Binomial Probability Distribution
Binomial Distribution 1
Binomial Distribution 2

Four Properties of a Binomial Experiment
1. The experiment consists of a sequence of n
identical trials.
2. Two outcomes, success and failure, are possible
on each trial.
3. The probability of a success, denoted by p, does
not change from trial to trial.
stationarity
assumption
4. The trials are independent.
Slide 1
Binomial Probability Distribution
Our interest is in the number of successes
occurring in the n trials.
We let x denote the number of successes
occurring in the n trials.
Slide 2
Binomial Probability Distribution

Example: Evans Electronics
Evans Electronics is concerned about a low
retention rate for its employees. In recent years,
management has seen a turnover of 10% of the
hourly employees annually.
Thus, for any hourly employee chosen at random,
management estimates a probability of 0.1 that the
person will not be with the company next year.
Choosing 3 hourly employees at random, what is
the probability that 1 of them will leave the company
this year?
Slide 3
Excel
𝒏!
𝒙! 𝒏 − 𝒙 !
FACT and COMBIN
n=10
x
FACT(n) FACT(x) FACT(n-x)FACT(n)/[FACT(x)*FACT(n-x)]
COMBIN(n,x)
1
3628800
1
362880
10
10
2
3628800
2
40320
45
45
3
3628800
6
5040
120
120
4
3628800
24
720
210
210
5
3628800
120
120
252
252
6
3628800
720
24
210
210
7
3628800 5040
6
120
120
8
3628800 40320
2
45
45
9
3628800 362880
1
10
10
10
3628800 3628800
1
1
1
Slide 4
Binomial Probability Distribution

Binomial Probability Function
n!
f (x) 
p x (1  p )( n  x )
x !(n  x )!
where:
x = the number of successes
p = the probability of a success on one trial
n = the number of trials
f(x) = the probability of x successes in n trials
n! = n(n – 1)(n – 2) ….. (2)(1)
Slide 5
Binomial Probability Distribution

Binomial Probability Function
n!
f (x) 
p x (1  p )( n  x )
x !(n  x )!
Number of experimental
outcomes providing exactly
x successes in n trials
Probability of a particular
sequence of trial outcomes
with x successes in n trials
Slide 6
Excel
n= 10
x
0
1
2
3
4
5
6
7
8
9
10
p= 0.3
COMB
px
(1-p) (n-x)
1
1
0.028247525
10
0.3
0.040353607
45
0.09
0.05764801
120
0.027
0.0823543
210
0.0081
0.117649
252
0.00243
0.16807
210
0.000729
0.2401
120
0.0002187
0.343
45
0.00006561
0.49
10 0.000019683
0.7
1
5.9049E-06
1
f(X) Formula
0.028247525
0.121060821
0.233474441
0.266827932
0.200120949
0.102919345
0.036756909
0.009001692
0.001446701
0.000137781
5.9049E-06
1
SUMP(X) Formula
0.028247525
0.149308346
0.382782786
0.649610718
0.849731667
0.952651013
0.989407922
0.998409614
0.999856314
0.999994095
1
Slide 7
Excel
Slide 8
n= 10
P(X=
P(X<
P(X<=
P(X>
P(X>=
P(X<>
P(2=<X<=3)
P(2<X<3)
Average np
Varriance np(p-1)
StdDev
3
3
3
3
3
3
p= 0.3
0.266828
0.382783
0.649611
0.350389
0.617217
0.733172
0.500302
0
3
2.10
1.45
Slide 9
I am going to have a 10 multiple choice question Quiz (with 4 answers )
tomorrow and have not studied at all. What is the chance of getting at
least 60% on the Quiz?
Slide 10
Excel
The employees of a corporation deposit their monthly retirement contribution
either in a mutual fund account or in a stock account.
Each employee has only one type account.
Assume
a) If
73%
12
have mutual fund account.
to calculate the probability that
exactly
4
to calculate the probability that
b) Calculate the probability that at least
6
c) Calculate the probability that at most
7
d) Calculate the probability of less than
11
have mutual fund account.
0.00357
employees have mutual fund account.
employees have mutual fund account.
having stock account.
0.98422
Slide 11
0.0509
1
Excel
When an old machine is functioning properly, 7% of the items produced are defective.
Assume that we will randomly select
3 parts.
We are interested in knowing the probabilities related to the number of defective parts.
a) use the proper probability distribution to calculate the probability that
exactly
2 parts are deffective.
0.01367
parts are deffective.
b) Calculate the probability that at least
1
0.98599
c) Calculate the probability that at most
d) Calculate the probability of less than
2
parts are deffective.
0.01401
2
parts are good.
0.01367
Slide 12
Excel
n= 5
x
0
1
2
3
4
5
p= 0.19
1
P(x=X)
P(x<=X) Prob(x<= 1) = 0.75762
0.34867844 0.34867844
0.34867844
0.40894385 0.75762229
0.40894385
0.191850201 0.949472491
0.045001899 0.99447439
0.005278001 0.99975239
0.00024761
1
Bin-Dist n=5 & P= 0.19
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
P(x=X)
Prob(x<= 1) = 0.75762
Slide 13
Poisson Probability Distribution
Poisson Distribution 1
Poisson Distribution 2
A Poisson distributed random variable is often
useful in estimating the number of occurrences
over a specified interval of time or space
It is a discrete random variable that may assume
an infinite sequence of values (x = 0, 1, 2, . . . ).
Slide 14
Poisson Probability Distribution
Examples of Poisson distributed random variables:
the number of knotholes in 14 linear feet of
pine board
the number of vehicles arriving at a toll
booth in one hour
Bell Labs used the Poisson distribution to model
the arrival of phone calls.
Slide 15
Poisson Probability Distribution

