an event A, test each sample point in the sample space S. If event A occurs, then that sample point is in the event A. For example, the event A, Observe an even number, in the die-toss experiment will occur if the sample point Observe a 2 occurs. By the same reasoning, the sample points Observe a 4 and Observe a 6 are also in event A.

To summarize, we have demonstrated that an event can be defined in words, or it can be defined as a specific set of sample points. This leads us to the following general definition of an event:

An event is a specific collection of sample points.

Example 3.4 Probability of an Event—Coin-Tossing Experiment

Problem

Consider the experiment of tossing two unbalanced coins. Because the coins are not balanced, their outcomes (H or T) are not equiprobable. Suppose the correct probabilities associated with the sample points are given in the accompanying table. [Note: The necessary properties for assigning probabilities to sample points are satisfied.] Consider the events

[&A: |cbo|~rom~Observe exactly one head.|cbc|~normal~ &]

[&B: |cbo|~rom~Observe at least one head.|cbc|~normal~ &]

\begin{array}{l} A : {Observe exactly one head .} \\ B : {Observe at least one head .} \end{array}

$\begin{array}{l} A : {Observe exactly one head .} \\ B : {Observe at least one head .} \end{array}$

Calculate the probability of A and the probability of B.

Sample Point	Probability
HH	$\frac{4}{9}$ $\frac{4}{9}$
HT	$\frac{2}{9}$ $\frac{2}{9}$
TH	$\frac{2}{9}$ $\frac{2}{9}$
TT	$\frac{1}{9}$ $\frac{1}{9}$

Solution

Event A contains the sample points HT and TH. Since two or more sample points cannot occur at the same time, we can easily calculate the probability of event A by summing the probabilities of the two sample points. Thus, the probability of observing exactly one head (event A), denoted by the symbol P(A), is

[&P|pbo|A|pbc||=|P|pbo|~rom~Observe~normal~ HT.|pbc||+|P|pbo|~rom~Observe~normal~ TH.|pbc||=|*frac*{2}{9}|+|*frac*{2}{9}|=|*frac*{4}{9} &]
$P (A) = P (Observe H T .) + P (Observe T H .) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}$ $P (A) = P (Observe H T .) + P (Observe T H .) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}$

Similarly, since B contains the sample points HH, HT, and TH, it follows that

[&P|pbo|B|pbc||=|*frac*{4}{9}|+|*frac*{2}{9}|+|*frac*{2}{9}|=|*frac*{8}{9} &]
$P (B) = \frac{4}{9} + \frac{2}{9} + \frac{2}{9} = \frac{8}{9}$ $P (B) = \frac{4}{9} + \frac{2}{9} + \frac{2}{9} = \frac{8}{9}$

Look Back

Again, these probabilities should be interpreted in the long run. For example, $P (B) = \frac{8}{9} \approx .89$ $P (B) = \frac{8}{9} \approx .89$ implies that if we were to toss two coins an infinite number of times, we would observe at least two heads on about 89% of the tosses.

Now Work Exercise 3.11

The preceding example leads us to a general procedure for finding the probability of an event A:

Probability of an Event

The probability of an event A is calculated by summing the probabilities of the sample points in the sample space for A.

Thus, we can summarize the steps for calculating the probability of any event, as indicated in the next box.

Steps for Calculating Probabilities of Events

Define the experiment; that is, describe the process used to make an observation and the type of observation that will be recorded.
List the sample points.
Assign probabilities to the sample points.
Determine the collection of sample points contained in the event of interest.
Sum the sample point probabilities to get the probability of the event.

Example 3.5 Applying the Five Steps to Find a Probability—Study of Divorced Couples

Problem

The American Association for Marriage and Family Therapy (AAMFT) is a group of professional therapists and family practitioners that treats many of the nation’s couples and families. The AAMFT released the findings of a study that tracked the post-divorce history of 100 pairs of former spouses with children. Each divorced couple was classified into one of four groups, nicknamed “perfect pals (PP),” “cooperative colleagues (CC),” “angry associates (AA),” and “fiery foes (FF).” The proportions classified into each group are shown in Table 3.2.

Table 3.2 Results of AAMFT Study of Divorced Couples

Group	Proportion
Perfect Pals (PP) (Joint-custody parents who get along well)	.12
Cooperative Colleagues (CC) (Occasional conflict, likely to be remarried)	.38
Angry Associates (AA) (Cooperate on children-related issues only, conflicting otherwise)	.25
Fiery Foes (FF) (Communicate only through children, hostile toward each other)	.25

Suppose one of the 100 couples is selected at random.

Define the experiment that generated the data in Table 3.2, and list the sample points.
Assign probabilities to the sample points.
What is the probability that the former spouses are “fiery foes”?
What is the probability that the former spouses have at least some conflict in their relationship?

Solution

The experiment is the act of classifying the randomly selected couple. The sample points—the simplest outcomes of the experiment—are the four groups (categories) listed in Table 3.2. They are shown in the Venn diagram in Figure 3.6.

