## Easy Mathematics Step-by-Step (2012)

### Chapter 18. Counting and Probability

In this chapter, you learn about counting and probability.

**Counting Methods**

You have different methods to find the number of ways to arrange or combine things. The most efficient method is to use the *counting principle*. The counting principle says that if you can do a first task in any one of *m* different ways, and after that you can do a second task in any one of *n* different ways, then you can do the first task followed by the second task in different ways.

When you use the counting principle, *multiply*, don’t add.

**Problem** A young boy is selecting a cap and a pair of tennis shoes to wear. He has a choice of a blue, red, orange, or yellow cap. He may select a black, brown, or white pair of tennis shoes. How many different ways can the boy choose one cap and one pair of tennis shoes?

**Solution**

*Step 1*. Determine how many tasks are involved.

Two tasks are involved: first, select a cap; next, select a pair of tennis shoes.

*Step 2*. Determine the number of ways to select a cap.

There are four colors to choose from, so the number of ways to select a cap is 4.

*Step 3*. Determine the number of ways to select a pair of tennis shoes.

There are three colors to choose from, so the number of ways to select a pair of tennis shoes is 3.

*Step 4*. Multiply the number of ways in step 2 by the number of ways in step 3.

(no. of ways to select a cap) × (no. of ways to select a pair of tennis shoes)

*Note:* “no.” is an abbreviation for “number.”

*Step 5*. State the answer.

There are 12 possible ways to select one cap and one pair of tennis shoes.

One popular method that you might have seen for working this problem is to list every possibility in a tree diagram. Here are the steps.

*Step 1*. List all the possibilities for the cap selection.

*Step 2*. Below each cap selection, draw three “branches,” one for each tennis shoe selection, and list the tennis shoe selections.

*Step 3*. Count the “leaves” at the base of the diagram to determine the number of possible ways to select one cap and one pair of tennis shoes.

*Step 4*. State the answer.

There are 12 leaves, so there are 12 possible ways to select one cap and one pair of tennis shoes.

Notice in the tree diagram that each leaf represents a choice. For instance, the first leaf represents the choice of selecting a blue cap and black tennis shoes. This choice is different from the choice represented by the fourth leaf, which is selecting a red cap and black tennis shoes.

A third commonly used method is to list every possibility in an organized table. For the above problem, you proceed systematically in building the table, as shown here.

*Step 1*. List a blue cap color with each of the tennis shoe colors that can be selected.

*Step 2*. List a red cap color with each of the tennis shoe colors that can be chosen.

*Step 3*. List an orange cap color with each of the tennis shoe colors that can be chosen.

*Step 4*. List a yellow cap color with each of the tennis shoe colors that can be chosen.

*Step 5*. Count the number of rows in the table.

There are 12 rows in the table.

*Step 6*. State the answer.

There are 12 possible ways to select one cap and one pair of tennis shoes.

Similar to the results of the tree diagram, each row in the table represents a choice. For instance, the 12th row represents the choice of selecting a yellow cap and white tennis shoes.

When you use a tree diagram or an organized table to count possibilities, be sure to proceed in a systematic manner, as illustrated in this chapter. Otherwise, you may overlook a possibility or count one more than once.

Tree diagrams and organized tables are useful counting methods when you have a limited number of choices, but they become increasingly unwieldy as the number of options increases. In contrast, the counting principle can be extended to accommodate any number of tasks, as shown in the following problem.

**Problem** How many different three-character codes consisting of one lowercase letter followed by two digits (0 to 9) are possible?

**Solution**

*Step 1.* Determine how many tasks are involved.

Three tasks are involved: first, select a lowercase letter; second, select a first digit; and third, select a second digit.

*Step 2*. Determine the number of ways to select a lowercase letter.

There are 26 lowercase letters, so the number of ways to select a lowercase letter is 26.

*Step 3*. Determine the number of ways to select the first digit.

There are 10 digits from which to choose, so the number of ways to select the first digit is 10.

*Step 4*. Determine the number of ways to select the second digit.

There are 10 digits from which to choose, so the number of ways to select the second digit is 10.

*Step 5*. Multiply the number of ways in step 2 by the number of ways in step 3 by the number of ways in step 4.

