1. STAT 111 Chapter 5 Special Probability Distributions 1

2. Often, the observations generated by different statistical experiments have the same general type of behavior. Consequently, discrete random variables associated with these experiments can be described by essentially the same probability distribution and therefore can be represented by a single formula. In fact, one needs only a handful of important probability distributions to describe many of the discrete random variables encountered in practice

3. The Discrete Uniform Distribution The simplest of all discrete probability distributions is one in which the random variable assumes each of its values with an equal probability. Such a probability' is called the discrete uniform distribution. 3

4. Definition (l)If the random variable X assumes the values x 1,x 2,...,x n (finite number of values), ( 2) with equal probabilities, then the discrete unifom distribution is given by 4 f (x) = P(X = x) = x = x 1, x 2,...,x n 1 n

5. Example when a fair die is tossad, each element of the sample space {1,,2,….,6} occurs with probability. therefore, we have a uniform distribution with 5 16 16 f(x) = x=1,2,………,6 16 16

6. Example let X uniformly distributed on 0,1,...,9. Calculate: 1. 2. 3. 4. 6 P(X≤2) 1 + 1 + 1 = 3 10 10 P(1<X<4) P(4<X<6 or 8<X ≤12) P(5≤X ≤ 9) = P(X=2)+P(X=1)+(X=0) = = P(X=2)+P(X=3)= 1 + 1 = 2 = 1 10 10 10 5 =P(4<X<6)+P(8<X ≤12) = 1 + 1 = 1 10 10 5 = 5 = 1 10 2

7. The Bernoulli Distribution An experiment of a particularly simple type is one in which there are only two possible outcomes, such as head or tail, success or failure, defective or not defective, It is convenient to designate the two possible outcomes of such an experiment as 0 and 1. 7

8. The following definition can then be applied to any this type. A random variable X has a Bernoulli distribution with parameter p (0≤ p ≤1) if X can take only the values 0 and 1 and the probabilities P(X=1)= p and P(X=0)=l-p If we let q=1-p, then the p.m.f of X can be written as follows; Note that that parameter of the Bernoulli distribution is p. f(x)= p x q 1-x x=0,1 0 otherwise {

9. Binomial Distribution A random variable X has a binomial distribution with parameters n and p, if X has a discrete distribution for which the p.m.f. is as follows; In this distribution n must be a positive integer and p must lie in the interval 0<p<l. p x q n-x q = 1- p x = o,1,...,n f(x) = o otherwise nxnx

10. The binomial distribution is of fundamental importance in probability and statistics because of the following result; Suppose that the outcome of an experiment can be either success or failure; that the experiment is performed n times independently; and that the probability of the success of any given trial is constant and is equal to p. If X denotes the number of successes among the n trials, then X has a binomial distribution with parameters n and p. We will often write X~ bin(n,p) to indicate that X is a binomial random variable based on n trials with success probability p. Note that the binomial distribution reduces to the Bernoulli distribution when n=l. 10

11. f (x) = p x q n-x x= 0,1,…n In summary, if 11 1. تتكرر التجربة n من المرات المستقلة 1-The experiment is performed n times independently 2.نتائج التجربة نتيجتين فقط (النتيجة التى تهمنا ونبحث عن الاحتمالات لها هى النجاح والفشل) 2-The outcome of an experiment can be either success(S) of failure(F) probability of a success = p, Probability of failure = q=1-p, If these conditions are satisfied then nxnx

12. f(6)= (0.9) 6 (0.1) 4 = 0.0112 Example An agricultural scientist plants 10 seeds of a certain variety of hybrid corn(ذرة مهجن). Past experience leads him to believe that the probability of a given seed germinating انبات)) is 0.9. 12 1. What is the probability that exactly 6 seeds germinate? X= number of seeds germ. n=10, S=seed germ. F=seed do not germ. P=0.9, q= 1-0.9=0.1 10 6

13. 13 f (4) = (0.9) 6 (0.1) 4 =0.0112 10 4 X=seed do not germ. n=l 0, S=seed do not germ. F= seed germ. p=0.1, q=0.9 3. What is the probability' that 4 seeds do not germinate? 2. What is the probability that all the seeds germinate? f (10) = (0.9) l0 (0.1) 0 = 0.349 10

14. f (x) = ( k ) x (1 - k ) n-x N N x=0,1,2,…,n Remark Consider sampling with replacement from an um containing M balls, K of which are defective. Let X represents the number of defective balls in a sample of size n. The experiment of taking a sample of size n with replacement consists of n repeated independent trials where p=P(success)=(k/N); so X has the binomial distribution 14 K N-K (S) (F) n nxnx