Figure 3.6
Venn diagram for AAMFT survey
If, as in Example 3.1, we were to assign equal probabilities in this case, each of the response categories would have a probability of one-fourth (1/4), or .25. But by examining Table 3.2, you can see that equal probabilities are not reasonable in this case because the response percentages are not all the same in the four categories. It is more reasonable to assign a probability equal to the response proportion in each class, as shown in Table 3.3.*
*Since the response percentages were based on a sample of divorced couples, these assigned probabilities are estimates of the true population response percentages. You will learn how to measure the reliability of probability estimates in Chapter 5.

Table 3.3 Sample Point Probabilities for AAMFT Survey

Sample Point Probability

PP .12

CC .38

AA .25

FF .25
The event that the former spouses are “fiery foes” corresponds to the sample point FF. Consequently, the probability of the event is the probability of the sample point. From Table 3.3, we find that $P (F F) = .25 .$ $P (F F) = .25 .$ Therefore, there is a .25 probability (or one-fourth chance) that the couple we select are “fiery foes.”
The event that the former spouses have at least some conflict in their relationship, call it event C, is not a sample point because it consists of more than one of the response classifications (the sample points). In fact, as shown in Figure 3.6, the event C consists of three sample points: CC, AA, and FF. The probability of C is defined to be the sum of the probabilities of the sample points in C:

[&P|pbo|~it~C|pbc||=|~normal~P|pbo|~rom~CC|pbc||+|~normal~P|pbo|~rom~AA|pbc||+|~normal~P|pbo|~rom~FF|pbc||=|~normal~.38|+|.25|+|.25|=|.88~norm~ &]
$P (C) = P (CC) + P (AA) + P (FF) = .38 + .25 + .25 = .88$ $P (C) = P (CC) + P (AA) + P (FF) = .38 + .25 + .25 = .88$

Sample Point	Probability
PP	.12
CC	.38
AA	.25
FF	.25

Thus, the chance that we observe a divorced couple with some degree of conflict in their relationship is .88—a fairly high probability.

Look Back

The key to solving this problem is to follow the steps outlined in the box. We defined the experiment (Step 1) and listed the sample points (Step 2) in part a. The assignment of probabilities to the sample points (Step 3) was done in part b. For each probability in parts c and d, we identified the collection of sample points in the event (Step 4) and summed their probabilities (Step 5).

Now Work Exercise 3.18

The preceding examples have one thing in common: The number of sample points in each of the sample spaces was small; hence, the sample points were easy to identify and list. How can we manage this when the sample points run into the thousands or millions? For example, suppose you wish to select 5 marines for a dangerous mission from a division of 1,000. Then each different group of 5 marines would represent a sample point. How can you determine the number of sample points associated with this experiment?

One method of determining the number of sample points for a complex experiment is to develop a counting system. Start by examining a simple version of the experiment. For example, see if you can develop a system for counting the number of ways to select 2 marines from a total of 4. If the marines are represented by the symbols $M_{1}, M_{2}, M_{3},$ $M_{1}, M_{2}, M_{3},$ and $M_{4},$ $M_{4},$ the sample points could be listed in the following pattern:

$(M_{1}, M_{2})$ $(M_{1}, M_{2})$	$(M_{2}, M_{3})$ $(M_{2}, M_{3})$	$(M_{3}, M_{4})$ $(M_{3}, M_{4})$
$(M_{1}, M_{3})$ $(M_{1}, M_{3})$	$(M_{2}, M_{4})$ $(M_{2}, M_{4})$
$(M_{1}, M_{4})$ $(M_{1}, M_{4})$

Note the pattern and now try a more complex situation—say, sampling 3 marines out of 5. List the sample points and observe the pattern. Finally, see if you can deduce the pattern for the general case. Perhaps you can program a computer to produce the matching and counting for the number of samples of 5 selected from a total of 1,000.

A second method of determining the number of sample points for an experiment is to use combinatorial mathematics. This branch of mathematics is concerned with developing counting rules for given situations. For example, there is a simple rule for finding the number of different samples of 5 marines selected from 1,000. This rule, called the combinations rule, is given in the box.

Combinations Rule

Suppose a sample of n elements is to be drawn without replacement from a set of N elements. Then the number of different samples possible is denoted by $(\begin{matrix} N \\ n \end{matrix})$ $(\begin{matrix} N \\ n \end{matrix})$ and is equal to

[&|3(|~MAT~[1%2%C%120%C%A]*MAT*{N}{n}|3)||=|*frac*{N|fract|}{n|fract||pbo|N|-|n|pbc||fract|} &]

(\begin{matrix} N \\ n \end{matrix}) = \frac{N!}{n! (N - n)!}

$(\begin{matrix} N \\ n \end{matrix}) = \frac{N!}{n! (N - n)!}$

where

[&n|fract||=|n|pbo|n|-|1|pbc||pbo|n|-|2|pbc||cdots||thn|*N*[-2%0]|pbo|3|pbc||pbo|2|pbc||pbo|1|pbc| &]

n! = n (n - 1) (n - 2) \dots (3) (2) (1)

$n! = n (n - 1) (n - 2) \dots (3) (2) (1)$

and similarly for $N!$ $N!$ and $(N - n)!$ $(N - n)!$ For example, $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.$ $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.$ [Note: The quantity 0! is defined to be equal to 1.]

