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Binomial Random Variables
Dr Tom IlventoDepartment of Food and Resource Economics
Overview• A special case of a Discrete Random Variable is the
Binomial
• This happens when the result of the experiment is a dichotomy
• Success or Failure
• Yes or No
• Cured or not Cured
• If the discrete random variable is a binomial, we have some easier ways to solve for probabilities
• Formula
• Probability Table
• And the solution for the Mean and Variance is much easier to solve
2
Binomial Random Variable
• In many cases the responses to an experiment are dichotomous
• Sucess/Failure
• Yes/No
• Alive/Dead
• Support/Don’t Support
• When our focus is conducting an experiment n times independently and observing the number x of times that one of the two outcomes occurs (Success)
• And the probability of success, p, remains the same from trial to trial
• This X is a Binomial Random Variable
• We can exploit this by using known formulas for a Binomial Probability Distribution
3
Conduct an experiment n times and observethe number x of times that Success occurs
Characteristics of a Binomial Distribution
• The experiment consists of n identical trials
• There are only two outcomes on each trial. Outcomes can be denoted as
• S for Success
• F for Failure
• The probability of S (success) remains the same from trial to trail
• Denoted as p the proportion
• The probability of F (failure)
• Denoted as q q=(1-p)
• The trials are independent of each other
• The binomial random variable x is the number of Successes in n trials 4
Example of a Binomial Random Variable: Marketing Survey
• Marketing survey of 100 randomly chosen consumers
• Record their preferences for a new and an old diet soda – ask them to choose their preference
• Let x be number of 100 who choose the new brand
• This is a binomial random variable
5
Conduct an experiment 100 times and observe the number x of times that the subject chooses the new brand
Example of a Binomial Random Variable: Fitness Example
• Heart Association says only 10% of adults over 30 can pass the fitness test
• Suppose 4 people over 30 are selected at random
• Let X be the number who pass the minimum requirements
• Find the probability distribution for X
6
Conduct an experiment 4 times and observe the number x of times that pass occurs
How to solve this using the strategy of a Discrete Random Variable
1. List the events
2. List the sample points that refer to that event
3. Calculate the probabilities
• p = .1 and q = (1.0 - .1) = .9
• I multiply through on the probabilities because each trial is independent of the others 7
Event X Sample Points Probability
All Fail FFFF (.9)(.9)(.9)(.9) = .6561
Can you solve it – the probability that exactly 1 person passes the test?
• Count the ways we could have only one pass, and three failures
• Assign probabilities to this event
8
• SFFF FSFF FFSF FFFS
• For each combination, the probabilities are:
• .1*.9*.9*.9
• And there are four ways to get one pass
• 4[.1*.9*.9*.9] = .2916
• Another way to write it is
• 4[.1*.93] = .2916
Let’s finish solving for the whole table
The number of times that an adult passes in a sample of four
9
Event X Sample Points Probability
0All Fail
FFFF (.9)(.9)(.9)(.9) = .6561
1One passes
SFFF FSFF FFSF FFFS 4[(.1)(.9)3] = .2916
2Two Pass
SSFF SFSF SFFS FSSF FSFS FFSS
6[(.1)2(.9)2] = .0486
3Three Pass
SSSF FSSS SFSS SSFS 4[(.1)3(.9)] = .0036
4Four Pass
SSSS (.1)(.1)(.1)(.1) = .0001
Probability Distribution
• When x = 0 All Fail P = .6561
• When x = 1 One Pass P = .2916
• When x =2 Two pass P =.0486
• When x=3 Three pass P = .0036
• When x=4 Four pass P = .0001
10
0
0.175
0.350
0.525
0.700
0 1 2 3 4
P(X
)
X 0 1 2 3 4
P(X) 0.6561 0.2916 0.0486 0.0036 0.0001
Fitness Example
• Find the probability that none of the adults pass the test
• P(x=0) = .6561
• Find the probability that 3 of 4 adults pass the test
• P(x=3) = .0036
• What is the probability that 2 or more adults pass the test?
• P(x=2) + P(x=3) + P(x=4) = .0486 + .0036 + .0001 = .0523
11
X 0 1 2 3 4
P(X) 0.6561 0.2916 0.0486 0.0036 0.0001
Binomial Probability Distribution Formula
• Sometimes the number of trials gets large
• We can also use the binomial probability distribution formula to generate the probabilities
• It uses factorial notation
• n! = n(n-1)(n-2)…(n-(n-1))
• 5! = 5x4x3x2x1 = 120
• 0! = 1, 1!=1, 2!=2x1=2, …
• The formula for any x in n trials is:
12
xnxqp
xnx
nxP
!
!= )()(
)!(!
!)(
What defines the Binomial Distribution?
• p = Probability of a success on a single trial
• q = (1-p) probability of failure
• n = number of trials
• x = number of successes in n trials
13
xnxqp
xnx
nxP
!