15. The probability that a binomial random variable X is less than or equal to some specified value x is given by the cumulative distribution function 15 F(x)=P(X≤x)= Σ x p i q n-i i=0 nini The binomial distribution has been tabulated (Table 1 in the Appendix) for various values of n and p by using the cumulative probability for samples of size n = 1,2,...20. and selected values of p from 0.1 to 0.9. We can also determine individual probabilities using this table since the binomial random variable is integer-valued. and thus the property f(x)=P(X=x)=F(x)-F(x-l)

16. Example 16 The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that 1-At least 10 survive? X=number of people that survive P(X ≥ 10) = 1 - P(X < 10) = 1 – P(X ≤ 9) = 1 – F(9) = 1 - 0.9662 = 0.0338 2-From 3 to 8 survive? P(3 ≤ X≤8) = F(8) - F(2) = 0.9050 -0.0271 = 0.8779 3-Exactly 5 survive? P(X=5)= F(5)-F(4)=0.4032-0.2173=0.1862 or P(X=5)=f(5) = (0.4) 5 (0.6) 10 = 0.1862 15 5

17. Example 17 cars coming to a dead-end intersection can turn either left or right. at successive cars choose a turning direction independently of one another and that p (left)=0.7. 1- Among the next 15 cars, what is the probability that at least 10 turn left? X= number of cars which turn left, n=15, S= cars which turn left, p=0.7 X " = cars which turn right, q=0.3 P(X≥ 10) = 1 - P(X < 10) = 1 - P(X ≤ 9) = 1 - F(9) = 1 - 0.278 = 0.722 ________________________________ 2.among the next 15 cars, what is the probability' that at least 10 turn in the direction? Let X " = number of cars turn right, n=15, S=cars which turn right, p=0.3,q=0.7 P( at least 10 turn in the same direction)=P(at least 10 turn left or at least 10 turn right) = P(X≥10 or X"≥10) =P(X≥10)+P(X"≥10) = 0.004+0.722

18. The binomial experiment becomes a multinomial experiment if we let each trial have more that two possible outcomes. Hence the classification of a manufactured product as being light, heavy, or acceptable constitutes a multinomial experiment 18

19. ƒ( x 1,x 2,…,x k )= p 1 p 2... p k With Definition If a given trial can result in the k outcomes, E 1,E 2,…,E k with probabilities p 1,p 2,...,p k, then the probability distribution of the random variables X 1,X 2,...,X k representing the number of occurrences for E 1,E 2,..,E k in n independent trials is 19 n x 1, x 2.., x k x 1 x 2 x k k i i=1 ∑ x i = n and ∑ p i = 1 k

20. ƒ( 3,3,3 ) = × × Example 20 A balanced die has two faces colored red, three faces colored white, and one colored blue. If it is thrown 9 times, what is the probability that each color appears 3 times. 9 3, 3, 3 2626 3 3636 3 1616 3

21. Exercise A pair of dice is tossed 6 times, what is the probability of obtaining a total of 7 or 11 twice, a matching pair once, and any other combination 3 times؟ 21 Answer f ( 2, 1, 3 ) = 0.1127

22. The Hypergeometric Distribution 22 The types of applications of the hypergeometric are very similar to those of the binomial distribution. The difference between the binomial distribution and the hypergeometric distribution lies in the way the sampling is done. In the case of the binomial, independence among trials is required (sampling with replacement). On the other hand, the hypergeometric distribution does not require independence and therefore, is based on the sampling done without replacement.

23. K N-K n (S) (F) N Consider a set of N elements of which k are successes and the other N-k are failures. We are interested in the probability of getting x successes in n trials, without replacement, X=number of success 23

24. Definition A random variable X has a hypergeometric distribution, if and only if its probability distribution is given by 24 K N - K x n – x x = 0,1,2,…,n f(x)= p(X =x)= N n 0 otherwise

25. Note that f(x) is equal to zero when x>k or when ( n-x )>N-k. Applications for the hypergeometric distribution are found in many areas where testing is done at the expense of the item being tested. The item is destroyed and hence cannot be replaced in the sample. Thus sampling without replacement is necessary. 25

26. Example A jar contains 5 red and 10 green marbles. Seven are selected without replacement. What is the probability that exactly 3 red are selected? That exactly 6 red are selected? 26 5 10 3 4 15 7 f(3) = = 0.326, f(6) = = zero 5 10 6 1 15 7

27. Example A committee of size 5 is to be selected at random from 3 chemists and 5 physicists. Find the probability distribution for the number of chemists on the committee? 27 f(0) = = —, f(1) = = —, f(2) = —, f(3) = — 3 5 4 5 8 5 1 56 3 5 1 4 8 5 30 56 10 56 15 56 x0123 f(x)1/5615/5630/5610/56 The probability distribution for X

28. When N is large and n is relatively small compared to N, the probability for each drawing will change only slightly and hence there is not much difference between sampling with replacement and sampling without replacement, and the formula for the binomial distribution with the parameters n and p = — may be used to approximate hypergeometric probabilities. 28 KNKN

29. Example 29 Among the 120 applicants for a job only 80 are actually qualified. If 5 of these applicants are randomly selected for an interview, find the probability that only 2 of the 5 will be qualified for the job by using 1. The hypergeometric distribution 80 40 2 3 120 5 X=2, n=5, N=120,k=80 f(2)= = 0.164 2.The binomial distribution as an approximation. 2 3 5 2 1 2 3 3 f(2) = — — = 0.165 As can be seen these results are close.

