# Probability Part C: Analyzing Binomial Probabilities (45 minutes)

In This Part: Making a Tree Diagram
A tree diagram is a helpful tool for determining theoretical or mathematical probabilities. Let’s begin by examining the problem of tossing a fair coin. We’ll focus on the number of heads that occur in a certain number of tosses. See Note 7 below.

A tree diagram for the toss of a single coin has two branches that represent the two possible outcomes of this random experiment. In this tree diagram, the red branch represents the outcome “heads” (H), and the blue branch represents the outcome “tails” (T): For a single toss, the outcome is either heads or tails. Since we’re looking at the number of heads that occur, the possible values from one toss are either 1 (heads) or 0 (tails).

We can extend the tree diagram to show more than one coin toss. Use the Interactive Activity to see how we construct the diagram. Try several rounds of two, three, and four tosses, and record your outcomes.

Let’s expand our tree diagram to two tosses of a fair coin. Again, each red branch represents the result heads, and each blue branch represents the result tails.

The tree diagrams below illustrate the four possible paths along the branches when you toss a coin twice:

Path 1: First Toss — Heads, Second Toss — Heads (Abbreviated HH): Path 2: First Toss — Heads, Second Toss — Tails (Abbreviated HT): Path 3: First Toss — Tails, Second Toss — Heads (Abbreviated TH): Path 4: First Toss — Tails, Second Toss — Tails (Abbreviated TT):  Video Segment
In this video segment, Professor Kader demonstrates how to construct a tree diagram. As you watch, ask yourself, What does a path on a tree diagram represent? View this segment after you’ve completed the Interactive Activity.
Note: In the experiment conducted by the onscreen participants, participants tried to guess whether dice would land on an even or an odd number. If their guess was correct, the outcome was labeled “C”; if incorrect, the outcome was labeled “I.”

In This Part: Probability Tables

If we assume that the coin is fair, each outcome (heads or tails) of a single toss is equally likely. This probability table summarizes the mathematical probability for the number of heads resulting from one toss of a fair coin: Let’s take a closer look at the tree diagram for two coin tosses. Each red branch represents the result heads (or H). Each blue branch represents the result tails (or T). The outcome associated with each path is indicated at the end of the path, together with the number of heads in that outcome. Since we are tossing a fair coin, each of the four outcomes (HH, HT, TH, TT) is equally likely. See Note 8 below.

Problem C1
Use this tree diagram to explain why the likelihood of getting exactly one head in two coin tosses is not the same as the likelihood of getting zero heads in two coin tosses. What is the probability of each possible outcome? The possible values for the number of heads from two tosses are two (HH), one (HT, TH), or zero (TT).

This probability table summarizes the mathematical probabilities for the number of heads resulting from two tosses of a fair coin: Problem C2
On a piece of paper, draw a tree diagram for three tosses of a fair coin. Label and tally all the possible outcomes as in the previous examples.

Problem C3
Complete the probability table for three tosses of a fair coin: When you toss a fair coin four times, there are 16 possible outcomes (2 x 2 x 2 x 2), and each is equally likely. Here is the tree diagram for four tosses: Problem C4
Complete the probability table for four tosses of a fair coin: Problem C5
Do you think a tree diagram would help you create a similar probability table for 10 tosses of a fair coin? Why or why not?

Problem C5 suggests that we may need to find a pattern to the probability tables in this section in order to use them for problems involving larger numbers.

In This Part: Binomial Experiments
Our coin tosses have been an example of a binomial experiment. A binomial experiment consists of n trials, where each trial is like a coin toss with exactly two possible outcomes. In each trial, the probability for each outcome remains constant.

In the previous section, we used a tree diagram to help us determine one particular outcome of a binomial experiment of n = 4 trials: the number of heads resulting from four tosses of a fair coin. These outcomes can be represented by the table you created in Problem C4: Let’s take a look at the patterns that emerge when you run this binomial experiment several times, each time increasing the number of trials:

One Toss Two Tosses Three Tosses Four Tosses If you display these possible outcomes in the following format, you’ll find that they form what’s known as Pascal’s Triangle: In This Part: Pascal’s Triangle There are two properties that define Pascal’s Triangle:
• Any number on the edge of Pascal’s Triangle is 1.
• Any other number is found by adding the two numbers above it. Video Segment
For example, if you add the 1 and the 3 that start the third row, you get 4, which you place below these two numbers. You’ve now started a fourth row. Then you add 3 and 3 to get 6, and 3 and 1 to get 4, which are the next two values in the fourth row, and so forth.

In this video segment, Professor Kader introduces Pascal’s Triangle. Watch this segment to review the process of generating Pascal’s Triangle. As you watch, keep in mind the following questions:
• What does each number in Pascal’s Triangle represent?
• Why is Pascal’s Triangle useful?

In this fashion, we can use successive rows of Pascal’s Triangle to predict the frequencies for the number of heads in binomial experiments with increasing numbers of trials.

