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# Basic Counting Rules

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Basic Counting Rules

In this Concept, you will learn about simple random samples and counting possibilities.

### Watch This

For an introduction to the Multiplication Rule for Counting, which is also called the Fundamental Counting Principle, see MathwithMrAlmeida, Fundamental Counting Principle (6:59).

### Guidance

Inferential Statistics is a method of statistics that consists of drawing conclusions about a population based on information obtained from samples. Samples are used because it can be quite costly in time and money to study an entire population. In addition, because of the inability to actually reach everyone in a census, a sample can be more accurate than a census.

The most important characteristic of any sample is that it must be a very good representation of the population. It would not make sense to use the average height of basketball players to make an inference about the average height of the entire US population. Likewise, it would not be reasonable to estimate the average income of the entire state of California by sampling the average income of the wealthy residents of Beverly Hills. The goal of sampling is to obtain a representative sample. There are a number of different methods for taking representative samples, and in this lesson, you will learn about simple random samples. It is important to know the size of your population in order to calculate probabilities of selecting an one individual from the population. For this reason, you will also be presented with the basic counting rules used to calculate probabilities.

Simple Random Sample

A simple random sample of size $n$ is one in which all samples of size $n$ are equally likely to be selected. In other words, if $n$ elements are selected from a population in such a way that every set of $n$ elements in the population has an equal probability of being selected, then the $n$ elements form a simple random sample.

Example : Suppose you randomly select 4 cards from a deck of playing cards, and all the cards selected are kings. Would you conclude that the deck is an ordinary deck, or would you conclude the deck is not an ordinary one and probably contains more than 4 kings?

The answer depends on how the cards were drawn. It is possible that the 4 kings were intentionally put on top of the deck, and hence, the drawing of the 4 kings was not unusual, and in fact, it was actually certain. However, if the deck was shuffled well, getting 4 kings is highly improbable.

#### Example A

Suppose a lottery consists of 100 tickets, and one winning ticket is to be chosen. What would be a fair method of selecting a winning ticket?

First, we must require that each ticket has an equal chance of winning. That is, each ticket must have a probability of $\frac{1}{100}$ of being selected. One fair way of doing this is to mix up all the tickets in a container and blindly pick one ticket. This is an example of random sampling.

However, this method would not be too practical if we were dealing with a very large population, such as, say, a million tickets, and we were asked to select 5 winning tickets. One method of picking a simple random sample is to give each element in the population a number. Then use a random number generator to pick 5 numbers. The people who were assigned one of the five numbers would then be the winners.

Some experiments have so many simple events that it is impractical to list them all. Tree diagrams are helpful in determining probabilities in these situations.

#### Example B

Suppose there are six balls in a box. They are identical, except in color. Two balls are red, three are blue, and one is yellow. We will draw one ball, record its color, and then set it aside. Next, we will draw another ball and record its color. With the aid of a tree diagram, calculate the probability of each of the possible outcomes of the experiment.

We first draw a tree diagram to aid us in seeing all the possible outcomes of this experiment.

The tree diagram shows us the two stages of drawing two balls without putting the first one back into the box. In the first stage, we pick a ball blindly. Since there are 2 red balls, 3 blue balls, and 1 yellow ball, the probability of getting a red ball is $\frac{2}{6}$ , the probability of getting a blue ball is $\frac{3}{6}$ , and the probability of getting a yellow ball is $\frac{1}{6}$ .

Remember that the probability associated with the second ball depends on the color of the first ball. Therefore, the two stages are not independent. To calculate the probabilities when selecting the second ball, we can look back at the tree diagram.

When taking the first ball and the second ball into account, there are eight possible outcomes for the experiment:

$RR$ : red on the $1^{\text{st}}$ and red on the $2^{\text{nd}}$

$RB$ : red on the $1^{\text{st}}$ and blue on the $2^{\text{nd}}$

$RY$ : red on the $1^{\text{st}}$ and yellow on the $2^{\text{nd}}$

$BR$ : blue on the $1^{\text{st}}$ and red on the $2^{\text{nd}}$

$BB$ : blue on the $1^{\text{st}}$ and blue on the $2^{\text{nd}}$

$BY$ : blue on the $1^{\text{st}}$ and yellow on the $2^{\text{nd}}$

$YR$ : yellow on the $1^{\text{st}}$ and red on the $2^{\text{nd}}$

$YB$ : yellow on the $1^{\text{st}}$ and blue on the $2^{\text{nd}}$

We want to calculate the probability of each of these outcomes. This is done as is shown below.

$P(RR) & = \frac{2}{6} \bullet \frac{1}{5}=\frac{2}{30}\\P(RB) & = \frac{2}{6} \bullet \frac{3}{5}=\frac{6}{30}\\P(RY) & = \frac{2}{6} \bullet \frac{1}{5}=\frac{2}{30}\\P(BR) & = \frac{3}{6} \bullet \frac{2}{5}=\frac{6}{30}\\P(YB) & = \frac{3}{6} \bullet \frac{2}{5}=\frac{6}{30}\\ P(YB) & = \frac{3}{6} \bullet \frac{1}{5}=\frac{3}{30}\\P(YB) & = \frac{1}{6} \bullet \frac{2}{5}=\frac{2}{30}\\P(YB) & = \frac{1}{6} \bullet \frac{3}{5}=\frac{3}{30}$

Notice that all of the probabilities add up to 1, as they should.

When using a tree diagram to compute probabilities, you multiply the probabilities as you move along a branch. In the above example, if we are interested in the outcome $RR$ , we note that the probability of picking a red ball on the first draw is $\frac{2}{6}$ . We then go to the second branch, choosing a red ball on the second draw, the probability of which is $\frac{1}{5}$ . Therefore, the probability of choosing $RR$ is $\left (\frac{2}{6}\right )\left (\frac{1}{5}\right )$ . The method used to solve the example above can be generalized to any number of stages.