Example 3.6 Using the Combinations Rule—Selecting 2 Marines from 4

Problem

Consider the task of choosing 2 marines from a platoon of 4 to send on a dangerous mission. Use the combinations counting rule to determine how many different selections can be made.

Solution

For this example, $N = 4, n = 2,$ $N = 4, n = 2,$ and

[&|3(|~MAT~[1%2%C%120%C%A]*MAT*{4}{2}|3)||=|*frac*{4|fract|}{2|fract|2|fract|}|=|*frac*{4|mdot|3|mdot|2|mdot|1}{|pbo|2|mdot|1|pbc||pbo|2|mdot|1|pbc|}|=|6 &]
$(\begin{matrix} 4 \\ 2 \end{matrix}) = \frac{4!}{2! 2!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{(2 \cdot 1) (2 \cdot 1)} = 6$ $(\begin{matrix} 4 \\ 2 \end{matrix}) = \frac{4!}{2! 2!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{(2 \cdot 1) (2 \cdot 1)} = 6$

Look Back

You can see that this answer agrees with the number of sample points listed on p. 126.

Now Work Exercise 3.13

Example 3.7 Using The Combinations Rule—Selecting 5 movies from 20

Problem

Suppose a movie reviewer for a newspaper reviews 5 movies each month. This month, the reviewer has 20 new movies from which to make the selection. How many different samples of 5 movies can be selected from the 20?

Solution

For this example, $N = 20$ $N = 20$ and $n = 5.$ $n = 5.$ Then the number of different samples of 5 that can be selected from the 20 movies is

[&|pbo|~MAT~[1%2%C%120%C%A]*MAT*{20}{5}|pbc||=|*N*[0.5%0]*frac*{20|fract|}{5|fract||pbo|20|-|5|pbc||fract|}|=|*frac*{20|fract|}{5|fract|15|fract|} &]
$(\begin{matrix} 20 \\ 5 \end{matrix}) = \frac{20!}{5! (20 - 5)!} = \frac{20!}{5! 15!}$ $(\begin{matrix} 20 \\ 5 \end{matrix}) = \frac{20!}{5! (20 - 5)!} = \frac{20!}{5! 15!}$

[&*AS**AP*|=|*frac*{20|mdot|19|mdot|18|mdot||cdotsns|*N*[-1%0]|mdot|3|mdot|2|mdot|1}{|pbo|5|mdot|4|mdot|3|mdot|2|mdot|1|pbc||pbo|15|mdot|14|mdot|13|mdot||cdotsns|*N*[-1.5%0]|mdot|3|mdot|2|mdot|1|pbc|}|=|15,504 &]
$\begin{aligned} = \frac{20 \cdot 19 \cdot 18 \cdot \dots \cdot 3 \cdot 2 \cdot 1}{(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) (15 \cdot 14 \cdot 13 \cdot \dots \cdot 3 \cdot 2 \cdot 1)} = 15, 504 \end{aligned}$ $\begin{aligned} = \frac{20 \cdot 19 \cdot 18 \cdot \dots \cdot 3 \cdot 2 \cdot 1}{(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) (15 \cdot 14 \cdot 13 \cdot \dots \cdot 3 \cdot 2 \cdot 1)} = 15, 504 \end{aligned}$

Look Back

You can see that attempting to list all the sample points for this experiment would be an extremely tedious and time-consuming, if not practically impossible, task.

The combinations rule is just one of a large number of counting rules that have been developed by combinatorial mathematicians. This counting rule applies to situations in which the experiment calls for selecting n elements from a total of N elements, without replacing each element before the next is selected.

Statistics in Action Revisited The Probability of Winning Lotto

In Florida’s state lottery game, called Pick-6 Lotto, you select six numbers of your choice from a set of numbers ranging from 1 to 53. We can apply the combinations rule to determine the total number of combinations of 6 numbers selected from 53 (i.e., the total number of sample points [or possible winning tickets]). Here, $N = 53$ $N = 53$ and $n = 6;$ $n = 6;$ therefore, we have

[&|3(|~MAT~[1%2%C%120%C%A]*MAT*{N}{n}|3)||=|*N*[0.475%0]*N*[0.475%0]*frac*{N|fract|}{n|fract||pbo|N|-|n|pbc||fract|}|=|*frac*{53|fract|}{6|fract|47|fract|} &]

(\begin{matrix} N \\ n \end{matrix}) = \frac{N!}{n! (N - n)!} = \frac{53!}{6! 47!}

$(\begin{matrix} N \\ n \end{matrix}) = \frac{N!}{n! (N - n)!} = \frac{53!}{6! 47!}$

