QBM117
Business Statistics
Probability Distributions
Binomial Distribution
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Objectives
•
To introduce the binomial distribution
•
Learn how to recognize an experiment whose
outcomes follow a binomial probability distribution
•
Learn how to calculate binomial probabilities
•
Find the mean and variance of a binomial
distribution
•
Learn how to use Excel to find binomial probabilities
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Binomial Probability Distribution
• The binomial probability distribution is a discrete
probability distribution.
• It is the most widely used distribution for discrete
random variables.
• The binomial distribution is used in situations where
there are only two possible outcomes.
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Binomial Experiment
A binomial experiment possesses the following
properties:
1. The experiment consists of a fixed number n of
trials.
2. The result of each trial can be classified into one
of two categories: success or failure.
3. The probability p of a success remains constant
for each trial.
4. Each trial of the experiment is independent of the
other trials.
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Binomial Random Variable
• The random variable in a binomial experiment is the
number of successes in the n trials.
• It is called the binomial random variable.
• It is a discrete random variable as the possible values
of the random variable are 0,1, 2,…, n.
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Binomial Distribution
• The distribution of a binomial random variable X is
the Binomial distribution with parameters n and p,
X~B(n,p)
• The parameter n is the number of observations.
• The parameter p is the probability of success on any
one observation.
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Example 1
A coin is tossed 10 times and the number of heads is
recorded.
- there are n = 10 trials (tosses)
- there are two possible outcomes on each toss:
head or tail
- the probability of a head is p = 0.5 on each toss
- each toss is independent of the others
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The properties of a binomial experiment are present.
Let X be the number of heads observed in 10 tosses.
X is a binomial random variable and has a binomial
distribution, X~B(10,0.05)
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Example 2
500 randomly selected computer chips were
produced at a manufacturing facility and each chip
was tested to determine whether it was defective.
Historically, 1% of the computer chips produced are
defective.
- there are n = 500 trials (computer chips tested)
- there are two outcomes per trial: the computer
chip is either defective or non-defective
- the probability of success (computer chip is
defective) is p = 0.01 for each computer chip
- the trials are independent since the computer
chips are randomly selected
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The properties of a binomial experiment are present.
Let X be the number of defective computer chips in
the 500 randomly selected computer chips.
X is a binomial random variable and has a binomial
distribution, X~B(500,0.01)
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Example 3
1500 people were asked whether they preferred the
taste of Coca-Cola over that of other colas. Previous
studies have shown that 20% of the population
prefers Coca-Cola.
- there are n = 1500 trials (people asked)
- there are two outcomes per trial: the person
prefers Coca-Cola or does not
- the probability of success (person prefers
Coca-Cola) is p = 0.20
- the trials are independent if the 1500 people are
randomly selected
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The properties of a binomial experiment are present.
Let X be the number of people who prefer Coca-Cola.
X is a binomial random variable and has a binomial
distribution, X~B(1500,0.20)
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Binomial Probability Distribution
• In a binomial experiment we are interested in
computing the probability of obtaining x successes in
n trials.
• To do this we need to know the probability
distribution.
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Example 4
Genetics says that children receive genes from their
parents independently. Each child of a particular pair
of parents has a probability 0.25 of having type O
blood. This particular pair of parents have 5 children
and the number who have type O blood is recorded.
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- There are n=5 trials, as there are 5 children.
- The trials are independent as children receive
genes from their parents independently.
- There are two possible outcomes for each child;
they either have type O blood or they don’t.
- The probability that each child has type O blood is
p=0.25.
Let X be the number of children who have type O
blood.
X is a binomial random variable and has a binomial
distribution, X~B(5,0.25).
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Each child born to a particular set of parents has a
probability 0.25 of having type O blood.
If these parents have 5 children, what is the
probability that exactly 2 of them have type O blood?
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If X is the number of children with type O blood
and is a binomial random variable with n = 5 trials
and probability p = 0.25 of a success on each trial.
We want to find
P(exactly 2 children have type O blood) = P(X=2)
Define the events and their probabilities:
S = event of a success
P(S) = 0.25
F = event of a failure
P(F) = 0.75
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Step 1
Find the probability that a specific 2 of the 5 trials
give successes, say the first and the second.
This is the outcome SSFFF.
The multiplication rule of independent events tells us
that
P(SSFFF) = P(S).P(S).P(F). P(F). P(F)
= 0.250.250.750.750.75
= (0.25)2(0.75)3
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Step 2
Find the number of different ways 2 successes and 3
failures can occur in 5 outcomes.
The possible combinations are
SSFFF SFSFF SFFSF SFFFS FSSFF
FSFFS FFSSF FFSFS FFFSS
FSFSF
Therefore there are 10 possible ways 2 successes
and 3 failures can occur in 5 outcomes.
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The overall probability of 2 successes is therefore
P(X=2) = 10.(0.25)2(0.75)3 = 0.2637
Hence the binomial probability is the number of
combinations of x successes in n trials, multiplied by
the probability of any specific combination of the x
successes.
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The possible values of X are 0, 1, 2, 3, 4 and 5.
We have calculated the probability that X = 2.
To obtain the probability distribution of X we would
have to calculate the probabilities that X = 0, X = 1,
X = 3, X = 4 and X = 5.
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Calculating Binomial Probabilities
• It is impractical to do this calculation each time we
wish to find a probability for a binomial random
variable, or to do all the calculations to obtain the
probability distribution for a binomial random variable.