#### Example C

A restaurant offers a special dinner menu every day. There are three entrées, five appetizers, and four desserts to choose from. A customer can only select one item from each category. How many different meals can be ordered from the special dinner menu?

Let’s summarize what we have.

Entrees: 3

Appetizer: 5

Dessert: 4

We use the Multiplicative Rule above to calculate the number of different dinners that can be selected. We simply multiply each of the numbers of choices per item together: $(3)(5)(4) = 60$ . Thus, there are 60 different dinners that can be ordered by the customers.

The Multiplicative Rule of Counting

The Multiplicative Rule of Counting states the following:

(I) If there are $n$ possible outcomes for event $A$ and $m$ possible outcomes for event $B$ , then there are a total of $nm$ possible outcomes for event $A$ followed by event $B$ .

Another way of stating this is as follows:

(II) Suppose you have $k$ sets of elements, with $n_1$ elements in the first set, $n_2$ elements in the second set, and $n_k$ elements in the $k^{\text{th}}$ set, and you want to take one sample from each of the $k$ sets. The number of different samples that can be formed is the product $n_1n_2n_3\ldots. n_k$ .

#### Example D

In how many different ways can you seat 8 people at a dinner table?

For the first seat, there are eight choices. For the second, there are seven remaining choices, since one person has already been seated. For the third seat, there are 6 choices, since two people are already seated. By the time we get to the last seat, there is only one seat left. Therefore, using the Multiplicative Rule above, we get $(8)(7)(6)(5)(4)(3)(2)(1) = 40, 320$ .

The multiplication pattern above appears so often in statistics that it has its own name, which is factorial , and its own symbol, which is '!'. When describing it, we say, “Eight factorial,” and we write, "8!."

Factorial Notation

$n!=n(n-1)(n-2)(n-3).....(3)(2)(1)$

### Vocabulary

Inferential statistics is a method of statistics that consists of drawing conclusions about a population based on information obtained from a subset or sample of the population.

A random sampling is a procedure in which each sample of a given size is equally likely to be selected.

The Multiplicative Rule of Counting states that if there are $n$ possible outcomes for event $A$ and $m$ possible outcomes for event $B$ , then there are a total of $nm$ possible outcomes for the series of events $A$ followed by $B$ .

The factorial sign, or ‘!’, is defined as $n!=n(n-1)(n-2)(n-3).....(3)(2)(1)$ .

### Guided Practice

How many positive integers less than 1,000 do not have 7 as any digit?

Solution:

The integers less than 1000 are those that have a values in the 1's, 10's and 100's places only. So, we can think of each place as a step:

Step 1: How many choices are there for the 1's place? There are 9, since we could put any number 0-9 in the ones place, except we can't use the 7.

Step 2: How many choices are there for the 10's place? Again, there are 9, since we could put any number 0-9 in the ones place, except we can't use the 7.

Step 3: How many choices are there for the 100's place? Again, there are 9, since we could put any number 0-9 in the ones place, except we can't use the 7.

Using the multiplication rule for counting:

(Possibilities at Step 1)(Possibilities at Step 2)(Possibilities at Step 3) $= 9\cdot 9\cdot 9=729.$

There are 729 integers that are less than 1,000 and do not have 7 as any digit?

### Practice

1. Determine the number of simple events when you toss a coin the following number of times. (Hint: As the numbers get higher, you will need to develop a systematic method of counting all the outcomes.)
1. Twice
2. Three times
3. Five times
4. $n$ times (Look for a pattern in the results of a) through c).)
2. Construct a tree diagram showing all possible outfits that can be made from 3 shirts (black, white and green), 2 pairs of pants (black and brown) and two pairs of shoes (black and white). If you randomly select an outfit, what is the probability that it will be all black?
3. Construct a tree diagram showing all possible results when four fair coins are tossed. What is the probability that at most one of the four coins landed on heads?
4. How many different three-digit numbers between 100 and 1,000 have 5 as the tens digit?
5. Flying into Los Angeles from Washington DC, you can choose one of three airlines and can also choose either first class or economy. How many travel options do you have?
6. Suppose an automobile license plate is designed to show a letter of the English alphabet, followed by a five-digit number. How many different license plates can be issued?
7. Find how many numbers between 2500 and 5000 can be formed using digits 1,2,3,4,5 and 7.
1. with no repeats.
2. with repeats.
8. Suppose you are going to make a sandwich and you have 2 choices of bread, 5 choices of meat (2 are fake meat options), and 4 choices of cheese. What are the number of possible sandwiches you can make assuming you choose one of each? What are the possible number of vegetarian sandwiches you can make, still assuming you choose one of each?
9. A music class of eight girls and nine boys is having a recital. If each member is to perform once, how many ways can the program be arranged if Tom must perform third?
10. If you go to the store to get 2 frozen items, 7 produce items, 2 drinks and 3 dry good items;
1. How many ways are their to order your selection of these items?
2. How many ways are there to order your selection of these items if you do it section by section?

Technology Notes:

Generating Random Numbers on the TI-83/84 Calculator

Press [MATH] , and then scroll to the right and choose PRB . Next, choose '1:rand' and press [ENTER] twice. The calculator returns a random number between 0 and 1. If you are taking a sample of 100, you need to use the first two digits of the random number that has been returned. If the calculator returns the same first two digits more than once, you can ignore them and press [ENTER] again.

Keywords

Event

Experiment

Factorial

Multiplicative Rule of Counting

Simple events

Simple random sample

Tree diagram