[&*AS**AP*|=|*frac*{|pbo|53|pbc||pbo|52|pbc||pbo|51|pbc||pbo|50|pbc||pbo|49|pbc||pbo|48|pbc||pbo|47|fract||pbc|}{|pbo|6|pbc||pbo|5|pbc||pbo|4|pbc||pbo|3|pbc||pbo|2|pbc||pbo|1|pbc||pbo|47|fract||pbc|} &]

\begin{aligned} = \frac{(53) (52) (51) (50) (49) (48) (47!)}{(6) (5) (4) (3) (2) (1) (47!)} \end{aligned}

$\begin{aligned} = \frac{(53) (52) (51) (50) (49) (48) (47!)}{(6) (5) (4) (3) (2) (1) (47!)} \end{aligned}$

[&*AS**AP*|=|22,957,480 &]

\begin{aligned} = 22, 957, 480 \end{aligned}

$\begin{aligned} = 22, 957, 480 \end{aligned}$

Now, since the Lotto balls are selected at random, each of these 22,957,480 combinations is equally likely to occur. Therefore, the probability of winning Lotto is

[&P|pbo|~rom~Win~normal~ 6|sol|53 ~rom~Lotto|pbc||=|~normal~1|sol||pbo|22,957,480|pbc||=|.00000004356~norm~ &]

P (Win 6 / 53 Lotto) = 1 / (22, 957, 480) = .00000004356

$P (Win 6 / 53 Lotto) = 1 / (22, 957, 480) = .00000004356$

This probability is often stated as follows: The odds of winning the game with a single ticket are 1 in 22,957,480, or 1 in approximately 23 million. For all practical purposes, this probability is 0, implying that you have almost no chance of winning the lottery with a single ticket. Yet each week there is almost always a winner in the Florida Lotto. This apparent contradiction can be explained with the following analogy:

Suppose there is a line of minivans, front to back, from New York City to Los Angeles, California. Based on the distance between the two cities and the length of a standard minivan, there would be approximately 23 million minivans in line. Lottery officials will select, at random, one of the minivans and put a check for $10 million in the glove compartment. For a cost of $1, you may roam the country and select one (and only one) minivan and check the glove compartment. Do you think you will find $10 million in the minivan you choose? You can be almost certain that you won’t. But now permit anyone to enter the lottery for $1 and suppose that 50 million people do so. With such a large number of participants, it is very likely that someone will find the minivan with the $10 million—but it almost certainly won’t be you! (This example illustrates an axiom in statistics called the law of large numbers. See the footnote at the bottom of p. 121.)

Exercises 3.1–3.37

Understanding the Principles

3.1 What is an experiment?
3.2 What are the most basic outcomes of an experiment called?
3.3 Define the sample space.
3.4 What is a Venn diagram?
3.5 Give two probability rules for sample points.
3.6 What is an event?
3.7 How do you find the probability of an event made up of several sample points?
3.8 Give a scenario where the combinations rule is appropriate for counting the number of sample points.

Learning the Mechanics

3.9 An experiment results in one of the following sample points: $E_{1}, E_{2}, E_{3}, E_{4},$ $E_{1}, E_{2}, E_{3}, E_{4},$ and $E_{5} .$ $E_{5} .$
1. Find $P (E_{3})$ $P (E_{3})$ if $P (E_{1}) = .1, P (E_{2}) = .2, P (E_{4}) = .1,$ $P (E_{1}) = .1, P (E_{2}) = .2, P (E_{4}) = .1,$ and $P (E_{5}) = .1$ $P (E_{5}) = .1$
2. Find $P (E_{3})$ $P (E_{3})$ if $P (E_{1}) = P (E_{3}), P (E_{2}) = .1,$ $P (E_{1}) = P (E_{3}), P (E_{2}) = .1,$ $P (E_{4})$ $P (E_{4})$ $= .2, and P (E_{5}) = .1$ $= .2, and P (E_{5}) = .1$
3. Find $P (E_{3})$ $P (E_{3})$ if $P (E_{1}) = P (E_{2}) = P (E_{4}) =$ $P (E_{1}) = P (E_{2}) = P (E_{4}) =$ $P (E_{5}) = .1$ $P (E_{5}) = .1$
3.10 The following Venn diagram describes the sample space of a particular experiment and events A and B:
1. Suppose the sample points are equally likely. Find P(A) and P(B).
2. Suppose $P (1) = P (2) = P (3) = P (4) = P (5) = \frac{1}{20}$ $P (1) = P (2) = P (3) = P (4) = P (5) = \frac{1}{20}$ and $P (6) = P (7) = P (8) = P (9) = P (10) = \frac{3}{20} .$ $P (6) = P (7) = P (8) = P (9) = P (10) = \frac{3}{20} .$ Find P(A) and P(B).
3.11 The sample space for an experiment contains five sample points with probabilities as shown in the table. Find the probability of each of the following events:

