When there are:

(i) The number of attempts is infinitely bigi.e., n ->âˆž

(ii) Constant probability of successful attempts is distinctively small i.e., p ->0,

(iii) np= Â a finite value.

Thus from the above points it is concluded that the Poisson distribution is acquired as a limiting incident of binomial distribution.

Therefore the probability of mass function is:

P [X = x]= e^{-Î»}Î»^{x}/x!, x=0,1,2,â€¦â€¦

Here A= parameter of this distribution.

**Proof.**

Let np = Ã°Â P=Â /n,Â q=1-/n

When n->âˆž

n!/ (n-x)!.x! = (n-k+1)(n-k+2)â€¦n/n^{k} = 1

lm(1- Î»Â /n)^{n-x} = (n-x) (-Î»/n) = -Î»

Therefore,

B (n, p) = e^{-Î»}Î»^{x}/x!,

**Properties**

(i)Â Â Â Â Â Â Â Î£^{âˆž}^{Â }_{x=0}Â Â P[X =x] = 1

(ii)Â Â Â Â Â Â Distribution function

F(x) = P[X <x] = e^{-Î»} .)Â Â Â Â Â Â Â Î£^{x}^{Â }_{k=0}Â Â Î»^{k}/ k!, x= 0,1,2â€¦..

(iii)Â Â Â Â Â Â First two moments about origin.

Âµâ€™_{1}= Î£^{âˆž}^{Â }_{x=0}Â x P[X=x] = Î£^{âˆž}^{Â }_{x=0}Â Â x e^{-Î»}Î»^{x}/x! = e^{-Î»}Î»^{x}. Î£^{âˆž}^{Â }_{x=1}Î»^{x-1}/x-1! = e^{-Î» }Î»e^{Î»} =Â Î»

When it is not an integer,

Mode = integral part

When it is an integer,

Mode = – 1

(i)Â In the queuing theory (that will be discussed in part B), it is said that under certain circumstances, it can be assumed that the probability of arrival of x customers in time t iswhich describes the Poisson distribution that consists of variables like a, t. This is exactly what called the Poisson process. As it consists the number of customers coming within a specific time t and is all that describes the Poisson distribution.

**Example 1.**If in a book of 520 pages there are 150 misprints. Then find out the probability that a single page will consist of maximum 2 misprints.

**Solution:** Here we knowÂ =150/520

Now following the poisons theory, we get,

Required probability= P[X â‰¤ 2]

=Â P[X=0] + P[X=1] + P[X=2]

=0.9968.

**Â **

**Example 2.** Let us assume the chances that a person who suffers from bad reaction from an injection is 0.001. So calculate the probability that out of 3000 peoples what would be the number that only two people will undergo through bad reactions, provided when:Â (a) exactly 3, (b) more than 2 individuals will suffer a bad reaction?

**Solution.**Â Â HereÂ we know from the above question isÂ =3000 * 0.001 = 3

(a)Â Â Â Required probability= P[X = 3] = 27.e^{-3}/6= 0.22

(b) Required probability = P[X > 2] = 1 – P[X â‰¤ 2]

=1-(P[X = 0] + P[X = 1] + P[X = 21]) = 1- 17e^{-3}/2 = 0.58

**Example 3.**Calculate the probability of taking out 2 defective pieces from a bundle of 100 items provided that all the manufacturing system contains is 0.2% defect. Calculate by using:

(a) Binomial distribution

(b) Poisson approximation.

**Solution.** (a) p = Probability of defective = 0.002, q = 1 – p = 0.998

n = 100

Probability of finding 2 or more defective items

=1 – [Probability of zero and one defective]

=1- (P[X = 0] + P[X = 1])

=1-[(0.998)^{100} + 100 (0.002) (0.998)^{99}]

=1- 0.983 = 0.017

(b) Here Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = np = 100 x 0.002 = 0.2

Probability of finding 2 or more defective items

=1-[P[X=O] + P[X=1]]

=1 – [e^{-0.2} + (0.2) e^{-0.2}]

=1 – 0.982 = 0.018.