Note: The first number in a row of Pascal’s Triangle corresponds to the frequency of zero heads in a number of coin tosses, not one; therefore, the third number in a row corresponds to the frequency of two, and not three, heads. This is a very common error.

Problem C6
Use the properties of Pascal’s Triangle to generate the fifth and sixth rows.

Problem C7
Use Pascal’s Triangle to determine the frequencies and probabilities for five and six tosses of a fair coin. If you wish, confirm your results with a tree diagram.

TAKE IT FURTHER

Problem C8
Write a formula for determining the number of possible outcomes of n tosses of a fair coin. Think about the tree diagram and start with n = 1 coin toss. Then look at n = 2 tosses. Then look at n = 3 tosses. Note that as you increase the number of tosses by one, you double the number of outcomes. Video Segment
In this video segment, Professor Kader discusses the usefulness of Pascal’s Triangle for determining probabilities. He then asks participants to use Pascal’s Triangle to determine the probabilities for a binomial experiment when n = 5. Watch this segment after completing Problems C7 and C8.

Note that the binomial experiment conducted by the onscreen participants involved predicting the outcomes when rolling a pair of dice.

TAKE IT FURTHER

Problem C9
Extend Pascal’s Triangle to the 10th row. Using the 10th row, determine the probability of tossing exactly five heads out of 10 coin tosses.

Tip: There will be 1,024 total outcomes in the 10th row, so the probability will be the frequency (found in Pascal’s Triangle) divided by 1,024.

We’ve been investigating the binomial probability model. In a random experiment with two possible outcomes, this model can be used to describe the probability of either result. Consider, for example, a True-False test. If a test has four True-False questions, and you make an independent guess on each question, how many will you get correct? (Of course, the only thing you can say for sure is that you will get either zero, one, two, three, or four questions correct!)

Problem C10
Use the binomial probability model to determine the following:
a. What is the most probable score you’ll get?
b. What are the least probable scores you’ll get?
c.
What is the probability of getting at least two answers correct?
d. What is the probability of getting at least three answers correct? Remember that “at least” includes the score itself, so the probability of getting “at least two” answers correct includes two, three, four, and five.

TAKE IT FURTHER

Problem C11
Find the probability of getting at least two questions right on a 10-question True-False test (where you must guess on each question).

### Notes

Note 7
The purpose of Part C is to investigate the idea and use of the binomial probability model.

Part C analyzes the question of how many times you will get heads when you toss a coin a given number of times. The outcome of a coin toss is a random event, so any answer to the question requires the use of probability.

Tree diagrams are used to analyze the coin toss problem. A tree diagram is a useful tool for analysis and also an effective pedagogical device.

Note 8
The tree diagram for two tosses of a fair coin helps to relate the four possible outcomes to the number of heads for each outcome: The key idea is to determine how many outcomes (paths through the tree) contain zero heads, one head, or two heads, respectively.

Some people have great difficulty with the notion that some values occur more often than others. They think that if there are three possible values (zero, one, and two), then each must be equally likely. The tree diagram may clarify the different frequencies of occurrence for different possible outcomes.

### Solutions

Problem C1
Two branches of the tree end with one head out of two tosses (HT and TH), and only one branch ends with zero heads (TT). Therefore, it is more likely to get one head than no heads.

Problem C2
Here is the tree diagram for three tosses of a fair coin: Problem C3
Here is the completed probability table: Problem C4
Here is the completed probability table: Problem C5
No, it would not be feasible to plot the necessary branches. There would be a total of 210= 1,024 branches to this tree diagram, and it would be far too cumbersome to count all the outcomes.

Problem C6
The fifth row is 1, 5, 10, 10, 5, 1, and the sixth row is 1, 6, 15, 20, 15, 6, 1, as shown below: Problem C7
Here are the completed probability tables:
Five Tosses Six Tosses Problem C8
If you say that P = the number of possible outcomes, and n is the number of tosses, then P = 2n.

Problem C9
The seventh row is 1, 7, 21, 35, 35, 21, 7, 1. The eighth row is 1, 8, 28, 56, 70, 56, 28, 8, 1. The ninth row is 1, 9, 36, 84, 126, 126, 84, 36, 9, 1. The 10th row is 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1. The frequency of five heads in 10 coin tosses is the sixth number in this row, which is 252 (note that it is the center number in the row). Since there are 210 = 1,024 possible outcomes in this row, the probability of getting five heads out of 10 tosses is 252/1,024, or about 24.6%.

Problem C10
a.
The most probable score is two correct. It has a probability of 6/16.
b. The least probable scores are zero correct and four correct. Each has a probability of 1/16.
c. The probability of getting at least two answers correct is 6/16 + 4/16 + 1/16 = 11/16.
d. The probability of getting at least three answers correct is 4/16 + 1/16 = 5/16.

Problem C11
The simplest way to approach this problem is to find the probability of getting less than two correct, then subtracting this from one. The probability of getting less than two correct is 1/1,024 + 10/1,024 = 11/1024, so the alternate probability is 1 – 11/1,024 = (1,024/1,024) – 11/1,024 = 1,013/1,024, or approximately 98.9%.