The Binomial Distribution

Many types of probability problems have only two outcomes, or they can be reduced to two outcomes.  For example, when a coin is tossed, it can land heads or tails.  when a baby is born, it will be either male or female.  In a basketball game, a team either wins or loses.  A true-false item can be answered in only two ways, true or false.  Other situations can be reduced to two outcomes.  For example, a medical treatment can be classified as effective or ineffective, depending on the results.  A person can be classified as having normal or abnormal blood pressure, depending on the measure of the blood pressure gauge.  A multiple-choice question, even though there are four or five answer choices, can be classified as correct or incorrect.  Situations like these are called binomial experiments.

A binomial experiment is a probability experiment that satisfies the following four requirements:

1.  Each trial can have only two outcomes or outcomes that can be reduced to two outcomes.  These outcomes can be considered as either success or failure.

2.  There must be a fixed number of trials.

3.  The outcomes of each trial must be independent of each other.

4.  The probability of a success must remain the same for each trial.

    A binomial experiment and its results give rise to a special probability distribution called the binomial distribution.

The outcomes of a binomial experiment and the corresponding probabilities of these outcomes are called a binomial distribution.

    In binomial experiments, the outcomes are usually classified as successes or failures.  For example, the correct answer to a multiple-choice item can be classified as a success, but any of the other choices would be incorrect and hence classified as a failure.  The notation that is commonly used for binomial experiment and the binomial distribution is defined next.


Notation for the Binomial Distribution

P(S)        The symbol for the probability of success

P(F)        The symbol for the probability of failure

p            The numerical probability of a success

q            The numerical probability of a failure

P(S) = p    and     P(F) = 1 - p = q

n            The number of trials

X            The number of successes

Note that 0Xn.


    The probability of a success in a binomial experiment can be computed with the following formula.


Binomial Probability Formula

In a binomial experiment, the probability of exactly X successes in n trials is

An explanation of why the formula works will be given in the following example.


Example 1:

A coin is tossed three times.  Find the probability of getting exactly two heads.


This problem can be solved by looking that the sample space.  There are three ways to get two heads.


The answer is or 0.375.

    Looking at the problem in the previous example from the standpoint of a binomial experiment, one can show that it meets the four requirements.

1.  There are only two outcomes for each trial, heads or tails.

2.  There is a fixed number of trials (three).

3.  The outcomes are independent of each other (the outcome of one toss in no way affects the outcome of another toss).

4.  The probability of a success (heads) is 1/2 in each case.


In this case, n = 3, X = 2, p = 1/2, and q = 1/2.  Hence, substituting in the formula gives

which is the same answer obtained by using the sample space.

    The same example can be used to explain the formula.  First, not that there are three ways to get exactly two heads and one tail from a possible eight ways.  They are HHT, HTH, and THH.  In this case, then, the number of ways of obtaining two heads from three coin tosses is , or 3.  In general, the number of ways to get X successes from n trials with out regard to order is

This is the first part of the binomial formula.  (Some calculators can be used for this.)

    Next each success has a probability of 1/2, and can occur twice.  Likewise, each failure has a probability of 1/2 and can occur once, giving the part of the formula.  To generalize, then, each success has a probability of p and can occur X times, and each failure has a probability of q and can occur (n-X) times.  Putting it all together yields the binomial formula.


    If a student randomly guesses at five multiple-choice questions, find the probability that the student gets exactly three correct.  Each question has five possible choices.



    In this case n = 5, X = 3, and p = 1/5, since there is one chance in five of guessing a correct answer.  Then,


    A survey from Teenage Research Unlimited (Northbrook, Ill.) found that 30% of teenage consumers receive their spending money from part-time jobs.  If five teenagers are selected at random, find the probability that at least three of them will have part-time jobs.



    To find the probability that at least three have a part-time job, it is necessary to find the individual probabilities for either 3, 4, or 5 and then add them to get the total probability.


    P(at least three teenagers have part-time jobs) = 0.132 + 0.028 + 0.002 = 0.162


Mean, Variance, and Standard Deviation for the Binomial Distribution

    The mean, variance, and standard deviation of a variable that has the binomial distribution can be found by using the following formulas.

These formulas are algebraically equivalent to the formulas for the mean, variance, and standard deviation of the variables for probability distributions, but because they are for variables of the binomial distribution, they have been simplified using algebra.  The algebraic derivation is omitted here, but their equivalence is shown in the next example.


    A coin is tossed four times.  Find the mean, variance, and standard deviations of the number of heads that will be obtained.


    With the formulas for the binomial distribution and n = 4, p = 1/2, and q = 1/2, the results are


    From the previous example, when four coins are tossed many, many times, the average of the number of heads that appears is two, and the standard deviation of the number of heads is one.  Note that these are theoretical values.

    As stated previously, this problem can be solved by using the expected value formulas.  The distributions is shown as follows:

No. of heads, X 0 1 2 3 4
Probability, P(X) 1/16 4/16 6/16 4/16 1/16

Hence, the simplified binomial formulas give the same result.