!= )()(
)!(!
!)(
Note: it uses the Combinatorial Rule as
the first part of the formula
This part reflects the probabilities with each
combination
For x=3 in the fitness example, n=4, p=.1
• The four is how many combinations of 3 success in 4
• The last part of the formula is the probability associated with each of these combinations
• The probability, .0036, is the exact same one we calculated earlier
14
xnxqp
xnx
nxP
!
!= )()(
)!(!
!)(
343 )9(.)1(.)!34(!3
!4)3( !
!=P
P(3) =4 ! 3 ! 2 !1
3 ! 2 !1( ) 1( )(.1)
3(.9)
4"3
P(3) =24
6(.1)
3(.9)
4!3
P(3) = 4(.1)3(.9)
4!3
P(3) = 4(.001)(.9)
P(3) = 4(.0009) = .0036
Your Try it: For x=2 in the fitness example, n=4, p=.1
• I will get you started
15
xnxqp
xnx
nxP
!
!= )()(
)!(!
!)(
!
P(2) =4!
2!(4 " 2)!(.1)
2(.9)
4"2
P(2) = 6(.0081) = .0486
Mean and Variance for a Binomial Random Variable
• Since a binomial is only a dichotomy, the formulas for the mean and the standard deviation will simplify
• From µ = !(x!P(x))
• To µ = n!p
• Our fitness example: µ = 4*.1 = .4
• The Variance changes from
• From "2 = ![(x-µ)2!P(x))]
• To !2 = n*p*q
• Our fitness example: !2 = 4*.1* .9 = .36
• and ! = .60 16
I could have solved for the mean using the formula for discrete random variables
• To solve for the mean I would use this formula from the discrete random variable lecture:
• E(x) = (0)(.6561) + (1)(.2916) + (2)(.0486) + (3)(.0036) + (4)(.0001)
• E(x) = .4
• Binomial approach
• E(x) = n·p = 4·(.1) = .4
• The Binomial approach is much easier
17
µ=!="=
n
i
iixPxxE
1
)()(
If I know my Discrete Random Variable is distributed as a binomial random variable, it will make things much easier
I could have solved for the variance using the formula for discrete random variables
• To solve for the variance I would have:
• E(x-µ)2 = (0 -.4)2(.6561) + (1-.4)2
(.2916) + (2-.4)2(.0486) + (3-.4)2
(.0036) + (4-.4)2(.0001)
• !2 = .36
• Binomial approach
• E(x) = n·p·q = 4·(.1)(.9) = .36
• The Binomial approach is much easier
18
If I know my Discrete Random Variable is distributed as a binomial random variable, it will make things much easier
2
1
22 )()(])[( !µµ ="=" #=
n
i
iixPxxE
Return to the Nitrous Oxide Example
• Suppose we were recording the number of dentists that use nitrous oxide (laughing gas) in their practice
• We know that 60% of dentists use the gas.
• p = .6 and q = .4
• Let X = number of dentists in a random sample of five dentists use use laughing gas.
• n = 5
• This is a Binomial Random Variable!
19
Conduct an experiment 5 times and observe the number x of times that use Nitrous Oxide
Nitrous Oxide Example
• How to solve for these probabilities?
20
xnxqp
xnx
nxP
!
!= )()(
)!(!
!)(
X 0 1 2 3 4 5
P(X) 0.0102 0.0768 0.2304 0.3456 0.2592 0.0778
Probability Distribution for the Nitrous Oxide Example
• µ = 3.00
• !2 = 1.20
• ! = 1.01
• µ = 5*.6 = 3.00
• !2 = 5*.6*.4 = 1.20
• ! = SQRT(1.20) = 1.0121
X 0 1 2 3 4 5
P(X) 0.0102 0.0768 0.2304 0.3456 0.2592 0.0778
0
0.1
0.2
0.3
0.4
0 1 2 3 4 5
Probability Distribution of X
P(X
)
Nitrous Oxide Example using Excel
• Open up the file, BINOM.xls
• Click on the Worksheet Problem
• This worksheet is designed to solve problems up to n=50, for any value of p
• You enter in:
• p = .6
• n = 5
• The spreadsheet will do the rest!