• Instead we need a general formula for the probability
of obtaining x successes in the n trials, when the
probability of success is p.
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• We will observe x successes in the n trials whenever
we observe x successes and (n-x) failures.
• The probability of success is p and the probability of
failure is q = 1-p.
• Applying the multiplication rule for independent
events, the probability that x successes and (n-x)
failures occur is pxqn-x.
• We now need to multiply this probability by the
number of ways we can get x successes in n trials.
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Counting Rule
• The number of different ways we can choose x
objects from n objects is
n!
C 
x !(n  x)!
n
x
where n! is read as ‘n factorial’ and is given by
n!  n(n  1)(n  2)...2.1
and 0! Is defined to be 1.
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Example 4 revisited
The number of ways we can choose 2 children from 5
is
5!
5!
C 

 10
2!(5  2)! 2!3!
5
2
as we calculated before.
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Using your calculator
• Combinations C xn can be calculated using your
calculator, using the nCr button (Casios).
• Factorials x! can also be calculated using the x!
button. It is a 2nd function on most calculators.
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Calculating Binomial Probabilities
• The probability that x successes and (n-x) failures
occur is pxqn-x.
• We now need to multiply this probability by the
number of ways we can get x successes in n trials.
• Hence the probability of obtaining x successes in n
trials is
n x n x
P  X  x   Cx p q
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Binomial Probability Distribution
• If the random variable X is the total number of
successes in the n trials of a binomial experiment that
has a probability p of a success on any given trial, the
probability distribution of X is given by
x n x
P( X  x)  C p q
n
x
(note that p = 1 – q is the probability of failure on any
given trial)
• This is known as the binomial probability distribution.
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Example 5
National Oil Company conducts exploratory oil drilling
operations. In order to fund the operation, investors
form partnerships, which provide the financial support
necessary to drill a fixed number of oil wells. Each
well drilled is classified as a producer well or a dry
well. Past experience shows that this type of
exploratory operation provides producer wells for
15% of wells drilled. A newly formed partnership has
provided the financial support for drilling at 12
exploratory locations.
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- there are n = 12 trials (wells)
- there are two outcomes per trial: the well is either
a producer well or it is dry
- the probability of success (well is a producer well)
is p = 0.15
- We assume that the trials are independent
The properties of a binomial experiment are present.
Define X as the number of producer wells.
X is a binomial random variable with n = 12 and
p = 0.15, X~B(12, 0.15).
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What is the probability that all 12 wells will be
producer wells?
P( X  12)  C1212 (0.15)12 (0.85)0
 1.2974  1010
 0.000000000012974
0
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What is the probability that all 12 wells will be dry?
The probability that all 12 wells are dry is the
probability that none of the wells are producer wells.
P( X  0)  C012 (0.15)0 (0.85)12
 0.1422
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What is the probability that exactly 1 well will be a
producer well?
P( X  1)  C112 (0.15)1 (0.85)11
 0.3012
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In order to make the partnership venture profitable, at
least 3 of the 12 exploratory wells must be producer
wells. What is the probability that the venture will be
profitable?
P( X  3)  P( X  3)  P( X  4)  ...  P( X  12)
 1-  P ( X  0)  P( X  1)  P( X  2) 
 1  0.1422  0.3012  C212 (0.15) 2 (0.85)10 
 1   0.1422  0.3012  0.2924
 1- 0.7358
 0.2642
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Binomial Tables
• An alternative of calculating binomial probabilities is
to use the Table 1 in Appendix C of the text
(pg 585 full text, pg 583 abridged).
• The probabilities given in this table are cumulative
probabilities:
P( X  k )  P( X  0)  P( X  1)  ...  P( X  k )
k
  P( X x)
x 0
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Example 6
A large bank that issues many loans for General
Motors estimates that 70% of the loans are
approved within 24 hours of application. If a
consumer group takes a random sample of 25
recent General Motors loan applications, what is the
probability that
a. at most 20 were approved within 24 hours?
b. at least 15 were approved within 24 hours?
c. at least 12, but no more than 18 are approved
within 24 hours?
d. exactly 22 were approved within 24 hours?
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a. The probability that at most 20 were approved within
24 hours is
P( X  20)  0.910
b. The probability that at least 15 were approved within
24 hours is
P( X  15)  1  P( X  14)
 1  0.098
 0.902
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c. The probability that at least 12, but no more than 18
are approved within 24 hours is
P(12  X  18)  P( X  18)  P( X  11)
 0.659  0.006
 0.653
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d. The probability that exactly 22 were approved within
24 hours is
P( X  22)  P( X  22)  P( X  21)
 0.991  0.967
 0.024
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Mean and Variance of Binomial Random
Variables
• If X is a binomial random variable, the mean and
variance of X are
E ( X )    np
V ( X )   2  npq
where n is the number of trials, p is the probability of
success in any trial, and q=1-p is the probability of
failure in any trial.
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Example 6 revisited
A large bank that issues many loans for General
Motors estimates that 70% of the loans are
approved within 24 hours of application. If a
consumer group takes a random sample of 25
recent General Motors loan applications, what is the
expected number of loans that will be approved?
E ( X )  np
 25  0.7
 17.5
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What is the variance of the number of loans that will
be approved?
V ( X )  npq
 25  0.7  0.3
 5.25
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Reading for next lecture
• Chapter 5 Section 5.5
Exercises
• 5.31
• 5.32
• 5.35
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