(no. of ways to select a letter) × (no. of ways to select a digit) × (no. of ways to select a digit)

*Step 6*. State the answer.

There are 2600 possible three-digit codes consisting of one lowercase letter followed by two digits.

**Basic Probability Concepts**

*Probability* is a measure of the chance that an event will happen. In simple words, an event is something that happens subject to chance, for instance, heads showing face up when you flip a coin. You use the notation *P* (*E*) as shorthand for “the probability that the event *E* will happen.” If all possible outcomes are *equally likely*, then the probability that an event *E* will occur is determined this way:

In a probability problem, the number of possible outcomes is always greater than or equal to the number of outcomes favorable to the event. Always check to make sure that the denominator is *larger* *than or equal to* the numerator when you substitute values into the formula.

Do not use the formula

unless all possible outcomes are equally likely. Doing so is a common error.

“Equally likely” outcomes are outcomes that have the same chance of happening. “Favorable” outcomes are the outcomes that will result in the event *E* occurring. These are the outcomes you are looking for. Naturally, you can use other letters, such as *A*, *B*, and *C*, to represent events. It is customary to use uppercase letters for this purpose.

The following problem illustrates the important role that counting plays in computing probabilities.

**Problem** Find the indicated probability.

** a**. A fair coin is flipped one time, and the face-up side of the coin is observed. Find the probability that the up face is heads.

** b**. A bag contains five tiles, all identical in size and shape, that are numbered 10, 20, 30, 40, and 50. If a person picks out a single tile from the bag without looking, what is the probability that the number on the tile will be divisible by 20?

** c**. Suppose a three-character digital code for a lock box consists of one lowercase letter followed by two digits (0 to 9). Find the probability that a person can randomly guess the correct code for the lock box.

**Solution**

** a**. A fair coin is flipped one time, and the face-up side of the coin is observed. Find the probability that the up face is heads.

*Step 1*. Count the number of possible outcomes.

The possible outcomes of one flip of a coin are heads on the face up (H) or tails on the face up (T), so the number of possible outcomes is 2.

*Step 2*. Count the number of favorable outcomes.

There is one favorable outcome, H.

*Step 3*. Check whether all possible outcomes are equally likely.

The coin is a fair coin, so H and T are equally likely outcomes.

*Step 4*. Compute the probability.

*Step 5*. State the answer.

On one flip of a fair coin, the probability of getting heads on the face-up side is .

** b**. A bag contains five tiles, all identical in size and shape, that are numbered 10, 20, 30, 40, and 50. If a person picks out a single tile from the bag without looking, what is the probability that the number on the tile will be divisible by 20?

*Step 1*. Count the number of possible outcomes.

The possible outcomes are a 10, 20, 30, 40, or 50 on the tile, so there are five possible outcomes.

*Step 2*. Count the number of favorable outcomes.

There are two favorable outcomes for this event: drawing a tile numbered 20 or drawing a tile numbered 40.

*Step 3*. Check whether all possible outcomes are equally likely.

Given that the tiles are identical in size and shape and that the person draws a tile without looking, all possible outcomes are equally likely.

*Step 4*. Compute the probability.

*Step 5*. State the answer.

The probability of drawing a tile whose number is divisible by 20 is .

** c**. Suppose a three-character digital code for a lock box consists of one lowercase letter followed by two digits (0 to 9). Find the probability that a person can randomly guess the correct code for the lock box.

*Step 1*. Count the number of possible outcomes.

The number of possible codes (outcomes) = (no. of ways to select a letter)(no. of ways to select a digit)(no. of ways to select a digit)

*Step 2*. Count the number of favorable outcomes.

There is only one favorable outcome, the correct three-character digital code.

*Step 3*. Check whether all possible outcomes are equally likely.

Given that the person is trying to randomly guess the code, all possible outcomes are equally likely.

*Step 4*. Compute the probability.

*Step 5*. State the answer.

The probability that a person can randomly guess the correct code for the lock box is .

Probabilities can be expressed as fractions, decimals, or percents. In problem b above, the probability of drawing an even-numbered tile can be expressed as , 0.4, or 40%.