$A : {Either 1, 2, or 3 occurs .}$ $A : {Either 1, 2, or 3 occurs .}$

$B : {Either 1, 3, or 5 occurs .}$ $B : {Either 1, 3, or 5 occurs .}$

$C : {4 does not occur .}$ $C : {4 does not occur .}$

Sample Points Probabilities

1 .05

2 .20

3 .30

4 .30

5 .15
3.12 Compute each of the following:
1. $(\begin{matrix} 9 \\ 4 \end{matrix})$ $(\begin{matrix} 9 \\ 4 \end{matrix})$
2. $(\begin{matrix} 7 \\ 2 \end{matrix})$ $(\begin{matrix} 7 \\ 2 \end{matrix})$
3. $(\begin{matrix} 4 \\ 4 \end{matrix})$ $(\begin{matrix} 4 \\ 4 \end{matrix})$
4. $(\begin{matrix} 5 \\ 0 \end{matrix})$ $(\begin{matrix} 5 \\ 0 \end{matrix})$
5. $(\begin{matrix} 6 \\ 5 \end{matrix})$ $(\begin{matrix} 6 \\ 5 \end{matrix})$
3.13 Compute the number of ways you can select n elements from N elements for each of the following:
1. $n = 2, N = 5$ $n = 2, N = 5$
2. $n = 3, N = 6$ $n = 3, N = 6$
3. $n = 5, N = 20$ $n = 5, N = 20$
3.14 Two fair dice are tossed, and the up face on each die is recorded.
1. List the 36 sample points contained in the sample space.
2. Assign probabilities to the sample points.
3. Find the probability of observing each of the following events:
  - $A : {A 3 appears on each of the two dice .}$ $A : {A 3 appears on each of the two dice .}$
  - $B : {The sum of the numbers is even .}$ $B : {The sum of the numbers is even .}$
  - $C : {The sum of the numbers is equal to 7 .}$ $C : {The sum of the numbers is equal to 7 .}$
  - $D : {A 5 appears on at least one of the dice .}$ $D : {A 5 appears on at least one of the dice .}$
  - $E : {The sum of the numbers is 10 or more .}$ $E : {The sum of the numbers is 10 or more .}$
3.15 Two marbles are drawn at random and without replacement from a box containing two blue marbles and three red marbles.
1. a. List the sample points.
2. b. Assign probabilities to the sample points.
3. c. Determine the probability of observing each of the following events:
  - $A : {Two blue marbles are drawn .}$ $A : {Two blue marbles are drawn .}$
  - $B : {A red and a blue marble are drawn .}$ $B : {A red and a blue marble are drawn .}$
  - $C : {Two red marbles are drawn .}$ $C : {Two red marbles are drawn .}$
3.16 Simulate the experiment described in Exercise 3.15, using any five identically shaped objects, two of which are one color and three another. Mix the objects, draw two, record the results, and then replace the objects. Repeat the experiment a large number of times (at least 100). Calculate the proportion of times events A, B, and C occur. How do these proportions compare with the probabilities you calculated in Exercise 3.15? Should these proportions equal the probabilities? Explain.

Sample Points	Probabilities
1	.05
2	.20
3	.30
4	.30
5	.15

Applet Exercise 3.1

Use the applet entitled Simulating the Probability of Rolling a 6 to explore the relationship between the proportion of sixes rolled on several rolls of a die and the theoretical probability of rolling a 6 on a fair die.

1. To simulate rolling a die one time, click on the Roll button on the screen while $n = 1.$ $n = 1.$ The outcome of the roll appears in the list at the right, and the cumulative proportion of sixes for one roll is shown above the graph and as a point on the graph corresponding to 1. Click Reset and repeat the process with $n = 1$ $n = 1$ several times. What are the possible values of the cumulative proportion of sixes for one roll of a die? Can the cumulative proportion of sixes for one roll of a die equal the theoretical probability of rolling a 6 on a fair die? Explain.
2. Set $n = 10$ $n = 10$ and click the Roll button. Repeat this several times, resetting after each time. Record the cumulative proportion of sixes for each roll. Compare the cumulative proportions for $n = 10$ $n = 10$ with those for $n = 1$ $n = 1$ in part a. Which tend to be closer to the theoretical probability of rolling a 6 on a fair die?
3. Repeat part b for $n = 1, 000,$ $n = 1, 000,$ comparing the cumulative proportions for $n = 1, 000$ $n = 1, 000$ with those for $n = 1$ $n = 1$ in part a and for $n = 10$ $n = 10$ in part b.
4. On the basis of your results for parts a, b, and c, do you believe that we could justifiably conclude that a die is unfair because we rolled it 10 times and didn’t roll any sixes? Explain.

Applet Exercise 3.2

Use the applet entitled Simulating the Probability of a Head with a Fair Coin to explore the relationship between the proportion of heads on several flips of a coin and the theoretical probability of getting heads on one flip of a fair coin.

1. Repeat parts a–c of Applet Exercise 3.1 for the experiment of flipping a coin and the event of getting heads.
2. On the basis of your results for part a, do you believe that we could justifiably conclude that a coin is unfair because we flipped it 10 times and didn’t roll any heads? Explain.

