MA305 Mathematics for ICE 1

MA305Binomial Distribution and

Poisson DistributionBy: Prof. Nutan Patel

Asst. Professor in MathematicsIT-NUA-203

patelnutan.wordpress.com

MA305 Mathematics for ICE 2

Binomial DistributionDefinition:

Let X be the random variable for a binomial distribution with n repeated trials, with p the probability of success, q the probability of failure and P(X=r)=, r= 0, 1, 2, 3,…, n.With Mean = np.

Variance = npq.Standard Deviation =

MA305 Mathematics for ICE 3

Ex. Team A has probability 2/3 of winning whenever it plays. If A plays 4 games, find the probability that A wins (i) exactly 2 games (ii) at least 1 game (iii) more than half of the games.Ans: n=4, p=2/3, q=1/3.i. P(2)=8/27.ii. 80/81.iii. P(3)+P(4)=16/27.

MA305 Mathematics for ICE 4

Ex: In sampling a large number of parts manufactured by a machine, the mean number of defectives in a sample of 20 is 2. out of 1000 such samples, how many would be expected to contain at least 3 defective parts?Ans: n=20, p=0.1, P(X≥3)=1-{ P(0) + P(1) + P(2) }

= 1- { 0.12157 + 0.27017 + 0.28517 } = 0.323.Expected number=1000*0.323=323.

MA305 Mathematics for ICE 5

Ex: If X be binomially distributed random variable with E(X)=4.5 and Var(X)=1.125, find the distribution of X.Ans: E(X)=mean=np=4.5,Var(X)=npq=1.125.Now,

Hence, The Binomial distribution is X P(X=x) P(X)0 0.0002441 0.0043952 0.0329593 0.1318364 0.2966315

0.3559576

0.177979 0 1 2 3 4 5 60

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

P(r)

MA305 Mathematics for ICE 6

Poisson DistributionBinomial Distribution is used when p and n are known, but when p is

small (p ≤ 0.1) and n is very large or n is infinite, the use of Binomial distribution is not logical,

In Such situations, we use the limiting form of Binomial distribution which is known as Poisson distribution.

Poisson distribution is the discrete probability distribution of a discrete random variable, which has no upper bound. It is defined for non-negative values of x as follows:

, for x= 0, 1, 2, 3, …Here, is called the parameter of Poisson distribution.*Simmeon Denis Poisson (1781-1840) French Mathematician.

MA305 Mathematics for ICE 7

• Note that in binomial distribution the number of successes out of a total definite number of n trials is determined, whereas in Poisson Distribution (P.D.) the number of successes at random point of time and space is determined.• P.D. is suitable for ‘rare’ events for which the probability of occurrence

p is very small and the number of trials n is very large.• In B.D. when and such that = np = constant.𝜆

MA305 Mathematics for ICE 8

Application of Poisson DistributionExample of rare events:Number of accidents on a highway.Number of printing mistakes per page. Number of deaths per day or per week due to a rare disease in a big

city.Number of defectives in a production centre.The count of Bacteria per c.c. in blood.

9

Properties of Poisson Distribution P. D. is a distribution of discrete random variable.In P. D. mean(𝜇) = variance () = . Hence its standard deviation () is . 𝜆

This is the acid test to be applied to any data which might appear to conform to P. D..

Recurrence relation between successive probabilities: , where x= 1, 2, 3, … , .

MA305 Mathematics for ICE

MA305 Mathematics for ICE 10

Ex: A certain screw making machine produces on an average 2 defective screws out of 100, and packs them in boxes of 300 (25 dozen). find the probability that a box contains 5 defective screws.Ans: Probability that a bulb is defective = p = 0.02.Now in a box, there are 300 bulbs. So, n=300.So, =np =0.02*300=6.𝜆By Poisson Distribution, the probability that x bulbs are defective is and =6.𝜆 =0.1606.

MA305 Mathematics for ICE 11

Ex: 40 electrical bulbs are found to be defective in a lot of 2000 bulbs. Find the probability that at the most 3 bulbs are defective in a box of 200 bulbs.Solution: Probability that a bulb is defective = p = 0.02.Now in a box, there are 200 bulbs. So, n=200.So, =np =0.02*200=4𝜆By Poisson Distribution, the probability that x bulbs are defective is and =4.𝜆For, at the most 3 bulbs defective x= 0 or 1 or 2 or 3.P(X ≤3) = P(0) + P(1) + P(2) + P(3)

= + + + =(1+4+8+10.6667)=0.01832 (23.6667)=0.4335

MA305 Mathematics for ICE 12

• EX: If 10% of the rivets produced by a machine are defective, find the probability that out of 5 rivets chosen at random (i) none will be defective, (ii) one will be defective, and (iii) at least two will be difective.• Ans: 0.5905, 0.32805, 0.08146.

MA305 Mathematics for ICE 13

Ex: if X is Poisson Distribution with P(x = 2) =0.25 and P(x=3)=0.125,Find P(x=0), P(x=1).Ans: The reccurence relation

,

so, =1.5,𝜆Hence,

MA305 Mathematics for ICE 14

• Comparison for Binomial and Poisson DistributionEx: Suppose 3% of bolts made by a machine are defective, the defects occurring at random during production. If bolts are packaged 50 per box, find probability that a given box will contain 5 defectives bolts by Binomial and Poisson Distribution.Solution: p=0.03, n=50, x=5,By Binomial Distribution: By Poisson Distribution: =np=0.03*50=1.5𝜆

MA305 Mathematics for ICE 15

Graphical Comparison for Binomial and Poisson DistributionEx: For n=100, p=0.04, so, =4.𝜆

x B.D. P(x) P.D. P(x)

0 0.01687 0.018316

1 0.070293 0.073263

2 0.144979 0.146525

3 0.197333 0.195367

4 0.199388 0.195367

5 0.159511 0.156293

6 0.105233 0.104196

7 0.05888 0.05954

8 0.02852 0.02977

9 0.012147 0.013231

10 0.004606 0.005292

0 1 2 3 4 5 6 7 8 9 100

0.05

0.1

0.15

0.2

0.25

B.D. v/s P.D.

B.D. P(x) P.D. P(x)

0 2 4 6 8 10 120

0.05

0.1

0.15

0.2

0.25

B.D. v/s P.D.

B.D. P(x) P.D. P(x)