The law of distribution of Poisson
A discrete random variable X has the law of distribution of Poisson with the parameter l > 0 if it takes on values 0, 1, 2, …, m, …(infinite countable set of values) with probabilities
The series of distribution of the Poisson law has the following form:
| xi | … | m | … | |||
| pi | e-l | le-l | l2e-l/2! | … | lme-l/m! | … |
Since the sum of the series
the basic property of distribution series holds, and consequently the Poisson law is well-defined.
The Poisson probability distribution was introduced by S.D. Poisson in a book he wrote regarding the application of probability theory to lawsuits, criminal trials, and the like.
Theorem. The mathematical expectation and the dispersion of a random variable distributed under the Poisson law coincide and are equal to the parameter l of the law, i.e. M(X) = l, D(X) = l.
The Poisson random variable has a tremendous range of applications in diverse areas because it may be used as an approximation for a binomial random variable with parameters (n, p) when n is large and p is small enough so that np is a moderate size. In other words, if n independent trials, each of which results in a success with probability p, are performed, then, when n is large and p small enough to make np moderate, the number of successes occurring is approximately a Poisson random variable with parameter l = np.
Some examples of random variables that usually obey the Poisson probability law follow:
1. The number of misprints on a page (or a group of pages) of a book.
2. The number of people in a community living to 100 years of age.
3. The number of wrong telephone numbers that are dialed in a day.
4. The number of packages of dog biscuits sold in a particular store each day.
5. The number of customers entering a post office on a given day.
6. The number of vacancies occurring during a year in the federal judicial system.
7. The number of a-particles discharged in a fixed period of time from some radioactive material.
Each of the preceding, and numerous other random variables, are approximately Poisson for the same reason – namely, because of the Poisson approximation to the binomial. For instance, we can suppose that there is a small probability p that each letter typed on a page will be misprinted. Hence the number of misprints on a page will be approximately Poisson with l = np, where n is the number of letters on a page.
Example. Suppose that the number of typographical errors on a single page of this book has a Poisson distribution with parameter l = 1/2. Calculate the probability that there is at least one error on this page.
Solution: Letting X denote the number of errors on this page, we have
.
Example. Suppose that the probability that an item produced by a certain machine will be defective is 0,1. Find the probability that a sample of 10 items will contain at most 1 defective item.
Solution: The desired probability is
whereas the Poisson approximation yields the value
Geometric distribution
A discrete random variable X has the geometric distribution with the parameter p if it takes on values 1, 2, …, m, … (infinite countable set of values) with probabilities
P(X = m) = pqm–1
where 0 < p < 1, q = 1 – p.
The series of the geometric distribution has the following form:
| xi | … | m | … | |||
| pi | P | Pq | pq2 | … | pqm-1 | … |
It is easy to see that the probabilities pi form the geometric progression with the first member p and denominator q (therefore, the law is said to be geometric).
Since the geometric distribution is well-defined.
A random variable X having the geometric distribution represents the number m of trials which have been carried out under Bernoulli circuit with probability p of occurrence of the event in each trial till the first positive outcome.
Theorem. The mathematical expectation of a random variable X having the geometrical distribution with the parameter p is M(X) = 1/p, and its dispersion D(X) = q/p2 where q = 1 – p.
Example. Testing a big batch of details up to detection of a rejected detail (without restriction of the number of tested details) is carried out. Compose the law of distribution of the number of tested details. Find its mathematical expectation and dispersion if it is known that the probability of reject for each detail is equal to 0,1.
Solution: The random variable X – the number of tested details up to detection of rejected detail – has geometrical distribution with the parameter p = 0,1. Therefore, the series of distribution has the following form:
| xi | … | m | … | ||||
| pi | 0,1 | 0,09 | 0,081 | 0,0729 | … | 0,9m -1 × 0,1 | … |
M(X) = 1/p = 1/0,1 = 10; D(X) = q/p2 = 0,9/(0,1)2 = 90.
Hypergeometric distribution
Hypergeometric distribution is widely used in practice of statistical acceptance control by quality of industrial production, in the problems connected to the organization of sampling inspections, and other areas.
A discrete random variable X has the hypergeometric distribution with the parameters n, M, N if it takes on values 0, 1, 2, …, m, …, min(n, M) with the probabilities
where M £ N, n £ N; n, M, N are natural numbers.
Let there be M standard details in a batch of N details. From the batch one randomly select n details (each detail can be extracted with the same probability), and the selected detail is not replaced in the batch before selection of the next detail (therefore Bernoulli formula here is not applicable). Then the random variable X which is the number m of standard details among n selected details has hypergeometric distribution.
Theorem. The mathematical expectation of a random variable X having the hypergeometric distribution with the parameters n, M, N is and its dispersion .
Example. In a lottery «Sportloto 6 of 45» monetary prizes are received by participants who have guessed 3, 4, 5 and 6 kinds of sports from randomly selected 6 kinds of 45 (the size of a prize increases with an increasing the number of guessed kinds of sports). Find the law of distribution of a random variable X – the number of guessed kinds of sports among randomly selected 6 kinds. What is the probability of receiving a monetary prize? Find the mathematical expectation and the dispersion of the random variable X.
Solution: Obviously, the number of guessed kinds of sports in the lottery “6 of 45” is a random variable having hypergeometric distribution with the parameters n = 6, M = 6, N = 45. The series of its distribution has the following form:
| xi | 5 | ||||||
| pi | 0,40056 | 0,42413 | 0,15147 | 0,02244 | 0,00137 | 0,00003 | 0,0000001 |
The probability of receiving a monetary prize
M(X) = n × M/N = 6 × 6/45 = 0,8; D(X) = 6 × 39/44 (1 – 39/45)(1 – 6/45) = 0,6145.
