Geometric Distribution

    
 
 
 

 


  

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Distributions of Probability : Introduction :

Geometric Distribution There are various number of distributions in probability; each with its own characteristics and conditions, with different number of outcomes .However , these probabilities relate to each other but with different no.of trials and inputs.These distributions minimize the area of calculation by categorizing them in a different way.

Poisson Distribution Binomial Distribution Hyper geometric Distr. Negative Binomial Distr.

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In the case of a BINOMIAL RANDOM VARIABLE , the number of trials is fixed beforehand , and the BINOMIAL VARIABLE ‘X’ counts the number of successes in that fixed number of trials. If there are ‘N’ trials then the possible values of ‘ X’ are x =0,1,2,3,……….,n By way of comparison, there are situations in which the goal is to obtain a fixed number of successes. In particular, if the goal is to obtain one success, a random variable X can be defined that counts the number of trials needed to obtain that first success. A random variable that satisfies the above description is called a geometric, and the distribution produced by this random variable is called a geometric distribution. The possible values of a geometric random variable are 1, 2, 3, ..., that is, an infinite set, because it is theoretically possible to proceed indefinitely without ever obtaining a success. 4

    
 
 
 

 


  

Geometric Distribution : Definition : A geometric distribution is a discrete probability distribution of a random variable x that satisfies the following conditions. 1.) A trial is repeated until a success occurs. 2.) The repeated trials are independent of each other. 3.) The probability of success p is constant for each trail. 4.) The result of each trial is either success or failure.

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Geometric Distribution :

In a series of Bernoulli trials (independent trials with constant probability p of a success), let the random variable X denote the number of trial until the first success. Then ‘ X ’ is a geometric random variable with parameter 0 < p <1

g(x; p)= p(q)x-1

x= 1,2,3,......,n

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Rules for calculating…  If X has a geometric distribution

with probability p of success and (1−p) of failure on each observation, the possible values of X are 1, 2, 3, ...; if n is any one of these values, the probability that the first success occurs on the nth trial is: P(n) = p(q)n-1  When you want to know the

number of Bernoulli trials until a specified event occurs for the first time, use the geometric distribution.

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 Graphical Behaviour 8

    
 
 
 

 


  

Geometric Distribution : Geometric Distribution can be considered the discrete analog of exponential distribution Geometric Distribution is also the  special case of the Negative Binomial Distribution b*(x; k, p) = ( x-1Ck-1)pk q x-k x= k, k+1,k+2,...... with k = 1 .  g(x;1,p) = p(q)x-1 x= 1,2,3,......,n  The range of geometric random variable x is : R (x) = {0,1,2,3,....∞} 

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Distribution Functions !!! PDF : g(x; p) = P(X=x)= p(q)x-1

x= 1,2,3,......,n

CDF : P(X ≤ x) = P(X = 1) + P(X = 2) · · · P(X = x) = p + qp + q2p · · · + qx−1p = p[1 − qx]/(1 − q) = 1 − qx

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Geometric distribution : Mean

µ

Characteristics Of Geometric Distribution Variance

σ2

Standard Deviation

σ

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Characteristics … Mean (µ) : E(x) = µ = 1/p

Variance (σ 2) : V(x) = σ 2 = (1-p)/p 2

Standard Deviation (σ) : S.D = σ = √(q)/ p 2

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Applications ...

 In applications, the Geometric Distribution is often used to model uncertain frequencies, such as frequency of claims on an auto insurance policy.

 Extensive use in modeling networks traffic Randomness exists - must deal with it

Area of mathematics highly related to computer networking

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---------------------------------------------------------------------------------------------A radio transmission tower is designed for a “50-year wind,” that is a wind velocity that has a 1/50 chance of being exceeded. What s the chance that it will be exactly 5 years before such velocity is exceeded (it will be exceeded in 5th year)? Soln:

g(5; 0.02)= (1/50)*(49/50)5-1= .018 16

    
 
 
 

 


  

From experience, you know that the probability that you will make a sale on any given telephone call is 0.23. Find the probability that your first sale on any given day will occur on your fourth or fifth sales call. Solution To find the probability that your first sale will occur on the fourth or fifth call, first find the probability that the sale will occur on the fourth call and the probability that the sale will occur on the fifth call. Then find the sum of the resulting

probabilities. a) Using p=0.23, q=0.77, and x= 4 you have P(x=4)=(0.23)*(0.77)3 = 0.105003 b) Using p=0.23, q=0.77 and x=5 you’ve P(x=5)= (0.23)*(0.77)4 = 0.080852

So, the probability that your first sale will occur on the fourth or fifth sale call is P(sale on the fourth or fifth call) = P(x=4) + P(x=5) =0.105003+0.080852 =0.186 17

    
 
 
 

 


  

Summary :  1.) A trial is repeated until a success occurs.  2.) The repeated trials are independent of each

other.  3.) The probability of success p is constant for each trail.  4.) The result of each trial is either success or failure. NOTE : Geometric Distributions are frequently called “ waiting time “ simulations because you continue to conduct trial until success is observed !

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T H A N K Y O U !!!

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