**Example 4.**Choose any suitable Poisson distribution for the following set of observations:

X | 0 | 1 | 2 | 3 | 4 |

f | 57 | 41 | 28 | 8 | 1 |

**Solution.Â Â Â Â Â Â Â Â **

Mean=Â =125/135=Â 0.926

The mean of the Poisson distribution isÂ = 0.926

Therefore the expected frequencies are given by

- e
^{-0.926}.(0.926)^{x}/x!, x= 0,1,2,3,4

Therefore the fitted distribution is given by

x | 0 | 1 | 2 | 3 | 4 |

Expected f | 53 | 50 | 23 | 7 | 2 |

**PROBLEMS**

1.Calculate the P[X > 2] taking X as the random variable of the Poisson distributionwhen, P[X = 1] = 3P[X = 2].

- From a large number of medicine stock, there were only 100 batches were being supplied. Amon these medicines total 50 items was marked as defective. So calculate the probability that each batch has: (a) no defective item; (b) at least three defective items.
- If 2 per cent of electric bulbs factory-made by a certain company are not working, then find the probability that in a set of 200 bulbs (a) less than 2 bulbs are defective; (b) more than 3 bulbs are defective.
- Let X follows Poisson distribution, find the value of the mean of the distribution of P[X = 1]= 3P[X = 2].
- In a certain factory, the blades are manufactured in a set of 10 packets. There is a probability that 0.1% of the blade to be defective. Calculate approximatelyUsing Poisson distribution the number of packets that contains two defective blades in a package of 10000 packets.
- Fit a Poisson distribution to the following:

x | 0 | 1 | 2 | 3 | 4 | 5 |

f | 20 | 16 | 11 | 7 | 4 | 2 |

- Fit a Poisson distribution to the following:

x | 0 | 1 | 2 | 3 | 4 | 5 |

f | 120 | 82 | 52 | 22 | 4 | 0 |

8.Calculate Using the Poisson distribution to determine the probability that among 20000 cars driven over this bridge, not more than one will have a flat tire when the Records show that the probability of a car to get a flat tire while driving over a certain bridge is 0.00002.

- A typ1st kept a record of mistakes made per day during 300 working days of a year:

Mistakes per day | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

No. of days | 143 | 90 | 42 | 12 | 9 | 3 | 1 |

Fit an appropriate Poisson distribution to the data.

- The probability that a Poisson variant X takes a positive value is ( 1 â€“ e
^{-1.5}). Find the variance and also the probability that X lies between -1.5 and 1.5. - A manufacturer, who produces medicine bottles, packs 500 bottles in a box and finds 0.1% of the bottles are defective. Now a drug manufacturer bought 100 boxes from the manufacturer of bottles. Calculate Using Poisson distribution that how many boxes will contain (i) no defectives, (ii} at least two defectives.
- What is the probability that in a company of 1000 people only one person will have birth day on New Yearâ€™s Day? (Assume that a year has 365 days).

**Â **

**ANSWERS**

**Â **

- 0.0302Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2. (a) 0.61Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 0.01
*(a)*0.09 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 0.58Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â**4.**2/3- Two packets

(The EF corresponding to x = 1, has been adjusted to make âˆ‘ EF = 60)

7.

x | 0 | 1 | 2 | 3 | 4 | 5 |

EF | 108 | 102 | 49 | 16 | 4 | 1 |

2

- 0.9380

x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

EF | 123 | 110 | 49 | 14 | 3 | 1 | 0 |

- Variance = 1.5, P(-1.5 <. X < 1.5) = 2.5 e
^{– l.5} - (i) 61Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (ii) 9 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 12. 0.18.

**Links of Previous Main Topic:-**

- Introduction to statistics
- Knowledge of central tendency or location
- Definition of dispersion
- Moments
- Bivariate distribution
- Theorem of total probability addition theorem
- Random variable
- Binomial distribution

**Links of NextÂ Statistics Topics:-**

- What is sampling
- Estimation
- Statistical hypothesis and related terms
- Analysis of variance introduction
- Definition of stochastic process
- Introduction operations research
- Introduction and mathematical formulation in transportation problems
- Introduction and mathematical formulation
- Queuing theory introduction
- Inventory control introduction
- Simulation introduction
- Time calculations in network
- Introduction of game theory