Binomial Probability Distribution
To understand binomial distributions and binomial probability, it helps to
understand binomial experiments and some associated notation; so we
cover those topics first.
Binomial Experiment
A binomial experiment
(also known as a Bernoulli trial)
is a
statistical experiment that has the following properties:

The trials are independent;
that is, the outcome on one trial does not affect the outcome on other trials.
Consider the following statistical experiment. You flip a coin 2 times and count
the number of times the coin lands on heads. This is a binomial experiment
because:
 The trials are independent; that is, getting heads on one trial does not affect
whether we get heads on other trials.
Notation
The following notation is helpful, when we talk about binomial probability.
 _{n}C_{r}: The number of
combinations of n things, taken r at a time.
Binomial Distribution
A binomial random variable is the number of successes x in
n repeated trials of a binomial experiment. The
probability distribution of a binomial random variable is called
a binomial distribution
(also known as a Bernoulli distribution).
Suppose we flip a coin two times and count the number of heads (successes). The
binomial random variable is the number of heads, which can take on values of 0,
1, or 2. The binomial distribution is presented below.
Number of heads

Probability 
0

0.25 
1

0.50 
2

0.25 
The binomial distribution has the following properties:
Binomial Formula and Binomial Probability
The binomial probability refers to the probability that a
binomial experiment results in exactly x successes. For example,
in the above table, we see that the binomial probability of getting exactly one
head in two coin flips is 0.50.
Given x, n, and P, we can compute the binomial probability
based on the binomial formula:
Binomial Formula. Suppose a binomial
experiment consists of
n trials and results in
x successes. If
the probability of success on an individual trial is
P, then the
binomial probability is:
b(x; n, P) = _{n}C_{x} * P^{x} * (1  P)^{n  x}
or
b(x; n, P) = { n! / [ x! (n  x)! ] } * P^{x} * (1  P)^{n  x}
Example 1
Suppose a die is tossed 5 times. What is the probability of getting exactly 2
fours?
Solution: This is a binomial experiment in which the number of trials is
equal to 5, the number of successes is equal to 2, and the probability of
success on a single trial is 1/6 or about 0.167. Therefore, the binomial
probability is:
b(2; 5, 0.167) = _{5}C_{2} * (0.167)^{2} * (0.833)^{3}
b(2; 5, 0.167) = 0.161
Cumulative Binomial Probability
A cumulative binomial probability refers to the probability
that the binomial random variable falls within a specified range (e.g.,
is greater than or equal to a stated lower limit and less than or
equal to a stated upper limit).
For example, we might be interested in the cumulative binomial probability of
obtaining 45 or fewer heads in 100 tosses of a coin (see Example 1 below). This
would be the sum of all these individual binomial probabilities.
b(x < 45; 100, 0.5) =
b(x = 0; 100, 0.5) + b(x = 1; 100, 0.5) + ... +
b(x = 44; 100, 0.5) + b(x = 45; 100, 0.5)
Binomial
Calculator
As you may have noticed, the binomial formula requires many timeconsuming
computations. The Binomial Calculator can do this work for you  quickly,
easily, and errorfree. Use the Binomial Calculator to compute binomial
probabilities and cumulative binomial probabilities. The calculator is free. It
can be found under the Stat Tables tab, which appears in
the header of every Stat Trek web page.
Example 1
What is the probability of obtaining 45 or fewer heads in 100 tosses of a coin?
Solution: To solve this problem, we compute 46 individual probabilities,
using the binomial formula. The sum of all these probabilities is the answer we
seek. Thus,
b(x < 45; 100, 0.5) = b(x = 0; 100, 0.5) + b(x = 1; 100, 0.5) + . . .
+ b(x = 45; 100, 0.5)
b(x < 45; 100, 0.5) = 0.184
Example 2
The probability that a student is accepted to a prestigious college is 0.3. If
5 students from the same school apply, what is the probability that at most 2
are accepted?
Solution: To solve this problem, we compute 3 individual probabilities,
using the binomial formula. The sum of all these probabilities is the answer we
seek. Thus,
b(x < 2; 5, 0.3) = b(x = 0; 5, 0.3) + b(x = 1; 5, 0.3) + b(x = 2; 5,
0.3)
b(x < 2; 5, 0.3) = 0.1681 + 0.3601 + 0.3087
b(x < 2; 5, 0.3) = 0.8369
Example 3
What is the probability that the world series will last 4 games? 5 games?
6 games? 7 games? Assume that the teams are evenly matched.
Solution: This is a very tricky application of the binomial
distribution. If you can follow the logic of this solution, you have
a good understanding of the material covered in the tutorial, to this
point.
In the world series, there are two baseball teams. The series
ends when the winning team wins 4 games. Therefore, we define a success as a
win by the team that ultimately becomes the world series champion.
For the purpose of this analysis, we assume that the teams are evenly matched.
Therefore, the probability that a particular team wins a particular game is
0.5.
Let's look first at the simplest case. What is the probability that the series
lasts only 4 games. This can occur if one team wins the first 4
games. The probability of the National League team winning 4 games
in a row is:
b(4; 4, 0.5) =
_{4}C_{4} * (0.5)^{4} * (0.5)^{0} = 0.0625
Similarly, when we compute the probability of the American League team
winning 4 games in a row, we find that it is also 0.0625.
Therefore, probability that the series ends in
four games would be 0.0625 + 0.0625 = 0.125; since the series would end if
either the American or National League team won 4 games in a row.
Now let's tackle the question of finding probability that the world series
ends in 5 games. The trick in finding this solution is to recognize that
the series can only end in 5 games, if one team has
won 3 out of the first 4 games. So let's first find the probability
that the American League team wins exactly 3 of the first 4 games.
b(3; 4, 0.5) =
_{4}C_{3} * (0.5)^{3} * (0.5)^{1} = 0.25
Okay, here comes some more tricky stuff, so listen up. Given that the
American League team has won 3 of the first 4 games, the American League team
has a 50/50 chance of winning the fifth game to end the series.
Therefore, the probability of the American League team winning the
series in 5 games is 0.25 * 0.50 = 0.125. Since the National League
team could also win the series in 5 games, the probability that
the series ends in 5 games would be 0.125 + 0.125 = 0.25.
The rest of the problem would be solved in the same way. You should find
that the probability of the series ending in 6 games is 0.3125; and
the probability of the series ending in 7 games is also 0.3125.