The Binomial Probability Distribution
Criteria for a Binomial Probability Experiment:
1. Experiment is performed a fixed number of times, with each repetition called a
trial.
2. Trials are independent, which means that the outcome of one trial will not
affect the outcome of another trial.
3. For each trial, there are two mutually exclusive outcomes—success or failure.
4. Probability of success is fixed for each trial of the experiment.
Notation Used in the Binomial Probability Distribution:
 n=number of independent trials of the experiment.
 Let p = probability of success and (1-p) = probability of failure.
 X=number of successes in n independent trials of the experiment.
So, 0 < x < n.
EXAMPLE—Identifying Binomial Experiments
Which of the following are binomial experiments?
(a) A player rolls a pair of fair die 10 times. The number X of 7’s rolled is recorded.
(b) The 11 largest airlines had an on-time percentage of 84.7% in November, 2001,
according to the Air Travel Consumer Report. In order to assess reasons for delays, an
official with the FAA randomly selects flights until she finds 10 that were not on time.
The number of flights X that need to be selected is recorded.
(c) In a class of 30 students, 55% are female. The instructor randomly selects 4 students.
The number X of females selected is recorded.
Example: Binomial Probability Distribution for Coin Toss
Problem: A fair coin is tossed twice and the number of heads, X, is recorded. Construct a
probability distribution for the random variable X.
1
sol
H1
H2
H1H2
=  
T2
H1T2
=  
H2
T1H2
=  
T2
T1T2
=  
1
2
1
2
1
4
1
2
1
2
1
4
1
2
1
2
1
4
1
2
1
2
1
4
2nd Toss
1st Toss
T1
2nd Toss
Let X be the number of heads.
1 1 1
 
2 2 4
P( X  1)  P( H 1T 2)orP (T 1H 2)  P( H 1T 2)  P(T 1H 2) 
P( X  0)  P(T 1T 2)  P(T 1)  P(T 2) 
P( H 1)  P(T 2)  P(T 1)  P( H 2) 
Probability Distribution of X:
x
0
1
2
1 1 1 1 1 1 1
     
2 2 2 2 4 4 2
P(X=x)
0.25
0.50
0.25
Binomial Probability Distribution Function:
The probability of obtaining x successes in n independent trials of a binomial
experiment where the probability of success is p is given by
P( X  x )  C x p x (1  p ) n  x , x  0,1,2,....., n
n
Apply the Binomial Probability Distribution Function to the coin toss problem on the
previous page with n=2 and p=1/2.
2
C
x
0
n
x

2!
Px
x! ( 2  x )!
1
0.50
=1.00
1
2
0.51
=0.50
2
1
0.52  0.25
(1  P) 2 x
P( X  x )  C x P x (1  P ) n  x
n
(1  0.50) 2  0.25 P( X  0)  C 2 (0.5) 0 (1  0.5) 2  1  1  0.25  0.25
0
(1  0.50)1  0.50 P( X  1)  C 2 (0.5)1 (1  0.5)1  2  0.5  0.5  0.5
1
(1  0.50) 0  1.00
P( X  2)  C 2 (0.5) 2 (1  0.5) 0  1  0.25  1  0.25
2
Example(2):
A test consists of 10 multiple choice questions with five choices for each
question. As an experiment, you GUESS on each and every answer without even reading
the questions.
What is the probability of getting exactly 6 questions correct on this test?
Solution
n = 10
r=6
n–r=4
p = 0.20 = probability of guessing the correct answer on a question
q = 1 - p = 0.80 = probability of not guessing the correct answer on a question
Example(3):
When rolling a die 100 times, what is the probability of rolling a "4" exactly 25 times?
Solution
n = 100
r = 25
n – r = 75
p = 1/6 = probability of rolling a "4"
q = 1 - p = 5/6 = probability of not rolling a "4"
3
Example(4):
At a certain intersection, the light for eastbound traffic is red for 15 seconds, yellow for 5
seconds, and green for 30 seconds. Find the probability that out of the next eight
eastbound cars that arrive randomly at the light, exactly three will be stopped by a red
light.
Solution
n=8
r=3
n–r=5
p = 15/50 = probability of a red light
q = 1 - p = 35/50 = probability of not a red light
Example(5)
Binomial Cumulative Distributions
When dealing with a binomial setting, you are often asked to find probabilities
dealing with “greater than” and “less than” situations. These are handled easily on
your calculator as long as you know what the calculator is doing and input the
correct information. Using the binomial cumulative distribution function of the
calculator means that you are summing probabilities from 0 to x. So if a problem
asks for the probability of 6 or fewer things happening, you need to have the
calculator sum P(0) + P(1) + . . . + P(6).
However, if a problem asks you to find the probability of more than 5 things
happening, this cannot be input directly into the calculator without some
4
preliminary work. P(more than 5) = P(5) + P(6) + . . . + P(n) where n is the number of
observations in the problem. But remember that the calculator can only computer
probabilities less than or equal to a certain value. So this situation becomes
1 - [P(0) + P(1) + P(2) + P(3) + P(4)] which is input in the calculator’s binomial
cumulative distribution screen with an x value of 4. Then make sure to subtract
the value you get from 1. It is always helpful to write out what you have to
compute first BEFORE jumping to the calculator.
Problem:
1. A basketball player has a free throw percentage of 75%. In a single game, the
player shoots 12 free throws. Compute the following probabilities.
a. What is the probability that he makes less than 9?
b. What is the probability that he makes at least 8?
c. What is the probability that he makes between 7 and 10?
This is a binomial situation so we can proceed with the binomial cumulative
distribution computations. However, let’s figure out “x” in each case before we
begin.
In part (a) we need P(# make < 9) which is P(0) + P(1) + . . . P(8) so x = 8. For part
(b) we are looking for P(at least 8) which means P(8) +P(9) +P(10) +P(11) + P(12).
This means we need to use x = 7 in the calculator and subtract our answer from 1.
Part (c) needs P(8) +P(9) so we have to use x = 7 and x = 10 and subtract the two
Example:
If X is binomially distributed with 6 trails and probability of success equal to
each attempt , what is the probability of :
a) Exactly 4 successes
b) At least one success.
Solution
1
4
3
4
1
9
135
 
 0.033
256 16 4096
3
729
 0.822
b) P(X  1) = 1  P( X  0)  1  ( ) 6  1 
4
4096
a) P( X  4)  C 4 ( ) 4 ( ) 2  15 
6
Example:
When an unbiased coin tossed 8 times what is the probability of obtaining:
a) Less than four heads.
b) More than 5 heads.
5
1
at
4
Solution
Let H= the number of heads
a) P(H  4)  P( H  3)  0.3633
Alternative solution
P(H  4)  P( H  0)  P( H  1)  P( H  2)  P( H  3)
1
1
1
8 1
8 1
2
2 2
2 2
b) P(H  5)  1  P( H  5)  1  0.8555  0.1445
1
2
1
2
= ( ) 8  C 1 ( )1 ( ) 7  C 2 ( ) 2 ( ) 6  C 3 ( ) 3 ( ) 5 
Or
8
P( H  5)  P( H  6)  P( H  7)  P( H  8) 
37
8 1 6 1 2
8 1 7 1 1
8 1 8
C 6 ( 2 ) ( 2 )  C 7 ( 2 ) ( 2 )  C8 ( 2 )  256  0.1445
6
93
 0.3633
256