Applying the Concepts—Basic

3.17 Do social robots walk or roll? Refer to the International Conference on Social Robotics (Vol. 6414, 2010) study of the trend in the design of social robots, Exercise 2.7 (p. 38). Recall that in a random sample of 106 social (or service) robots designed to entertain, educate, and care for human users, 63 were built with legs only, 20 with wheels only, 8 with both legs and wheels, and 15 with neither legs nor wheels. One of the 106 social robots is randomly selected, and the design (e.g., wheels only) is noted.
1. List the sample points for this study.
2. Assign reasonable probabilities to the sample points.
3. What is the probability that the selected robot is designed with wheels?
4. What is the probability that the selected robot is designed with legs?
3.18 Crop damage by wild boars. The level of crop damage by wild boars in southern Italy was investigated in Current Zoology (Apr. 2014). In the study, crop damage occurred when the wild boars partially or completely destroyed agricultural crops in privately owned lands. The researchers identified 157 incident of crop damage in the study area caused by wild boars over a five-year period. The accompanying table gives the types of crops destroyed and corresponding percentage of incident. Consider the type of crop damaged by wild boars for one randomly selected incident.

Type Percentage

Cereals 45%

Orchards 5

Legumes 20

Vineyards 15

Other crops 15

TOTAL 100%
1. List the possible outcomes of this experiment.
2. Assign reasonable probabilities to the outcomes.
3. What is the probability that cereals or orchards are damaged?
4. What is the probability that a vineyard is not damaged?
3.19 Colors of M&Ms candies. Originally, M&Ms Plain Chocolate Candies came in only a brown color. Today, M&Ms in standard bags come in six colors: brown, yellow, red, blue, orange, and green. According to Mars Corporation, today 24% of all M&Ms produced are blue, 20% are orange, 16% are green, 14% are yellow, 13% are brown, and 13% are red. Suppose you purchase a randomly selected bag of M&Ms Plain Chocolate Candies and randomly select one of the M&Ms from the bag. The color of the selected M&M is of interest.
1. Identify the outcomes (sample points) of this experiment.
2. Assign reasonable probabilities to the outcomes, part a.
3. What is the probability that the selected M&M is brown (the original color)?
4. In 1960, the colors red, green, and yellow were added to brown M&Ms. What is the probability that the selected M&M is either red, green, or yellow?
5. In 1995, based on voting by American consumers, the color blue was added to the M&M mix. What is the probability that the selected M&M is not blue?
3.20 Rare underwater sounds. A study of underwater sounds in a specific region of the Pacific Ocean focused on scarce sounds, such as humpback whale screams, dolphin whistles, and sounds from passing ships (Acoustical Physics, Vol. 56, 2010). During the month of September (non-rainy season), research revealed the following probabilities of rare sounds: $P (whale scream) = .03$ $P (whale scream) = .03$ , $P (ship sound) = .14$ $P (ship sound) = .14$ , and $P (r a i n) = 0$ $P (r a i n) = 0$ . If a sound is picked up by the acoustical equipment placed in this region of the Pacific Ocean, is it more likely to be a whale scream or a sound from a passing ship? Explain.
3.21 USDA chicken inspection. The United States Department of Agriculture (USDA) reports that, under its standard inspection system, one in every 100 slaughtered chickens passes inspection for fecal contamination.
1. If a slaughtered chicken is selected at random, what is the probability that it passes inspection for fecal contamination?
2. The probability of part a was based on a USDA study that found that 306 of 32,075 chicken carcasses passed inspection for fecal contamination. Do you agree with the USDA’s statement about the likelihood of a slaughtered chicken passing inspection for fecal contamination?
3.22 African rhinos. Two species of rhinoceros native to Africa are black rhinos and white rhinos. The International Rhino Federation estimates that the African rhinoceros population consists of 5,055 white rhinos and 20,405 black rhinos. Suppose one rhino is selected at random from the African rhino population and its species (black or white) is observed.
1. List the sample points for this experiment.
2. Assign probabilities to the sample points on the basis of the estimates made by the International Rhino Federation.

Type	Percentage
Cereals	45%
Orchards	5
Legumes	20
Vineyards	15
Other crops	15
TOTAL	100%

Applying the Concepts—Intermediate

3.23 STEM experiences for girls. Refer to the 2013 National Science Foundation (NSF) study on girls’ participation in informal science, technology, engineering, and mathematics (STEM) programs, Exercise 2.12 (p. 39). Recall that the researchers sampled 174 young women who recently participated in a STEM program. Of the 174 STEM participants, 107 were in urban areas, 57 in suburban areas, and 10 in rural areas. If one of the participants is selected at random, what is the probability that she is from an urban area? Not a rural area?
3.24 Health risks to beachgoers. According to a University of Florida veterinary researcher, the longer a beachgoer sits in wet sand or stays in the water, the higher the health risk (University of Florida News, Jan. 29, 2008). Using data collected at 3 Florida beaches, the researcher discovered the following: (1) 6 out of 1,000 people exposed to wet sand for a 10-minute period will acquire gastroenteritis; (2) 12 out of 100 people exposed to wet sand for two consecutive hours will acquire gastroenteritis; (3) 7 out of 1,000 people exposed to ocean water for a 10-minute period will acquire gastroenteritis; and (4) 7 out of 100 people exposed to ocean water for a 70-minute period will acquire gastroenteritis.
1. If a beachgoer spends 10 minutes in the wet sand, what is the probability that he or she will acquire gastroenteritis?
2. If a beachgoer spends two hours in the wet sand, what is the probability that he or she will acquire gastroenteritis?
3. If a beachgoer spends 10 minutes in the ocean water, what is the probability that he or she will acquire gastroenteritis?
4. If a beachgoer spends 70 minutes in the ocean water, what is the probability that he or she will acquire gastroenteritis?

