Binomial Probability Distribution Web Page
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Begin Text Explaining the Binomial Probability Distribution
In the last lecture we studied the Normal Probability Distribution. The Binomial is another very important probability distribution which is deeply involved in the theory of statistics. We will begin this topic by reviewing what we mean by a Bernoulli Trial. We touched on Bernoulli Trials previously in the Probability Lecture. It's an idea which is not generally familiar to people and it involves some jargon, so it's worth going through again.
Abduction: It's a girl. Let's take an infinite process, the birth of a human child. There are many dependent measurement operations that are commonly applied to a birth. We could measure the weight at birth or the length of the child in inches. We could measure the number of hours of labor, and so on. There's an infinite number of things we could measure about a birth. None of them nor all of them together, of course, would fully capture the experience of the woman or child during birth.
One operation we seem to be utterly fascinated with is to categorize the child by gender. So one of the things we do, both conversationally and in research, is reduce the person to their gender, male or female.
Once we have categorize the birth by gender, we can model gender in terms of probability. We say things like "the probability of a girl is .5."
Abduction: It's a head. Another thing that we like to do is flip coins. As we've noted before flipping a coin is an infinite process. We could measure the amount time the coin is in the air, we could count the number of times the coin turns over, we could make any number of measurements about this process. Of course we generally just look at the side facing up when the coin comes to rest. By convention we name the two sides Heads and Tails. So we reduce the process of a coin flip to one of two categories: head or tail.
As with birth, we model the result of the coin flip in terms of probability. We say things like "the probability of a head is .5."
Bernoulli Trials. There are processes in nature which are reduced by human operations to two categories and then modeled in terms of probability. This is what we mean by a Bernoulli Trial. A Bernoulli Trial is a process that can have only two outcomes.
Now, as I keep stressing, most processes are infinite and the two outcomes are a result of our measurement operation. So obviously our child at birth is an infinite process, and the two outcomes exist only because we choose to talk about or do research on gender. This is fine; but for both human and scientific reasons, it is important to remember the infinity behind the operations.
Success and Failure. Traditionally in the jargon of probability theory one of the two possible outcomes of a Bernoulli Trial is called a success and the other one a failure. There's no value judgment implied by the use of these words in the probability context. Success and failure are being used as conventions. They are merely names which indicate the two outcomes of the Bernoulli Trial. Whatever you decide to call a success and a failure is completely arbitrary.
A head could be called a success. In that case we would say that the probability of a success is .5. Of course that makes a tail a failure. And the probability of a failure would also be .5.
p and q. The probability of a success is typically denoted by a small p. And the probability of a failure is denoted by a small q.
Okay, so probability of a success is equal to p, probability of a failure is equal to q. That's the main information here.
Since there are two outcomes, we might as well notice that p plus q must be equal to 1 (because the total probability in any system must be 1). [Remember that the sure event in flipping a coin is that you're going to get a head or a tail, and the probability of the sure event is 1. So p plus q must be equal to 1.]
That means that if you have p you can find q because q will be 1 - p.
We should note that p and q do not always have to be .5. We just happen to have used two examples where p and q are both .5. But they can be other probabilities. In general, p might very well be .8, and then q would be .2.
Now we're going to define the binomial distribution.
Binomial Distribution
N Independent Bernoulli Trials. To define a binomial distribution, we need two things. First, we need some number, N, independent Bernoulli trials. By independent I mean events that are not correlated with each other. When we flip two coins, the outcome of one flip has no effect on the outcome of the other. If the first coin is a head, the probability of a head on the second coin is still .5. There is no connection or correlation between the results of the two flips. So flipping two coins can be thought of as 2 Bernoulli Trials. The outcome of the second trial is independent of the outcome of the first trial. In this case N = 2.