The probability that an event is certain to happen is 1 or 100%. For instance, if you have a bag containing tiles numbered 10, 20, 30, 40, and 50 only, the probability of drawing a tile that has a multiple of 10 on it from the bag is 1 (because the numbers 10, 20, 30, 40, and 50 are all multiples of 10).

If an event cannot happen, then it has a probability of 0. For instance, the probability of drawing a tile with the number 60 on it from a bag containing tiles numbered 10, 20, 30, 40, and 50 only is 0 (because none of the tiles has 60 on it). Thus, the lowest probability is 0, and the highest probability is 1. All other probabilities fall between 0 and 1. Therefore, if you work a probability problem, and your answer is greater than 1 or your answer is negative, you’ve made an error! Go back and check your work.

Always remember to check whether the outcomes are equally likely before using the formula for probability. This situation is illustrated in the following problem.

**Problem** Suppose you spin a spinner that is one-fourth red, one-fourth yellow, and one-half green, as shown below. Find the probability that the spinner lands on red or green.

**Solution**

*Step 1*. Count the number of possible outcomes.

There are three possible outcomes because the spinner can land on red, yellow, or green.

*Step 2*. Count the number of favorable outcomes.

There are two favorable outcomes: landing on red or green.

*Step 3*. Check whether all possible outcomes are equally likely.

The three outcomes are not equally likely because the green section is larger than the other two sections.

*Step 4*. Compute the probability.

Given that the green and red sections occupy of the spinner, then

*Step 5*. State the answer.

The probability that the spinner lands on red or green is .

**Probability When Drawing with or Without Replacement**

Sometimes a probability problem involves a two-step selection process. For instance, drawing a marble from a box of colored marbles, and then drawing a second marble from the same box. In this problem, it is important to consider whether the drawing is done “with replacement” or “without replacement.” “With replacement” means that the first marble drawn is put back in the box before the second drawing takes place. “Without replacement” means that the first marble is not put back before the second drawing takes place. After accounting for “with replacement” or “without replacement,” you multiply the probabilities of making each of the two selections, as illustrated in the following problem.

**Problem** Suppose a box contains 10 marbles: 3 red marbles, 5 blue marbles, and 2 green marbles, all identical except for color.

** a**. If you draw two marbles, one at a time, from the box without looking, what is the probability that you will draw two red marbles if your first draw is done

*without*replacement?

** b**. If you draw two marbles, one at a time, from the box without looking, what is the probability that you will draw two red marbles if your first draw is done

*with*replacement?

**Solution**

** a**. If you draw two marbles, one at a time, from the box without looking, what is the probability that you will draw two red marbles if your first draw is done

*without*replacement?

*Step 1*. Find the probability of drawing a red marble on the first draw.

Initially, there are 3 red marbles in the box of 10 marbles, so

*Step 2*. Find the probability of drawing a red marble on the second draw. After a red marble is drawn and not put back, there are 2 red marbles in a box of 9 marbles, so *P*(red on second draw after drawing red on the first draw

*Step 3*. Multiply the probability from step 1 by the probability from step 2.

*Step 4*. State the answer.

The probability of drawing two red marbles if the first draw is done *without* replacement is .

** b**. If you draw two marbles, one at a time, from the box without looking, what is the probability that you will draw two red marbles if your first draw is done

*with*replacement?

*Step 1*. Find the probability of drawing a red marble on the first draw.

Initially, there are 3 red marbles in the box of 10 marbles, so

*Step 2*. Find the probability of drawing a red marble on the second draw.

After a red marble is drawn and put back, you again have 3 red marbles in a box of 10 marbles, so *P* (red on second draw after drawing red on the first draw *with replacement*)

*Step 3*. Multiply the probability from step 1 by the probability from step 2.

*Step 4*. State the answer.

The probability of drawing two red marbles if the first draw is done *with* replacement is .

This process of computing the probability of a sequence of events can be extended to any number of events. You should always consider whether “with replacement” or “without replacement” is a concern as you go from one event to the next.

**Probability for Independent Events**

Of course, when you are flipping coins or tossing numbered cubes, replacement is not a concern. Each flip of the coin or toss of the numbered cube is *independent* of the other flips or tosses. *Independent events* do not affect the probabilities of one another, so you simply multiply the probabilities in these cases. There are numerous other situations that involve independent events as well. In most cases, you can use your judgment to decide whether the events are independent.