30. The hypergeometric distribution can be extended to treat the case where the N items can be partitioned into k cells A 1,A 2,...A k, with a 1 elements in the first cell, a 2 elements the second cell,..., a k elements in the kth cell. 30

31. Dethfinition (Multivariate Hypergeometric Distribution) 31 If N items can be partitioned into the k cells A 1,A 2,...A k, with a 1, a 2, ….., a k elements,, then the probability distribution of the random variables X 1,X 2,...X k, representing the number of element selected from A 1,A 2,...A k in a random sample of size n, is a 1 a 2 a k x 2 x 2 … x k f(x 1,x 2,…..,x k ) = N with n ∑ x i = n and ∑ a i = N k k i=1

32. Example A group of 10 individuals is being used in a biological case study. The group contains 3 people with blood type O 4 with blood type A, and 3 with blood type B. what is the probability that a random sample of 5 will contain 1 person with blood type O, 2 people with blood type A and 2 people with blood type B. 32 3 4 3 1 2 2 3 f(1,2,2) = = 10 14 5

33. The Geometric Distribution 33 Suppose that independent trials, each having a probability p, 0<p<l, being a success, are performed until the first success occurs. If we let X equal the number of trials required, then sample space=(S,FS,FFS,...), X= number of trials, and P(X=l)= P(S) = p P(X=2)= P(FS) = P(F)P(S) =(l-p)p P(X=3)= P(FFS) = (l-p) 2 p In general, P(X =k)= P(FF...FS) = (1 -p) k -1 p The probability of this intersection of k independent events gives the geometric probability distribution (note that the geometric distribution is well named since the values of the geometric probability mass function are the terms of geometric series).

34. Definition 34 Let 0<p<l. Then the real valued function f defined on R by p(1-p) x-1 x=1,2,… f(x)= P(X=x)= 0 otherwise Is called the geometric probability distribution. X ~ G(p) will mean that X has the Geometric distribution with parameter p

35. 35 A comparison of the binomial and geometric distribution will show that, although both serve as models for experiments involving independent, repeated Bernoulli trials, in the binomial case there is a fixed number of trials, and the random variable represents the number of successes which occur, In the case of the geometric, there is no fixed number of trials. The random variable represents the number of trials until the first success

36. In summary 36 The binomial Distribution 1.Independent trials 2.number of trials is n 3.Two outcome (success or failure) 4.X= number of successes among the n trials The Geometric Distribution 1.Independent trials 2.لا يوجد عدد محدود لتكرار التجربة 3.Two outcome(success or failure) 4.X= number of trials until the first success عدد مرات حصول التجربة حتى الحصول على اول نجاح

37. 37 Example Find the probability that a person flipping a balanced coin requires 4 tosses to get a head. X ~ G (0.5) f (4) = 0.5 (0.5) 3 = 1 16 Example In certain manufacturing process it is known that 1 in every 100 items is defective. What is the probability that the fifth item inspected is the first defective item found X ~ G (0.01) f(5) = 0.01 ( 0.99) 4 = 0.0096

38. Theorem 38 Let X be G(p) and k a nonnegative integer. Then P(X>k)=(l-p) k (1) From which, P(X ≥ k) = P(X>k-1 ) = ( 1 –p) k-1 (2) From (1) F(k)= P(X ≤ k) = 1 -P(X>k) = 1 -(1 -p) k From (2) P(X<k)= l-P(X ≥ k) = l-(l-p) k-1

39. Example 39 An urn contains 10 white and 8 black balls. Balls are randomly selected, one at a time, until a black one is obtained. if we assume that each selected ball is replaced before the next one is drawn, what is the probability that 1. Exactly 5 draws are needed. X = number of draw's needed to select a black ball P = 8 X~G( 8 ) 18 4 18 P(X=5)= 10 8 18 18 2. At least 4 draws are needed P (X ≥4) = (1 - 8 ) 3 = 10 3 18 18

40. The Negative Binomial Distribution 40 A random variable with a negative binomial distribution originates from a context that is very similar to the one that leads to the geometric distribution. A gain, we focus on independent and identical trials, each of which results on one of two outcomes, success or failure. The probability of success is p and stays the same from trial to trial. The geometric distribution handles the case where we are interested in the number of the trial on which the first success occurs. The negative binomial distribution is used when we are interested in the number of trial on which the rth success occurs (r=2,3,,..).