Flipping 5 coins can be thought of as 5 Bernoulli Trials. The result of each flip is independent of the results of the other flips. In this case N = 5. If we flip 10 coins then N = 10 because we have 10 Bernoulli Trials.
r Successes in N Trials. Second, we need to count some number of successes in the N Bernoulli Trials. We use r as the symbol for the number of successes. Suppose we flip 10 coins and call a head a success; then N = 10. We might wonder what the chances are of getting 8 successes (heads) in 10 flips. In this case r = 8 and N =10. We might wonder what the chances are of getting 4 successes in 10 flips. In this case r = 4 and N = 10.
Let's say that there is a neighborhood in which, of 20 children, 11 are girls. We might wonder what the probability of getting 11 girls in 20 births is. Here r = 11 and N = 20.
Binomial Probability Distribution. When you have a question that asks the probability of a certain number of successes (r) in a certain number (N) of Bernoulli trials then that probability can be calculated from what's called the Binomial Distribution.
The Binomial gives you the probability of r successes in N trials.
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P(r; p, N). To work with the Binomial we must specify three things: r, p, and N. We must know how many successes we are interested in; we must know the probability of a success: and we must know how many trials we are talking about. The symbols for these three are, of course, r, p, and N. The standard notation for expressing the probability of r successes in N trials with p as probability of a success is P(r: p, N).
For example, suppose we have 8 independent births and we define a success as a girl. Suppose p = .5. We might want to know the probability of 4 girls (successes) in 8 births. This would be expressed as P(4; .5, 8). The probability of 11 girls in 20 births would be P(11; .5, 20).
The standard notation allows you to write a big sentence in a little symbol.
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Between. Just as with the Normal Distribution, with the Binomial we will distinguish "probabilities between values" from "probabilities outside values." Suppose N = 8, p = .5 and we want to know what the chances are of getting between 2 and 5 girls in 8 births.
The convention is that "between" is inclusive. When I say between 2 and 5 girls I mean 2 or 3 or 4 or 5 girls. Both 2 and 5 are included in the potential number of successes.
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Outside. Conversely, the convention is that the probability of outside 2 and 5 girls excludes 2 and 5. If we have N = 8 births then outside 2 and 5 means 1 or 6 or 7 or 8 births. Both 2 and 5 are excluded.
Between is inclusive. Outside is exclusive.
The last graphic on this topic shows how the Binomial Distribution looks when drawn by StatCenter's Binomial Tool. We will learn how to use that tool in the next topic.
Notice that the number of successes (r) runs along the horizontal axis. The probability for each number of successes goes up the vertical axis--the higher the black area the higher the probability. And, circled in green, in the top left corner, the standard notation appears. In this case it says P(r; p = 0.5, N = 10).
In the example shown in the graphic N = 10 and p =.5. The number of successes (horizontal axis) runs from 0 to 10. This of course is all the possible numbers of successes we could get in 10 trials. Notice that the probability of 5 success is relatively high compared to, say, the probability of 2 successes. If that's not obvious yet (or you can't read the graphic very well), that's OK. We'll practice a lot with this distribution.
Suppose we flip a fair coin 8 times. Suppose we define a success as a head. Suppose also, the coin is fair, p = .5. The Binomial Distribution allows us to answer questions like what's the probability of 4 heads in 8 flips. Or, what's the probability of between 2 and 5 heads in 8 flips. Or, what's the probability of getting outside 2 and 5 heads in 8 flips. Use StatCenter's Binomial Tool to find these probabilities.
Finding "Binomial Tool". You can find this tool from either the Desk, the Ducks, or the Course Menu interfaces. From the Desk click on the Interactive Learning icon and look for Binomial Tool. From Ducks, just click on the Binomial Sample Tool link under the Binomial Distribution Lecture. And from the Menu, just open the Work and Learn folder and click on Interactive Learning and choose Binomial Tool.
Click on the middle button (Probability Tools for Binomial Distribution). The Binomial Tool will appear. Alternate between the Binomial Tool and this web page and, perhaps, the Authorware program. Best to minimize or close all other programs except these three. For example, if StatCenter's Main Menu is on the screen, minimize it.