Glossary
binomial– биномиальный; Poisson– Пуассон
lawsuit– судебный процесс; tremendous – огромный
moderate– небольшой, доступный; diverse areas – разнообразные области
particle – частица; to discharge – разряжать
circuit – схема; hypergeometric – гипергеометрический
Exercises for Seminar 9
9.1. A die is tossed three times. Write the law of distribution of the number of appearance of 6.
9.2. Find an average number (mathematical expectation) of typing errors on page of the manuscript if the probability that the page of the manuscript contains at least one typing error is 0,95. It is supposed that the number of typing errors is distributed under the Poisson law (typing error – опечатка; an average number – среднее число).
The answer: 3.
9.3. The switchboard of an enterprise serves 100 subscribers. The probability that a subscriber will call on the switchboard within 1 minute is equal 0,02. Which of two events is more probable: 3 subscribers will call or 4 subscribers will call within 1 minute? (Subscriber – абонент, switchboard – коммутатор).
9.4. A die is tossed before the first landing «six» aces. Find the probability that the first appearance of «six» will take place:
(a) at the second tossing the die;
(b) at the third tossing the die;
(c) at the fourth tossing the die.
The answer: (a) 5/36.
9.5. Suppose that a batch of 100 items contains 6 that are defective and 94 that are non-defective. If X is the number of defective items in a randomly drawn sample of 10 items from the batch, find (a) P(X = 0) and (b) P(X > 2).
9.6. There are 7 standard details in a set of 10 details. 4 details are randomly taken from the set. Find the law of distribution of the random variable X equal to the number of standard details among the taken details.
9.7. An urn contains 5 white and 20 black balls. 3 balls are randomly taken from the urn. Compose the law of distribution of the random variable X equal to the number of taken out white balls.
9.8. At horse-racing competitions it is necessary to overcome four obstacles with the probabilities equal 0,9; 0,8; 0,7; 0,6 respectively. At the first failure the sportsman in the further competitions does not participate. Compose the law of distribution of a random variable X – the number of taken obstacles. Find the mathematical expectation of the random variable X (obstacle – препятствие).
The answer: M(X) = 2,4264.
9.9. Two shooters make on one shot in a target. The probability of hit by the first shooter at one shot is 0,5, and by the second shooter – 0,4.
(а) Find the law of distribution of the random variable X – the number of hits in the target;
(b) Find the probability of the event X ³ 1.
The answer: b) 0,7.
9.10. A set of families has the following distribution on number of children:
| xi | x1 | x2 | ||
| pi | 0,1 | p2 | 0,4 | 0,35 |
Determine x1, x2, p2, if it is known that M(X) = 2, D(X) = 0,9.
Exercises for Homework 9
9.11. Compose the law of distribution of probabilities of the number of appearances of the event A in three independent trials if the probability of appearance of the event is 0,6 for each trial.
9.12. Let X be a random variable equal to the number of boys in families with five children. Assume that probabilities of births of both boy and girl are the same. Find the law of distribution of the random variable X. Find the probabilities of the following events:
(a) there are 2-3 boys in a family;
(b) no more than three boys;
(c) more than 1 boy.
The answer: a) 5/8; b) 13/16; c) 13/16.
9.13. A factory has sent 5000 suitable details to its warehouse. The probability that a detail is broken during a transportation is 0,0002. Find the probability that 3 non-suitable details will be arrived at the warehouse.
The answer: 0,06.
9.14. Suppose that the number of accidents occurring on a highway each day is a Poisson random variable with parameter l = 3.
(a) Find the probability that 3 or more accidents occur today.
(b) Repeat part (a) under the assumption that at least 1 accident occurs today.
The answer: a) 0,577; b) 0,627.
9.15. A hunter shoots on a game before the first hit, but he managed to make no more than four shots. The probability of hit by him at one shot is 0,9.
(а) Find the law of distribution of a random variable X – the number of misses;
(b) Find the probability of the following events: X < 2, X £ 3, 1 < X £ 3 (hunter – охотник; game – дичь).
The answer: b) 0,99; 0,9999; 0,0099.
9.16. There are 11 students in a group, and 5 of them are girls. Compose the law of distribution of the random variable X – the number of girls from randomly selected three students.
9.17. There are 8 pencils in a box, and 5 of them are green. 3 pencils are randomly taken from the box.
(a) Find the law of distribution of a random variable X equal to the number of green pencils among taken.
(b) Find the probability of the event: 0 < X £ 2.
The answer: b) 45/56.
9.18. There are 20 products in a set, and 4 of them are defective. 3 products are randomly chosen for checking their quality. Compose the law of distribution of a random variable X – the number of defective products contained in the sample.
9.19. The probability of successful passing an exam by the first student is 0,7, and by the second – 0,8. Compose the law of distribution of a random variable X – the number of the students successfully passed the exam if each of them can retake only once the exam if he didn’t pass it at the first time. Find the mathematical expectation of the random variable X.
The answer: M(X) = 1,87.
9.20. A discrete random variable X is given by the following law of distribution:
| xi | X2 | x3 | ||
| pi | 0,1 | P2 | 0,5 | 0,1 |
Determine x2, x3, p2, if it is known that M(X) = 4, M(X 2) = 20,2.
The answer: x2 = 2; x3 = 6 or x2 = 7; x3 = 3.
L E C T U R E 10