MOLARS 3.25 Cheek teeth of extinct primates. Refer to Consider the American Journal of Physical Anthropology (Vol. 142, 2010) study of the dietary habits of extinct mammals, Exercise 2.9 (p. 38). Recall that 18 cheek teeth extracted from skulls of an extinct primate species discovered in western Wyoming were analyzed. Each tooth was classified according to degree of wear (unworn, slight, light-moderate, moderate, moderate-heavy, or heavy). The 18 measurements are reproduced in the accompanying table. One tooth is randomly selected from the 18 cheek teeth. What is the probability that the tooth shows a slight or moderate amount of wear?

Data on Degree of Wear
Unknown	Slight
Unknown	Slight
Unknown	Heavy
Moderate	Unworn
Slight	Light-moderate
Unknown	Light-moderate
Moderate-heavy	Moderate
Moderate	Unworn
Slight	Unknown

3.26 Chance of rain. Answer the following question posed in the Atlanta Journal-Constitution: When a meteorologist says, “The probability of rain this afternoon is .4,” does it mean that it will be raining 40% of the time during the afternoon?
3.27 Railway track allocation. Refer to Consider the Journal of Transportation Engineering (May 2013) investigation of railroad track allocation at a railway station with 11 tracks, Exercise 2.16 (p. 40). A simple algorithm designed to minimize waiting time and bottlenecks is one in which engineers randomly assign one of the 11 tracks to a train entering the station.
1. What is the probability that track #7 is assigned to an entering train?
2. Unknown to the engineer, tracks #2, #5, and #10 will have maintenance problems. What is the probability that one of these problematic tracks is assigned to an entering train?
MMC 3.28 Museum management. Refer to the Museum Management and Curatorship (June 2010) study of the criteria used to evaluate museum performance, Exercise 2.22 (p. 41). Recall that the managers of 30 leading museums of contemporary art were asked to provide the performance measure used most often. A summary of the results is reproduced in the table. One of the 30 museums is selected at random. Find the probability that the museum uses big shows most often as a performance measure.

Performance Measure Number of Museums

Total visitors 8

Paying visitors 5

Big shows 6

Funds raised 7

Members 4
3.29 Choosing portable grill displays. A study of how people attempt to influence the choices of others by offering undesirable alternatives was published in the Journal of Consumer Research (Mar. 2003). In one phase of the study, the researcher had each of 124 college students select showroom displays for portable grills. Five different displays (representing five different-sized grills) were available, but only three would be selected. The students were instructed to select the displays to maximize purchases of Grill #2 (a smaller grill).
1. In how many possible ways can the three-grill displays be selected from the 5 displays? List the possibilities.
2. The next table shows the grill display combinations and number of each selected by the 124 students. Use this information to assign reasonable probabilities to the different display combinations.
3. Find the probability that a student who participated in the study selected a display combination involving Grill #1.
Grill Display Combination Number of Students

1–2–3 35

1–2–4 8

1–2–5 42

2–3–4 4

2–3–5 1

2–4–5 34

Based on Hamilton, R. W. “Why do people suggest what they do not want? Using context effects to influence others’ choices.” Journal of Consumer Research, Vol. 29, Mar. 2003, Table 1.
3.30 ESL students and plagiarism. The Journal of Education and Human Development (Vol. 3, 2009) investigated the causes of plagiarism among six English-as-a-second-language (ESL) students. All students in the class wrote two essays, one in the middle and one at the end of the semester. After the first essay, the students were instructed on how to avoid plagiarism in the second essay. Of the six ESL students, three admitted to plagiarizing on the first essay. Only one ESL student admitted to plagiarizing on the second essay. (This student, who also plagiarized on the first essay, claimed she misplaced her notes on plagiarism.)
1. If one of the six ESL students is randomly selected, what is the probability that he or she plagiarized on the first essay?
2. Consider the results (plagiarism or not) for the six ESL students on the second essay. List the possible outcomes (e.g., Students 1 and 3 plagiarize, the others do not).
3. Refer to part b. Assume that, despite the instruction on plagiarism, the ESL students are just as likely as not to plagiarize on the second essay. What is the probability that no more than one of the ESL students plagiarizes on the second essay?
3.31 Jai alai Quinella bet. The Quinella bet at the parimutuel game of jai alai consists of picking the jai alai players that will place first and second in a game, irrespective of order. In jai alai, eight players (numbered 1, 2, 3, … , 8) compete in every game.
1. How many different Quinella bets are possible?
2. Suppose you bet the Quinella combination 2–7. If the players are of equal ability, what is the probability that you win the bet?
3.32 Using game simulation to teach a course. In Engineering Management Research (May 2012), a simulation game approach was proposed to teach concepts in a course on production. The proposed game simulation was for color television production. The products are two color television models, A and B. Each model comes in two colors, red and black. Also, the quantity ordered for each model can be 1, 2, or 3 televisions. The choice of model, color, and quantity is specified on a purchase order card.
1. Using a tree diagram, list how many different purchase order cards are possible. (These are the simple events for the experiment.)
2. Suppose, from past history, that black color TVs are in higher demand than red TVs. For planning purposes, should the engineer managing the production process assign equal probabilities to the simple events in part a? Why or why not?

