# 25 Binomial Distributions

advertisement ```Bernoulli trial
• Bernoulli Trial ~ a trial with 2 outcomes
success or failure (arbitrary names)
• p denotes P (success)
• P (failure)= 1-p
A Binomial Experiment
• A binomial experiment is a series of Bernoulli
trials done to determine
• X ~ the number of successes.
• X is a discrete random variable.
• n denotes the number of Bernoulli trials.
• The Trials are independent
• discrete means that the set of values that the RV
can take on is countable or countably infinite.
(think a set of individual values)
Probability Distribution
n k
nk
p ( X  k )    p (1  p )
k 
Remember that X is a random variable being the number
of successes after n trials. p is the probability of success.
NOTE: this is another
common way to write
combinations.
n
   nCk
k 
Your Book’s Equation
• The Equation is written this way in the text book.
It is identical, they use q to represent 1-p and
express the combination as nCx as they use x to
represent the particular value of the random
variable X being considered.
p ( x) n C x p q
x
You are welcome to use either form.
n x
Expected value
n
E(X ) 
 x P(X
i
 x i )  np
i 1
The short cut to the expected value
of X in a binomial experiment is np
Example: 4 dice are rolled, how
many 1’s are likely to appear?
• This is a binomial experiment
• Each die roll is a Bernoulli Trial with
success~ 1 fail ~ 2 through 6
• X ~ the number of 1’s
• X={0,1,2,3,4}
•
Probability Distribution
• P(X=k)=(nCk)pk(1-p)n-k
•
•
•
•
•
P(X=0)=4C0(5/6)4=0.48
P(X=1)=4C1(1/6)(5/6)3=.386
P(X=2)=4C2(1/6)2(5/6)2=.116
P(X=3)=4C3(1/6)3(5/6)1=.015
P(X=4)=4C4(1/6)4=.00077
Expected Value E(X)
n
E(X ) 
 x P(X
i
 xi )
i 1
= x1P(X=x1)+ x2P(X=x2)+ x3P(X=x3)+…+ xnP(X=xn)
• E(X)=0*0.48+1*0.386+2*.116+3*.015
+4*.00077 = 0.666
But Remember E(X)=np
• 4 trials, each with a probability p=1/6 of
success
• E(X)=np=4*1/6=0.667
• (this answer is more accurate as we did
not round until the end)
A company is producing brake
callipers.
• Probability of a defect is 1.2%
• If 150 brake callipers are produced, what
is the probability that no more than 2 are
defective?
Analyze the Event
• If X represents the number of defective
callipers, the event that no more than 2
are defective is the event that X≤2.
• If X ≤ 2 then X=0 or X=1 or X=2
• P(X ≤2)=P(0) +P(1)+P(2)
P(X ≤2)=P(0) +P(1)+P(2)
• p=0.012
 150
p ( X  0 )  
 0
 150
p ( X  1)  
 1
1-p=0.988 n=150

0
150
 0 . 012 ( 0 . 988 )  16 . 35 %


1
149
 0 . 012 ( 0 . 988 )  29 . 79 %

 150
p ( X  2 )  
 2

2
148
 0 . 012 ( 0 . 988 )  26 . 95 %

• P(X ≤2)=P(0) +P(1)+P(2)
•
=16.35% + 29.79% + 26.95%
•
=73.09%
Re Cap
• When ever you recognize that a series of
Bernoulli trial are being done to determine the
number of successes this is a binomial
experiment.
• The outcomes probabilities are distributed
n
according to : p ( X  k )   k  p (1  p )
k
nk
• The expected value E(X) = np if X is the result of
a binomial experiment.
Practice
• Page 385 1 to 12, 14
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