Now let's use the Binomial Tool. Suppose we flip a fair coin 8 times. Define a head as a success, with p = .5. What is the probability of r = 4 successes in 8 trials when p = .5?
Set N. The graphic points where to set the number of trials, N. In this case enter 8, and click the Enter N button.
Set p. The graphic also points out where to set the probability of a success, p. In this case p should already read .5. If not, enter .5 and click the Enter Probability button.
Set Between. Click on the "Between Icon."
Set upper and lower scores. In this case we want to know the probability of exactly r = 4 successes in 8 trials. So we will set both the upper and the lower score to 4. Remember that between is inclusive. So if the both the lower and upper score are set to 4, the probability will include 4 (and only 4). This may seem a bit odd at first, but it works.
Read the Probability. The small white window in the lower right corner should read 0.2734. This is the probability of 4 successes in 8 trials when p = .5. Another way to write this is P(4; .5, 8) = 0.2734.
Black Area. As with the Normal Distribution, with the Binomial the black area represents the relevant probability.
That's all you have to do. After your experience with the Normal Tool, this procedure should seem familiar.
Practice. What is the probability of exactly 3 successes in 8 trials with p = .5? (Answer: .2188.) What is the probability of 3 successes in 8 trials when p = .4? (Answer: .2787.) What is P(4; .4, 8)? (Answer: .2322. You have to change the upper value first.) What is P(12; .8, 20)? (Answer: .0222.)
Now let's answer another kind of question. What's the probability of getting between 2 and 5 heads in 8 flips of a fair coin? Here again, you use the binomial tool.
Input information given in the question. Set N = 8, set p = .5, set the lower value of r =2, and set the upper value of r=5. Make sure the Between Icon is clicked.
Then simply read the probability which is .8203. The probability will again be represented by the black area. As usual with both the Normal and Binomial distributions, we're interpreting probability as an area.
Practice. What is the probability of getting between 0 and 3 heads in 8 flips of a fair coin? (Answer: .3633.) What is the probability of getting between 0 and 8 heads in 8 flips of a fair coin? (Answer: 1. Between 0 and 8 successes in 8 trials includes all possible outcomes. And all possible outcomes must have a probability of 1.) What is probability of between 3 and 14 heads in 20 flips of a fair coin? (Answer: .9791.) What is the probability of getting between 3 and 14 successes if p = .8? (Answer: .1958.)
Now let's find the probability of getting outside 2 and 5 heads in 8 tosses of a fair coin. This will be the same of the example we just finished, of course, except that you click on the Outside Icon instead of the Between Icon.
Input information from the question. Click on the Outside Icon. Enter p = .5. Enter N = 8. Enter lower and upper scores of 2 and 5. Remember that "outside" is exclusive and does not include 2 and 5.
Read the probability. In the probability output window in the lower right corner you will find the answer. The probability of getting outside 2 and 5 heads is .1797.
Black Area. Once again the black area represents probability. Click back and forth between the Outside Icon and the Between Icon so you can see the relationship between them both with the black area and the probability. Notice that the probability outside plus the probability between equals 1. That is, .8203+.1797=1.
Practice. What is the probability of getting outside 6 and 8 successes in 12 trials if p = .6? (Answer: .3835.) What is the probability of getting between 6 and 8 successes in 12 trials if p = .6: (Answer: .6165.)
Play with the Binomial Tool, entering various different parameters to discover what happens.
Catching Cold. Let's say that the probability of catching a cold in Salt Lake City in January and February is known to be .5. In this example we're keeping the probability at .5 so that it's like the familiar flip of a coin. Suppose also that some researchers develop a cold vaccine and therefore are on the threshold of becoming famous and perhaps rich. But before they get famous, they will have to convince people that their vaccine works. One way to convince them is to do some research. They run a study with 10 volunteers to test whether the vaccine is effective or not. They administer the vaccine to the 10 volunteers and then determine if each volunteer gets a cold or not during January and February. If the vaccine works, the chances of catching cold among the volunteers should be less than .5.