Coins (and other physical objects used in games of chance) do not have “memories,” so do not make the mistake of thinking that a certain outcome is “due.”

**Problem** Find the indicated probability.

** a**. A fair coin is flipped three times. Find the probability of getting three heads in a row.

** b**. A young boy wears a cap and a pair of tennis shoes to school every day. Suppose he randomly picks a blue, red, orange, or yellow cap, then randomly picks a black, brown, or white pair of tennis shoes. Find the probability that the boy will pick a red cap and black tennis shoes.

**Solution**

** a**. A fair coin is flipped three times. Find the probability of getting three heads in a row.

*Step 1*. Find the probability of obtaining heads on the first flip.

*Step 2*. Find the probability of obtaining heads on the second flip.

*Step 3*. Find the probability of obtaining heads on the third flip.

*Step 4*. The events are independent, so multiply the probability from step 1 by the probability from step 2 by the probability from step 3.

*Step 5*. State the answer.

When a fair coin is flipped three times, the probability of getting three heads in a row is .

** b**. A young boy wears a cap and a pair of tennis shoes to school every day. Suppose he randomly picks a blue, red, orange, or yellow cap, then randomly picks a black, brown, or white pair of tennis shoes. Find the probability that the boy will pick a red cap and black tennis shoes.

*Step 1*. Find the probability of picking a red cap.

*Step 2*. Find the probability of picking black tennis shoes.

*Step 3*. Because the boy is picking randomly and because the hat selection and shoe selection probabilities do not affect each other, the events are independent, so multiply the probability from step 1 by the probability from step 2.

*Step 4*. State the answer.

The probability the boy will pick a red cap and black tennis shoes is .

Notice that the answer to problem b above is consistent with what you previously found; that is, that there are 12 possible ways for the boy to select one cap and one pair of tennis shoes. In only one of those 12 ways would the boy select a red cap and black tennis shoes. Thus, using the probability formula, you have

**Exercise 18**

__1.__ Suppose you are playing a carnival game. You can pick door 1, door 2, or door 3, and then you select one of four curtains behind each door. How many different ways can you select one door and one curtain?

__2.__ Suppose a code is dialed by means of three disks, each of which is stamped with 15 letters. How many three-letter codes are possible using the three disks?

__3.__ A woman is ordering a sandwich with chips and a drink for lunch. She has a choice of three kinds of bread (white, whole wheat, or rye), four sandwich fillings (beef, chicken, ham, or turkey), two kinds of chips (potato chips or corn chips), and three kinds of drinks (cola, juice, or tea). How many different lunches can the woman order if she makes one bread selection, one sandwich filling selection, one chips selection, and one drink selection?

__4.__ You toss a cube, whose faces are numbered 1, 2, 3, 4, 5, and 6, one time. What is the probability that a 5 appears face up?

__5.__ You have a box of 50 marbles, all identical except for color. The box contains 20 blue, 10 red, 14 green, and 6 yellow marbles. If a person picks out a single marble from the box without looking, what is the probability that the marble will be green?

__6.__ You have a box of 50 marbles, all identical except for color. The box contains 20 blue, 10 red, 14 green, and 6 yellow marbles. If a person picks out a single marble from the box without looking, what is the probability that the marble will be yellow or blue?

__7.__ You have a box of 50 marbles, all identical except for color. The box contains 20 blue, 10 red, 14 green, and 6 yellow marbles. If a person picks out two marbles, one at a time, from the box without looking, what is the probability that the person will draw two red marbles if the first draw is done *without* replacement?

__8.__ You have a box of 50 marbles, all identical except for color. The box contains 20 blue, 10 red, 14 green, and 6 yellow marbles. If a person picks out two marbles, one at a time, from the box without looking, what is the probability that the person will draw two red marbles if the first draw is done *with* replacement?

__9.__ A fair coin is flipped five times. Find the probability of getting five heads in a row.

__10.__ A spinner for a board game has three red sections, two yellow sections, four blue sections, and one green section. The sections are all of equal size. What is the probability of spinning red on the first spin and green on the second spin?