Section 5-3
Binomial
Probability
Distributions
Binomial Probability Distribution
A binomial probability distribution results from a
procedure that meets all the following requirements:
1. The procedure has a fixed number of trials.
2. The trials must be independent. (The outcome of any
individual trial doesn’t affect the probabilities in the
other trials.)
3. Each trial must have all outcomes classified into two
categories (commonly referred to as success and
failure).
4. The probability of a success remains the same in all
trials.
Notation for Binomial Probability Distributions
S and F (success and failure) denote the two possible
categories of all outcomes; p and q will denote the
probabilities of S and F, respectively, so
P(S) = p
(p = probability of success)
P(F) = 1 – p = q
(q = probability of failure)
Notation (continued)
n
denotes the fixed number of trials.
x
denotes a specific number of successes in n trials, so x
can be any whole number between 0 and n, inclusive.
p
denotes the probability of success in one of the n trials.
q
denotes the probability of failure in one of the n trials.
P(x)
denotes the probability of getting exactly x successes
among the n trials.
Example 1: Determine whether or not the given procedure
results in a binomial distribution. If it does not binomial,
identify at least one requirement that is not satisfied.
Treating 863 subjects with Lipitor (Atorvastatin) and
recording whether there is a “yes” response when they are
asked if they experienced a headache (based on data from
Pfizer, Inc.)
1. The procedure has a fixed number of trials.
Yes, all four
requirements are met.
2. The trials must be independent. (The outcome of any
individual trial doesn’t affect the probabilities in the other
trials.)
3. Each trial must have all outcomes classified into two
categories (commonly referred to as success and failure).
4. The probability of a success remains the same in all trials.
Example 2: Determine whether or not the given procedure
results in a binomial distribution. If it does not binomial,
identify at least one requirement that is not satisfied.
Treating 152 couples with YSORT gender selection
method developed by the Genetics & IVF Institute and
recording the ages of the parents.
1. The procedure has a fixed number of trials.
No, there are more than
2. The trials must be independent. (The outcome of any
individual trial doesn’t affect the probabilities in the other
trials.)
two possible outcomes
3. Each trial must have all outcomes classified into two
categories (commonly referred to as success and failure).
for the ages of the parents.
4. The probability of a success remains the same in all trials.
Important Hints
 Be sure that x and p both refer to the same
category being called a success.
 When sampling without replacement, consider
events to be independent if n < 0.05N.
Methods for Finding
Probabilities
We will now discuss three methods for finding
the probabilities corresponding to the random
variable x in a binomial distribution.
Method 1: Using the Binomial
Probability Formula
n!
P( x) 
 p x q n x
 n  x ! x !
for x = 0, 1, 2, . . ., n
where
n = number of trials
x = number of successes among n trials
p = probability of success in any one trial
q = probability of failure in any one trial (q = 1 – p)
Binomial Probability Formula
n!
x
n x
P( x) 
 p q
 n  x ! x !
Number of
outcomes with exactly x
successes among n trials
The probability of x
successes among n trials
for any one particular order
Example 4: Assume that a procedure yields a binomial distribution
with a trial repeated n times. Use the binomial probability formula to
find the probability of x successes given the probability p of success
on a single trial.
n!
P  x 
a) n = 12, x = 10, p = 3/4
P( x  10) 
12!
10
2
 0.75  0.25 
 2!10!
 0.232
b) n = 20, x = 4, p = 0.15
P( x  4) 
20!
4
16
 0.15   0.85 
16!4!
 0.182
 n  x ! x !
p x qn x
Method 2: Using Minitab Table
Technology
Minitab Table results can be used to find binomial probabilities.
Minitab
Example 5: Refer to the Minitab display. (When blood donors were
randomly selected, 45% of them had blood that is Group O (based on
data from the Greater New York Blood Program).) The display shows
the probabilities obtained by entering the values of n = 5 and p = 0.45.
a) Find the probability that at least 1 of the 5 donors
has Group O blood.
P( x  1)  1  P( x  0)
 1  0.050328
 0.950
b) If at least 1 Group O donor is needed, is it
reasonable to expect that at least 1 will be obtained?
x
P(x)
0
0.050328
1
0.205889
2
0.336909
3
0.275653
4
0.112767
5
0.018453
Yes, it is reasonable to expect that at least one group O donors will be
obtained since 0.950 > 0.05.
Method 3: Using TI 83/84 Technology
We will be doing binomial probabilities using the calculator:
*Keyword “exact”
…find probability with
binompdf(number of trials, probability of success, number of successes)
*Keywords “less than or equal to”
…find probability with
binomcdf(number of trials, probability of success, number of successes)
**If it says “at least” or “greater than or equal to” use
1 – binomcdf with 1 less number for success
What you are doing here is 1 minus what you DO NOT WANT and that will give
you the probability that you do want.
If you don’t have a graphing calculator, use this link on your phone:
http://stattrek.com/online-calculator/binomial.aspx
Example 6: Use the TI 83/84 calculator to calculate the following
probabilities.
a) Calculate the probability of tossing a coin 20 times and getting exactly 9
heads.
n = 20, p = 0.5, x = 9
binompdf(20, 0.5, 9) = 0.160
b) Calculate the probability of tossing a coin 20 times and getting less than 6
heads.
n = 20, p = 0.5, x < 6 ≤ 5
binomcdf(20, 0.5, 5) = 0.021
c) Calculate the probability of tossing a coin 32 times and getting at least 14
heads.
n = 32, p = 0.5, x ≤ 13
1 – binomcdf(32, 0.5, 13) = 0.811
Example 6 continued:
d) About 1% of people are allergic to bee stings. What is the
probability that exactly 1 person in a class of 25 is allergic to bee
stings?
n = 25, p = 0.01, x = 1
binompdf(25, 0.01, 1) = 0.196
e) Refer to (d), what is the probability that 4 or more of them are
allergic to bee stings?
n = 25, p = 0.01, x ≤ 3
1 – binomcdf(25, 0.01, 3) = 0.00011
Example 7: Use the TI 83/84 calculator to calculate the following
probabilities.
a) At a college, 53% of students receive financial aid. In a random
group of 9 students, what is the probability that exactly 5 of them
receive financial aid?
n = 9, p = 0.53, x = 5
binompdf(9, 0.53, 5) = 0.257
b) Now using (a), what is the probability that fewer than 3 students in
the class receive financial aid?
n = 9, p = 0.53, x ≤ 2
binomcdf(9, 0.53, 2) = 0.064