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Binomial Distribution
Introduction:
• Binomial distribution has only two
outcomes or can be reduced to two
outcomes.
• There are a lot of examples in engineering
science as a binomial distribution such
that: electrical switch, a born baby it will
be either male or female, medical
treatment can be classified as effective or
ineffective, blood pressure normal or
abnormal blood pressure.
Binomial Distribution
•
1.
2.
3.
4.
Definition: A binomial experiment is a
probability experiment that satisfied the
following four requirements:
There must be a fixed number of trials.
Each trial can have only two outcomes or
outcomes that can be reduced to two
outcomes. These outcomes can be considered
as either success or failure.
The outcome of each trial must be independent
of one another.
The probability of success must remain the
same for each trial.
Binomial Distribution
Notations of Binomial distribution:
P(S): the probability of success.
P(F): the probability of failure.
P: the numerical probability of success.
Q: the numerical probability of failure.
P(S) = p, P(F) = 1- p = q
N: the number of trials.
X: the number of success of trial.
Binomial Distribution
• The probability of success in a
binomial experiment can be computed
with this formula is:
n!
P( X ) 
 p X q n X
(n  X )! X !
mean(  )  n. p
Variance ( 2 )  n. p.q
( ) 
n. p.q
Binomial Distribution
• Example 1:
A coin tossed 3 times finds the
probability of getting two heads.
• Solution:
• S: = HHH, HHT, HTH, THH, TTH, THT,
HTT, TTT.
3
P (2heads )   0.375
8
Binomial Distribution
•
We can solve the above problem by
binomial distribution as:
1. There are a fixed number of trial (Three times)
2. There are only two outcomes for each trial,
heads or tails.
3. The outcomes are independent of one another
[The outcome of one toss in no way affects the
outcome of another toss]
4. The probability of a success (heads)=0.5 in each
case.
Binomial Distribution
• Then we can use binomial distribution
as:
• n = 3, X = 2, p=0.5, q = 0.5 then
2
n!
3!
1 1 3
P(2heads) 
p X q n X 
       0.375
(n  X )! X !
(3  2)!2!  2   2  8
mean(  )  n. p  3 
1
 1.5
2
1 1 3
var iance ( 2 )  n. p.q  3   
2 2 4
  n. p.q 
3
4
Binomial Distribution
• Example 2:
A survey found that one out of five
Americans say that he or she has
visited a doctor in any given month, if
10 people are selected at random, find
the probability that exactly 3 will have
visited a doctor last month.
• Solution:
Binomial Distribution
• Solution:
• N = 10, X = 3,
3
P(3) 
p 
1
4
,q 
5
5
7
10!  1   4 
     0.201
(10  3)!3!  5   5 
1
mean(  )  n. p  10   2
5
1 4 8
var iance ( )  n. p.q  10   
5 5 5
8
 
5
2
Binomial Distribution
• Example 3:
A survey from teenage research
unlimited found that 30% of teenage
consumers receive their spending
money from part time jobs. If 5
teenagers are selected at random, find
the probability that at least 3 of them
will have part time jobs.
Binomial Distribution
• Solution:
• To find the probability at least 3 have
part time we should compute P (3), P (4)
and P (5).
• P = 0.3, q = 1- 0.3 = 0.7, n = 5, X = [3, 4,
5]
5!
P(3) 
(0.3) 3 (0.7) 2  0.132
(5  3)!3!
Binomial Distribution
• Solution:
5!
P(4) 
(0.3) 4 (0.7)1  0.028
(5  4)!4!
5!
P(5) 
(0.3) 5 (0.7) 0  0.002
(5  5)!5!
P (at least 3 have part time jobs) =
P (3) + P (4) + P (5) = 0.132 + 0.028 + 0.002 = 0.162
Binomial Distribution
• Example 4:
•
A die is rolled 360 times, find the
mean , variance and standard deviation
of the number 4’s that will be rolled.
• Solution:
• N=360,
p
1
5
,q 
6
6
Binomial Distribution
• Solution:
1
mean(  )  n. p  360   60
6
1 5
var iance( )  n. p.q  360    50
6 6
  50
2
Zaha Hadid
• Zaha hadid