Science. The scientific hypothesis is that the vaccine will improve health (that is, it will reduce the chances of a cold). The independent variable (IV) then is the vaccine, and the dependent variable (DV) is health, which they will measure by categorizing whether or not each volunteer gets a cold.
No Effect. The scientists expect the vaccine (IV) will affect the DV (health). But for the moment we're going to assume that (unknown to researchers) the vaccine is completely ineffective. Despite the researchers' belief that it's effective, it's actually worthless. Therefore the vaccine is going to have no effect on the chances of getting a cold. Therefore the 10 volunteers have the same probability of catching a cold as anyone else. P(Cold) = .5. That's the set up.
Abduction. We take volunteers and we reduce them via measurement operations to a variable, call it X. X will be equal to 0 if they don't get a cold during January and February. X will equal 1 if they do get a cold. During January and February each volunteer is going to have a lot of experiences. But we've reduced this person to a 0 or a 1. Perhaps the person will sign up for a really bad class and wish that their life wasn't ruined by it, or maybe he or she will meet someone really interesting. There's all kinds of things that are going to happen during January and February. But we've reduced her or him down to 0 and 1--get a cold and don't get a cold. That's a massive reduction but of some use for the question we're asking about vaccines.
Once we have X defined we can model it as a Bernoulli trial. We'll call getting a cold a success and not getting a cold a failure. (Remember, success and failure are just arbitrary names.) The probability of a success, p, is .5 because we are assuming that the vaccine is ineffective.
10 Bernoulli Trials. Now each person in the study has been modeled as a Bernoulli trial. Since we have 10 volunteers we have 10 Bernoulli trials with p = .5.
Binomial Distribution. If we want to know the probability of a certain number of successes (colds) in 10 volunteers we can use the Binomial Distribution.
Questions. Now we can answer questions like what are the chances of 0 colds in our sample of 10 subjects? That is, what is that probability that none of our subjects will get a cold? Or what are the chances that just 0 or 1 subject will get a cold? Or what are the chances of 0, 1, or 2 colds in our 10 subjects?
To answer the questions, open up the Binomial Tool. Set N = 10 because there are 10 subjects. Enter .5 as a probability of success.
If you want to know the probability of 0 colds in our 10 volunteers, you click the Between Icon, and enter 0 as both the lower and upper values. Read the probability which is .0010 which is .001 or one in a thousand. There is one chance in a thousand that you would get no colds out of 10 people in this two month period, assuming that the probability of a cold is one half.
That's the same probability as flipping no heads in ten flips of a fair coin. It could happen but the probability is low, just one in a thousand.
The next question is what is the probability of 0 or 1 colds in our 10 volunteers? Keep all other information the same and enter 1 as the upper value. The probability is .0107, which is pretty close to one in a hundred. There's about one in a hundred chance that we would get either 0 or 1 colds in our sample of 10 volunteers.
What's the probability of 0 or 1 or 2 colds in our sample? Just enter 2 as the upper value and read the probability, which is .0547. So the chances of getting 0 or 1 or 2 colds in this group of 10 subjects is .0547.
Wrapping up the Vaccine example. That's all we have to say about the Vaccine example for now. We will return to it later in the Hypothesis Testing lecture. Make sure you understand it well, because we will build on it at that time.
Normal approximation. An interesting proof in probability theory is that as N becomes large, and for mathematicians becoming large means approaching infinity, the shape of the binomial approaches the shape of the normal probability distribution. In the limiting case of an infinite number of Bernoulli trials, when N is infinite, there's no difference between the normal distribution and the binomial distribution. They are both normal.
In the current graphic I've set p = .5, and N equal 100. You can see that with N merely equal to a hundred (which is far below infinity) the binomial is giving us an approximation of what the normal distribution looks like.
Practice. Set p = .5 and N = 250. Look at the resulting Binomial Distribution. Is it even smoother and more like the Normal Distribution than when N was 100?