Performance Measure	Number of Museums
Total visitors	8
Paying visitors	5
Big shows	6
Funds raised	7
Members	4

Grill Display Combination	Number of Students
1–2–3	35
1–2–4	8
1–2–5	42
2–3–4	4
2–3–5	1
2–4–5	34

Applying the Concepts—Advanced

3.33 Lead bullets as forensic evidence. Chance (Summer 2004) published an article on the use of lead bullets as forensic evidence in a federal criminal case. Typically, the Federal Bureau of Investigation (FBI) will use a laboratory method to match the lead in a bullet found at a crime scene with unexpended lead cartridges found in the possession of a suspect. The value of this evidence depends on the chance of a false positive—that is, the probability that the FBI finds a match, given that the lead at the crime scene and the lead in the possession of the suspect are actually from two different “melts,” or sources. To estimate the false positive rate, the FBI collected 1,837 bullets that the agency was confident all came from different melts. Then, using its established criteria, the FBI examined every possible pair of bullets and counted the number of matches. According to Chance, the FBI found 693 matches. Use this information to compute the chance of a false positive. Is this probability small enough for you to have confidence in the FBI’s forensic evidence?
3.34 Matching socks. Consider the following question posed to Marilyn vos Savant in her weekly newspaper column, “Ask Marilyn”:

I have two pairs of argyle socks, and they look nearly identical—one navy blue and the other black. [When doing the laundry] my wife matches the socks incorrectly much more often than she does correctly…. If all four socks are in front of her, it seems to me that her chances are 50% for a wrong match and 50% for a right match. What do you think?

Source: Parade Magazine, Feb. 27, 1994.

Use your knowledge of probability to answer this question. [Hint: List the sample points in the experiment.]
3.35 Post-op nausea study. Nausea and vomiting after surgery are common side effects of anesthesia and painkillers. Six different drugs, varying in cost, were compared for their effectiveness in preventing nausea and vomiting (New England Journal of Medicine, June 10, 2004). The medical researchers looked at all possible combinations of the drugs as treatments, including a single drug, as well as two-drug, three-drug, four-drug, five-drug, and six-drug combinations.
1. How many two-drug combinations of the six drugs are possible?
2. How many three-drug combinations of the six drugs are possible?
3. How many four-drug combinations of the six drugs are possible?
4. How many five-drug combinations of the six drugs are possible?
5. The researchers stated that a total of 64 drug combinations were tested as treatments for nausea. Verify that there are 64 ways that the six drugs can be combined. (Remember to include the one-drug and six-drug combinations as well as the control treatment of no drugs.)
3.36 Dominant versus recessive traits. An individual’s genetic makeup is determined by the genes obtained from each parent. For every genetic trait, each parent possesses a gene pair, and each parent contributes one-half of this gene pair, with equal probability, to his or her offspring, forming a new gene pair. The offspring’s traits (eye color, baldness, etc.) come from this new gene pair, each gene of which possesses some characteristic.

For the gene pair that determines eye color, each gene trait may be one of two types: dominant brown (B) or recessive blue (b). A person possessing the gene pair BB or Bb has brown eyes, whereas the gene pair bb produces blue eyes.
1. Suppose both parents of an individual are brown eyed, each with a gene pair of type Bb. What is the probability that a randomly selected child of this couple will have blue eyes?
2. If one parent has brown eyes, type Bb, and the other has blue eyes, what is the probability that a randomly selected child of this couple will have blue eyes?
3. Suppose one parent is brown eyed with a gene pair of type BB. What is the probability that a child has blue eyes?
3.37 Drug testing of firefighters. Hillsborough County, Florida, has 42 fire stations that employ 980 career firefighters and 160 volunteers. Both types of firefighters are drug tested every six months. However, the career firefighters and volunteers are treated as separate populations. For drug testing, 50% of the career firefighters are randomly selected and 50% of the volunteers are randomly selected. Some firefighters at a particular station argue that due to the smaller number of volunteers, an individual volunteer is more likely to be selected in the current sampling scheme than if all the career firefighters and volunteers were combined into a single population and 50% sampled. Do you agree? Explain your reasoning.

$[Hint : (\begin{array}{l} N - 1 \\ n - 1 \end{array}) / (\begin{array}{l} N \\ n \end{array}) = n / N]$ $[Hint : (\begin{array}{l} N - 1 \\ n - 1 \end{array}) / (\begin{array}{l} N \\ n \end{array}) = n / N]$

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Table of Contents for 3.1 Events, Sample Spaces, and Probability

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Table of Contents for
3.1 Events, Sample Spaces, and Probability