41. Definition 41 A random variable X is said to have a negative binomial probability distribution if and only if x - 1 r - 1 p r (1 –p ) x- r x=r, r+1,r+2,.. f(x)=p(X=x)= 0≤ p ≤1 0 otherwise X~b*(r,p) will mean that X has the negative binomial distribution with parameters r and p.

42. 42 Example The probability that a person living in a certain city owns a car is 0.3. Find the probability that the 10 th person randomly interviewed in this city is the 5th one to own a car? 9 X~b * (5,0.3) f(10)= 4 (0.3) 5 (0.7) 5 = 0.0515 Example If the probability is 0.40 that a child exposed to a certain disease will cach it, what is the peobability that the tenth child exposed to the disease will be the third to cach it? 9 X~b * (3,0.4) f(10) = 2 (0.4) 3 (0.6) 7 = 0.0645

43. The Poisson Distribution 43 The Poisson distribution (named after Simeon Poisson, the nineteenth century French probabilities who described it) is another extremely useful discrete probability distribution in which the random variable represents the number of occurrences of independent events that take place over a given time interval or in a specified region. The given time interval may be of any length, such as a minute, a day, a week, a month, or even a year. Typical examples are the number of arrivals to a service facility in a given time, the number of telephone calls per hour received by an office, the number of days school is closed due to snow during the winter. The specified region could be a line segment, an area, a volume, or perhaps a piece of material. In this case X might represent the number of bacteria in a given culture, or the number of typing errors per page,

44. Definition 44 Let X be a random variable with a discrete distribution, and suppose that the value of X must be a nonnegative integer. It is said that X has a Poisson distribution with mean λ(λ>0) if the p.m.f of X is as follows: f(x) = P(X = x) = e –λ λ x x=0,1,2,…. x! 0 otherwise The parameter of the Poisson distribution is λ, the average number of occurrences of the random event per unit of time. X~P(λ) will mean that X has the Poisson distribution with parameter λ.

45. The average number of radioactive particles ( جزيئات اشعاع ذري) passing through acounter (عاكس ) during 1 millisecond in a laboratory experiment is 4. What is the probability that 6 particles enter the counter in given millisecond? 45 Example X~P(4) f(6) = e -4 4 6 = 0.1042 6!

46. 46 In Table 2 of The Appendix, the cumulative distribution function of the Poisson random variable is tabulated for selected values of x and λ. Several illustrations of the use of Table 2 now follow. Let λ =2.5. The probability that X is less than 3 is P (X<3)= P(X≤ 2)= F(2)= 0.5438 The probility that X is no less than 4 is P(X≥ 4) =1-P(X≤ 3) =1-F(3) =0.2424 The probability that X is exactly 2 is P(X=2) = F(2)-F(1)= 0.2565

47. 47 Example average number of oil tanker ( ) arriving each day at a certain port city is known to be 10. The facilities at the port can handle at most 15 tankers per day, What is the probability that on a given day tankers will have to be sent away? ناقلات النفط X= number of tankers arriving each day P(X >15) = 1- P(X≤15) =1- F(15) =1- 0.9513 =0.0487

48. 48 Example Suppose that the average number of telephone calls arriving at the switchboard of a small corporation is 30 calls per hour. 1. What is the probability that 3 calls will arrive in a 1-minute period 30 calls ------ 60 min ? Calls -------- 1min ? = 30 = 0.5 f (3) = e -0.5 0.5 3 60 3! 2.What is the probability that no calls will arrive in a 3-minute period 0.5 calls ---- 1 min ? Calls ----- 3 min ?= 0.5 * 3 = 1.5 f(0)= e -1. 5 1.5 0 = 0.223 0! 3. What is the probability that more than five calls will arrive in a 5 minutes interval ( Answer 0.042)

49. The Poisson Approximation to the Binomial Distribution 49 The poisson random variable can be used as an approximation to the binomial random variable with parameters (n,p). When n is large and p is small enough so that np is moderate size, the value of the p.m.f of the binomial distribution can be approximated, for x=0,l,2,..., by the value of the p.m.f of the Poisson distribution for which λ= np

50. 50 Suppose that in a large population the proportion of people that have a certin disease is 0.01. What is the probability that in a random group of 200 people at least four people will have the disease? Example X= number of people having the disease among the 200 people X~bin (200, 0.01 ) can be approximated by poisson with λ= np = 2 P(X ≥ 4)= 1- P(X<4)= 1 - P(X ≤ 3 )=0.1428

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