Geometric and Hypergeometric Probability PDF

FACULTY OF SCIENCE AND SCIENCE EDUCATION DEPARTMENT OF MATHEMATICAL SCIENCES FIRST SEMESTER 2021/2022 ACADEMIC SESSION COURSE CODE: STA 211 COURSE TITLE: PROBABILITY I (2UNITS) LECTURER: MR. O. C. AYENI T...

FACULTY OF SCIENCE AND SCIENCE EDUCATION DEPARTMENT OF MATHEMATICAL SCIENCES FIRST SEMESTER 2021/2022 ACADEMIC SESSION COURSE CODE: STA 211 COURSE TITLE: PROBABILITY I (2UNITS) LECTURER: MR. O. C. AYENI TOPIC: PROBABILITY DISTRIBUTION OF DISCRETE AND CONTINUOUS RANDOM VARIABLES. SUB TOPIC: GEOMETRIC AND HYPERGEOMETRIC DISTRIBUTION GEOMETRIC RANDOM VARIABLE Consider a Bernoulli trail with probability p of a success on one trail. The trail is continued until a success occurs. Let X denotes number of trails before the first success. For example, a student decides to continue taking JAMB examination until he passes. X in this case denotes number of times he takes the examination before the first success. The probability that the first (x – 1) trail is failures and the Xth trail is a success is given by To see this, the required probability is: P (F F F … F S) = P (F) P (F) … P (F) P (S) = (1 – P) (1 – P) (1 – P) … (1 – P) P = (1 – P)x-1 P. The probability that x trails is needed for the first success is the same as the probability that the first (x – 1) trail is failures and the Xth is a success. Thus, P(X = x) = { Definition: A discrete random variable X is called a geometric random variable if its probability density function is given by F(x) = P(X = x) = { Example 1: A fair coin is tossed until a head appears (a) what is the probability that three tosses are needed? (b) What is the probability that at most three tosses are needed? Solution Let X denotes the number of tosses until a success (a head) occurs. Since the coin is fair, ( ) ( ) ( ) P (at most three tosses are needed) = P(X ≤ 3) = P (X = 1) + P (X = 2) + P (X = 3) =( ) ( )( ) ( ) ( ) ( ) ( ) ( ) Example 2: A fair die is rolled until a six appears. What is the probability that (i) at most 4 rolls are needed? (ii) at least 3 rolls are needed? Solution (i) P (at most 4 rolls are needed)= P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) X is a geometric random variable with ( ) ( ) ( ) Thus ( ) ( ) ∑( ) Example 3: What is the probability of getting the first “6” on the fourth rolling using a six- sided die? Solution The sample space, { } First trial = fail Second trial = fail Third trial = fail Fourth trial = success Therefore, while We can also use the formula to do the calculation and still get the same answer. Students should check that out. Example 4: 15% of all cars passing along a certain road are blue. What is the probability that the seventh car will be the first blue car that you will see passing along this road? Solution THE MEAN AND VARIANCE OF GEOMETRIC DISTRIBUTION If , then Note: Example 5: 4% of the population in a small town work as a teacher. a. What is the probability that the 10th person that you encounter in this town is a teacher? b. Calculate the mean, variance and standard deviation? Solution Given that: The mean, The variance, The standard deviation, √ √ Example 6: 2% of all tires produced by company XYZ have a defect. A random sample of 100 tires is tested for quality assurance. a. What is the probability that the 8th tire selected has a defect? b. What is the probability that the first defect is identified among the first 5 samples? c. What is the probability that the first defect is detected among the first 10 samples? d. How many tires would you expect to test until you find the first defective one? Solution a. b. c. d. Expected to test until the first defective is discovered is the mean. HYPERGEOMETRIC DISTRIBUTION The main distinguishing feature of the binomial and the hyper-geometric distributions is that the latter is concerned with sampling without replacement while the former is concerned with sampling with replacement. Consequently, we have the entire population of interest consisting of items that can be partitioned into two classes in terms of “defective/ non-defective”, “success/ failure”, and so on. Suppose the entire items or objects in a given population is “N”, such that “N1” of these items processes a particular characteristics of interest (e.g. defective items), then the remaining “N - N1” do not possess the characteristic of interest (e.g. non-defective items). Further, a sample of size “n” is taken with “x” of the items in the sample processing the characteristic of interest while the remaining “n – x” items do not possess the characteristic of interest. The probability density function of the hyper- geometric distribution is given in the sub – section that immediately follows. Probability density function for hyper-geometric distribution The probability density function is given as equation (1) below:   N1  N  N1        x  n  x  , x  0, n  x  0 f ( x)   N........................................................................(1)     n  0, otherwise Other constraints that must be satisfied are: Example 7 Suppose a population consists of 200 individuals, of whom ten percent have high blood pressure. Find the probability of getting at most two individuals with high blood pressure when choosing five individuals at random. Solution From the information provided in the question, we have the following: Number of individuals in the population, N = 200 Number of individuals in the population with high blood pressure (10% of N) N1 = 20 Number of individuals in the sample drawn, n = 5. Let the random variable X denote the number of individuals with high blood pressure, then the required probability is calculated as:  2  20  180    x  5  x  P ( X  2)  x 0     200     5   20  180   20  180   20  180            0  5  0   1  5  1  2  5  2      200   200   200         5   5   5   20 180   20 180   20 180            0  5   1  4   2  3      200   200   200         5   5   5  1  1488847536 20  42296805 190  955860 1488847536 845936100 181613400       2535650040 2535650040 2535650040 2535650040 2535650040 2535650040  0.5872  0.3336  0.0716  0.9924 Example 8 Three out of 9 finalists in a West African beauty competition are Nigerians. If two winners are to be selected, find the probability that a. at least one of them would be a Nigerian; b. only one of them would be a Nigerian Solution From the information provided, we have the following: N = 9, n = 2, N1 = 3 a.  3  9  3   3  6         0  2  0   0  2   1  15  15  0.4167 P ( X  0)   9 9 36 36      2  2   3  9  3   3  6         1  2  1   1  1   3  6  18  0.5 b. P( X  1)   9 9 36 36      2  2 Exercise 1. A company short-lists 12 applicants for an interview out of which 4 are men. If there are only five vacant posts to be filled, find the probability that the list of successful applicants would contain a. 3 men; b. at least 2 men; c. only 1 man; and d. all women 2. A lot contains 30 items, 6 of which are defective. What is the probability that a random sample of five items from the lot will contain; a. no defective?; b. not more than one defective?; and c. more than two defectives? Mean and Variance of Hyper-geometric Variable For a hyper-geometric variable X, the mean and the variance are expressed as equation (2) and (3) below: Mean of X; E(X) = np ………………………………………………….(2) N n Variance of X; Var(X) = npq   ……………………………………….(3)  N  1  Where Example 9 Two chips are taken at random without replacement from a bowl that has six red chips and four blue chips. If X equals the number of red chips among the two drawn, determine;  a. the expected value of X; and b. the variance of X. Solution From the information provided above, we have the following: N = (6 + 4) = 10 chips; N1 = 6; n = 2; and a. b. * + * + CONCLUSION: We have been able to look at geometric and hyper-geometric distribution with their means and variances.