The Binomial Theorem
Binomial Theorem: For any two real numbers x and y, and for any positive integer n, we have:
Proof: (by induction on n)
Basis case:
For n = 1.
The left side for n = 1 is simply x + y.
The right side is
Induction case:
Assume the formula works for n = k and show it works for n = k + 1.
Assume that:
Show that :
Using the induction assumption, we can rewrite the left side of this equation as follows.
Using Pascal’s identity , we get that the right side of what we are trying to show can be expanded as follows. Note that we need to be careful since Pascal’s identity does not apply for k=0 or for k=n. In our sum, this means we need to split out the j=0 and j=k+1 terms before applying Pascal’s identity.
Now we have the problem of matching the two sums from the left side with the two from the right side. First, notice that the out front on the right side can be thought of as the j = 0 term in the first sum. Next, the out front on the right side can be thought of as the j = k+1 term in the second sum. This gives that the right side is equal to the following.
From above, .
The first sums are clearly equal since the only difference is that the exponent of x is changed from k+1-j to k-j+1. All that is left is to show that the second sums are equal.
To show the second sums are the same, take the second sum from the right side and change the summation index j to something that starts at zero. Keeping the same name is allowed, but it is easier to follow if we pick a different letter, so let q=j-1. This gives j=q+1 and 1-j=-q. Also, since j goes from 1 to k+1, q will go from 0 to k. Substituting these into the second sum for the right side gives the following.
This is the same as the second sum on the left side since the name of the summation index is the only difference between the two. Q. E. D.
Binomial Experiments
Which of the following are binomial experiments? Can any that are not be modified so that they will become binomial experiments?
a) Richard has just been given a 10-question multiple-choice quiz in his history class. Each question has 5 possible responses, only one of which is correct. Since Richard has not attended class recently, he does not know any of the answers. Assuming that Richard guesses on all 10 questions, is this a binomial experiment?
i)The experiment consists of 10 trials.
ii)The trials are identical to each other; each trial consists of Richard’s randomly guessing the answer to one of the questions.
iii)The trials are independent of each other due to random guessing – whether his guess is correct on one question gives no information about whether his guess is correct on any other question.
iv)Each trial has two possible outcomes: Success = {Richard guesses the correct answer to the question} or Failure = {Richard does not guess the correct answer to the question}.
v)P(Success) = 0.20 for each trial, since Richard is randomly picking one of the 5 possible responses to each question.
Yes, this is a binomial experiment, with n = 10 and p = 0.20.
b) One hundred consumers are surveyed to determine whether they like Sudsy Soap.
i)The experiment consists of 100 trials.
ii)The trials are identical to each other; each trial consists of asking one consumer whether he/she likes Sudsy Soap.
iii)We cannot say that the trials are independent of each other. Why?
iv)Each trial has two possible outcomes: Success = {the consumer says that he/she does like Sudsy Soap} or Failure = {the consumer says that he/she does not like Sudsy Soap}.
v)We cannot say that the probability of success is the same for each trial. Why?
How could this experiment be modified to make it a binomial experiment?
c) A card is drawn from a well-shuffled deck. We are betting on whether the card is a heart.
i)The experiment consists of one trial.
ii)The trial consists of drawing a card from a well-shuffled deck.
iii)Independence is not an issue in this case, since there is only a single trial.
iv)The trial has two possible outcomes: Success = {card drawn is a heart} or Failure = {card drawn is not a heart}.
v)P(Success) = 0.25 for the single trial, since the deck is well-shuffled.
Yes, this is a binomial experiment, with n = 1 and p = 0.25.
Binomial Distribution, Example 1
Assume that today is October 15, 2008. We want to conduct a poll to predict the outcome of the 2008 Presidential election. We will assume that there are only two candidates running, Senator John McCain of Arizona and Senator Barack Obama of Illinois. We will randomly select a simple random sample of size 1068 from the population of eligible, registered, and likely voters (There were over 125 million people who actually voted in the 2008 Presidential election). We will ask the preference of each voter. According to an earlier poll, if the race were between Senator McCain and Senator Obama, 46% of the voters would support Senator McCain, and 53% would support Senator Obama. Assume that the level of support for Senator Obama is actually 53%, and let X be the number of voters in the sample who say they will vote for Senator Obama.
Is this a binomial experiment? Why or why not?
i)The experiment consists of 1068 trials.
ii)The trials are identical to each other; each trial consists of randomly selecting a voter and asking, “Do you support Senator Obama for President?”
iii)The trials are independent of each other, due to random sampling.
iv)Each trial has two possible outcomes: Success = {voter says he/she supports Senator Obama} or Failure = {voter says he/she supports Senator McCain}.
v)P(Success) = 0.53 for each trial, due to random sampling.
Then X ~ Binomial( n = 1068, p = 0.53).
How likely is it that exactly half of the voters in the sample will say they support Senator Obama?
.
How likely is it that at least 50% of the voters in the sample will say they support Senator Obama?
.
What is the expected (mean) number of supporters of Senator Obama in our sample?
Since X ~ Binomial(n = 1068, p = 0.53), then
.
The standard deviation of the distribution is
.
If we subtract 2 standard deviations from the mean, we get
,
which is just below 50% of the sample. In particular, two standard deviations of this binomial distribution represents , or about 3% of the sample. Does this number look familiar?
Binomial Distribution, Example 2
Assume that the population of interest is the population of all adult (at least 18 years old) males in the U. S. Assume that a certain proportion, p, of this population are over 6 feet tall, where
0 p 1. We will select a random sample of 10 people from this population, and ask each person in the sample, “Are you over 6 feet tall?”
Is this a binomial experiment? Why or why not?
i)The experiment consists of 10 trials.
ii)The trials are identical to each other; each trial consists of randomly selecting a member of the population of all adult males in the U.S. and finding that person’s height.
iii)The trials are independent of each other, due to random sampling.
iv)Each trial has two possible outcomes: Success = {person selected says he is over 6 feet tall} or Failure = {person selected says he is not over 6 feet tall}.
v)P(Success) = p for each trial, due to random sampling.
Let X = number of people in the sample who are over 6 feet tall. Let us assume, for convenience, that p = 0.30. Then X ~ Binomial( n = 10, p = 0.30).
How likely is it that exactly 3 of the men in the sample will say they are over 6 feet tall.
.
How likely is it that at least 50% of the men in the sample will say they are over 6 feet tall?
.
What is the expected (mean) number of men over 6 feet tall in the sample?
Since X ~ Binomial(n = 10, p = 0.30), then
.
Interpretation of the mean: The random experiment is to select a random sample of 10 men from the population of all U.S. adult males, and ask each one whether he is over 6 feet tall. If I repeat this experiment very many times, each time counting the number of “Yes” responses, I will find that the average of the counts will get closer and closer to 3.
The variance of the distribution is
.
The standard deviation of the distribution is
.