22
Reverse p = 0.6000
X p(X) Cum p(X) Cum p(>=X) q = 0.4000
0 0.0102 0.0102 1.0000 n = 5
1 0.0768 0.0870 0.9898
2 0.2304 0.3174 0.9130 Mean 3.0000
3 0.3456 0.6630 0.6826 Variance 1.2000
4 0.2592 0.9222 0.3370 Std Dev 1.0954
5 0.0778 1.0000 0.0778
Binomial Formula using Excel
• In Excel, the formula for the Binomial Distribution function is:
• BINOMDIST(X,N,P,cumulative)
• X is the number of successes
• N is the number of independent trials
• P is the probability of success on each trial
• Cumulative is an argument - you enter TRUE or FALSE
• Entering TRUE gives a cumulative probability up to and
including X successes (or 1)
• Entering FALSE gives the exact probability of X successes
in N trials (or 0)
23BINOMDIST(3,5,.6,TRUE)
Binomial Formula using Excel
• For our example of dentists
• BINOMDIST(2,5,.6,TRUE)
• cumulative probability up to and including 2 successes
• = .3174
• BINOMDIST(2,5,.6,FALSE)
• the exact probability of X successes in N trials
• = .2304
24
Binomial Table
• Another way to get probabilities form Binomial Random Variables is via a table
• In exams, I will give you a table which contains cumulative probabilities for n= 5, 6, 7, 8, 9, 10, 15, 20, and 25
• Each table lists values of P across the top
• P = .01, .05, .1, .2, .3, …, .95, .99
• x = # of successes as the rows
• It is a Cumulative Table
25
Binomial Table for n = 5
• The probability associated with p=.3 and x = 4 is .998
• This means that the cumulative probability, or P(x " 4) = .998
• The actual probability of P(x = 4) = .998 - .969 = .029
• You must subtract two values to get the actual probability of x
26
The values shown are cumulative probabilities for the probability of x (denoted as k in the table)
PROBABILITIES (p)
x 0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0.99
0 0.951 0.774 0.590 0.328 0.168 0.078 0.031 0.010 0.002 0.000 0.000 0.000 0.000
1 0.999 0.977 0.919 0.737 0.528 0.337 0.188 0.087 0.031 0.007 0.000 0.000 0.000
2 1.000 0.999 0.991 0.942 0.837 0.683 0.500 0.317 0.163 0.058 0.009 0.001 0.000
3 1.000 1.000 1.000 0.993 0.969 0.913 0.813 0.663 0.472 0.263 0.081 0.023 0.001
4 1.000 1.000 1.000 1.000 0.998 0.990 0.969 0.922 0.832 0.672 0.410 0.226 0.049
5 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000
• The table is arranged cumulatively• For each probability, the value in the cell is the
cumulative probability up to and including X• The last row (in this case for x = 5), the
cumulative probability is 1.000
Nitrous Oxide Example
• Use the n = 5 Table for p = .6
• Solve the probability for x = 3
• P(x " 3) = .663
• P(x " 2) = .317
• P(x=3) = .663 - .317 = .346
• This is the same value (with some rounding error) that we calculated using the formula (.3456) 27
The values shown are cumulative probabilities for the probability of x (denoted as k in the table)
PROBABILITIES (p)
x 0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0.99
0 0.951 0.774 0.590 0.328 0.168 0.078 0.031 0.010 0.002 0.000 0.000 0.000 0.000
1 0.999 0.977 0.919 0.737 0.528 0.337 0.188 0.087 0.031 0.007 0.000 0.000 0.000
2 1.000 0.999 0.991 0.942 0.837 0.683 0.500 0.317 0.163 0.058 0.009 0.001 0.000
3 1.000 1.000 1.000 0.993 0.969 0.913 0.813 0.663 0.472 0.263 0.081 0.023 0.001
4 1.000 1.000 1.000 1.000 0.998 0.990 0.969 0.922 0.832 0.672 0.410 0.226 0.049
5 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000
Nitrous Oxide Example
• Use the n = 5 Table for p = .6
• Solve the probability for x > 3
• P(x " 3) = .663
• P(x>3) = 1 - .663 = .337
• Solve the probability for x " 2
• P(x " 2) = .317
28
The values shown are cumulative probabilities for the probability of x (denoted as k in the table)
PROBABILITIES (p)
x 0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0.99
0 0.951 0.774 0.590 0.328 0.168 0.078 0.031 0.010 0.002 0.000 0.000 0.000 0.000
1 0.999 0.977 0.919 0.737 0.528 0.337 0.188 0.087 0.031 0.007 0.000 0.000 0.000
2 1.000 0.999 0.991 0.942 0.837 0.683 0.500 0.317 0.163 0.058 0.009 0.001 0.000
3 1.000 1.000 1.000 0.993 0.969 0.913 0.813 0.663 0.472 0.263 0.081 0.023 0.001
4 1.000 1.000 1.000 1.000 0.998 0.990 0.969 0.922 0.832 0.672 0.410 0.226 0.049
5 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000
The Rare Event Approach
• What if we had 5 dentists selected randomly and none of them used nitrous oxide?
• Given p=.6, this would be a very rare event
• P(x=0) = .010
• This is possible, but not probable
• Was this just by chance????
• Or was the assumption wrong – that p =.6?
29
Summary
• The Binomial is a special form of the discrete random variable
• There are other discrete random variables - poisson
• If you know it is a Binomial Random Variable it makes it easy to solve for probabilities, the mean and the variance
• For probabilities you can use:
• The Binomial Formula
• The Binomial Tables
• Excel also has functions to solve for binomials
30