Two Properties of a Poisson Experiment
1. The probability of an occurrence is the same
for any two intervals of equal length.
2. The occurrence or nonoccurrence in any
interval is independent of the occurrence or
nonoccurrence in any other interval.
Slide 16
Poisson Probability Distribution

Poisson Probability Function
f ( x) 
 x e
x!
where:
x = the number of occurrences in an interval
f(x) = the probability of x occurrences in an interval
 = mean number of occurrences in an interval
e = 2.71828
x! = x(x – 1)(x – 2) . . . (2)(1)
Slide 17
Poisson Probability Distribution

Poisson Probability Function
Since there is no stated upper limit for the number
of occurrences, the probability function f(x) is
applicable for values x = 0, 1, 2, … without limit.
In practical applications, x will eventually become
large enough so that f(x) is approximately zero
and the probability of any larger values of x
becomes negligible.
Slide 18
Poisson Probability Distribution

Example: Mercy Hospital
Patients arrive at the emergency room of Mercy
Hospital at the average rate of 6 per hour on
weekend evenings.
What is the probability of 4 arrivals in 30 minutes
on a weekend evening?
Slide 19
Poisson Probability Distribution

Example: Mercy Hospital
 = 6/hour = 3/half-hour, x = 4
3 4 (2.71828)3
f (4) 

4!
Using the
probability
function
.1680
Slide 20
Using Excel to Compute
Poisson Probabilities

Excel Formula Worksheet
A
1
2
3
4
5
6
7
8
9
10
B
3 = Mean No. of Occurrences ()
Number of
Arrivals (x )
0
1
2
3
4
5
6
… and so on
Probability f (x )
=POISSON.DIST(A4,$A$1,FALSE)
=POISSON.DIST(A5,$A$1,FALSE)
=POISSON.DIST(A6,$A$1,FALSE)
=POISSON.DIST(A7,$A$1,FALSE)
=POISSON.DIST(A8,$A$1,FALSE)
=POISSON.DIST(A9,$A$1,FALSE)
=POISSON.DIST(A10,$A$1,FALSE)
… and so on
Slide 21
Using Excel to Compute
Poisson Probabilities

Excel Value Worksheet
A
1
2
3
4
5
6
7
8
9
10
B
3 = Mean No. of Occurrences ()
Number of
Arrivals (x )
0
1
2
3
4
5
6
… and so on
Probability f (x )
0.0498
0.1494
0.2240
0.2240
0.1680
0.1008
0.0504
… and so on
Slide 22
Poisson Probability Distribution
Example: Mercy Hospital
Poisson Probabilities
0.25
Probability

0.20
Actually,
the sequence
continues:
11, 12, 13 …
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9
10
Number of Arrivals in 30 Minutes
Slide 23
Using Excel to Compute
Cumulative Poisson Probabilities

Excel Formula Worksheet
A
1
2
3
4
5
6
7
8
9
10
B
3 = Mean No. of Occurrences ( )
Number of
Arrivals (x )
0
1
2
3
4
5
6
… and so on
Cumulative Probability
=POISSON.DIST(A4,$A$1,TRUE)
=POISSON.DIST(A5,$A$1,TRUE)
=POISSON.DIST(A6,$A$1,TRUE)
=POISSON.DIST(A7,$A$1,TRUE)
=POISSON.DIST(A8,$A$1,TRUE)
=POISSON.DIST(A9,$A$1,TRUE)
=POISSON.DIST(A10,$A$1,TRUE)
… and so on
Slide 24
Using Excel to Compute
Cumulative Poisson Probabilities

Excel Value Worksheet
A
1
2
3
4
5
6
7
8
9
10
B
3 = Mean No. of Occurrences ( )
Number of
Arrivals (x)
0
1
2
3
4
5
6
… and so on
Cumulative Probability
0.0498
0.1991
0.4232
0.6472
0.8153
0.9161
0.9665
… and so on
Slide 25
Poisson Probability Distribution
A property of the Poisson distribution is that
the mean and variance are equal.
=s2
Slide 26
Poisson Probability Distribution

Example: Mercy Hospital
Variance for Number of Arrivals
During 30-Minute Periods
=s2=3
Slide 27
Airline passengers arrive randomly and independently at the passenger-screening
facility at a major international airport.
The mean arrival rate is 10 passengers per minute.
1.Compute the probability of no arrivals in a one-minute period.
2.Compute the probability that three or fewer passengers arrive in a one-minute period.
3.Compute the probability of no arrivals in a 15-second period.
4.Compute the probability of at least one arrival in a 15-second period.
5.Compute the probability of at least
6
arrival in 30
a
second period.
0.00005
0.0103
0.0821
0.9179
0.384
=POISSON.DIST(0,F3,0)
=POISSON.DIST(3,F3,1)
=POISSON.DIST(0,F3/4,0)
=1-P6
=1-POISSON.DIST(I8-1,F3/2,1)
 = s = 10  s = SQRT(10
2
Slide 28
Patients arrive at the emergency room of Mercy Hospital at the average rate of
7
per hour on weekend evenings.
enables many B&B to attract quests
a) Compute the probability of
4
arrivals in 30 minutes on a weekend evening?
b) Compute the probability of
2
or less arrivals in 30 minutes on a weekend evening?
c) Compute the probability of
3
or more arrivals in 30 minutes on a weekend evening? 0.188812
d) Compute the probability of
12
or more arrivals in two hous on a weekend evening? 0.320847
e) Compute the probability that the interarrival between two consequtive customersis less than 0.679153
0.73996
10 minutes.
Slide 29
End of Chapter 5
Slide 30