3.2 Binomial and Hypergeometric Probabilities

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1 3.2 Binomial and Hypergeometric Probabilities Ulrich Hoensch Wednesday, January 23, 2013

2 Example An urn contains ten balls, exactly seven of which are red. Suppose five balls are drawn at random and with replacement from the urn. Consider it a success S if a red ball is drawn. The number of draws is n = 5; the probability of a success on a single draw is p = 7/10; the draws are independent and p is constant since the balls are drawn with replacement. Let X be the number of red balls drawn. Suppose we want to find the probability that exactly three of the five balls drawn are red (three successes), i.e. we want to compute P(X = 3).

3 Example Suppose the five balls are (red, red, red, not red, not red), chosen in that order, or (S, S, S, F, F ). Since the draws are independent, P(S, S, S, F, F ) = (7/10)(7/10)(7/10)(3/10)(3/10) = (7/10) 3 (3/10) 2 = Any other combination of three S s and two F s has the same probability, and the combinations can be found by elementary methods to be (S, S, S, F, F ), (S, S, F, S, F ), (S, S, F, F, S), (S, F, S, S, F ), (S, F, S, F, S), (S, F, S, S, F ), (F, S, S, S, F ), (F, S, S, F, S), (F, S, F, S, S), (F, F, S, S, S). Thus, P(X = 3) = (10)(7/10) 3 (3/10) 2 =

4 Example I The number of different combinations of three S s and two F s can also be found as follows: select three of the five slots to put the S s without replacement, and the order of selection of the slots does not matter. Hence, the number of selections is 5 = Remark. The binomial probabilities in this example can be computed by using: I 2ND DISTR binompdf(5, 7/10, 3) on a TI-83/84 Plus calculator. I In Mathematica: In[1]:= Out[1]= 0.7D, 3D

5 Binomial Probabilities A binomial experiment is a random experiment with the following properties: 1. The experiment consists of n identical trials. 2. Each trial results in exactly one of two outcomes: a success S or a failure F. 3. The probability of a success on a single trial is p, 0 < p < 1, and this probability is the same for every trial. 4. The outcomes of the trials are independent. Let X be the number of successes in n trials. We say that X has a binomial probability distribution, and write X B(n, p) in this situation. Theorem If X B(n, p), the probability of having exactly k successes is ( ) n P(X = k) = p k (1 p) n k, k = 0, 1, 2,..., n. k

6 Example A manufacturer of inkjet printers claims that only 5% of their printers require repairs within the first year. If a random sample of 18 of the printers is collected, and if the manufacturer s claim is true, compute the probability that four or more of the 18 printers require repairs with the first year. The number X of printers requiring repair is binomially distributed with n = 18 trials, and probability of success on a single trial p = 0.05; that is, X B(18, 0.05). Then, we need to compute 18 ( ) 18 P(X 4) = (0.05) k (0.95) 18 k k k=4 3 ( ) 18 = 1 (0.05) k (0.95) 18 k k k=0 = %.

7 Example The probability P(X 3) = computed by using: P3 k=0 18 k (0.05)k (0.95)18 k can be I Using a TI-83/84 Plus calculator as 2ND DISTR binomcdf(18, 0.05, 3). I In Mathematica: In[2]:= Out[2]= 0.05D, 3D

8 Sampling and Binomial Probabilities Suppose the proportion of individuals in a population of size N with a certain trait is p. If a sample of size n is collected from this population, then the probability that there are k individuals with the trait in the sample is ( ) n P(X = k) = p k (1 p) n k, k in the following two situations: 1. The sample is collected at random and with replacement. 2. The sample is collected at random and without replacement, but the size of the sample is small compared to the size of the population; i.e. n N.

9 Example Suppose an urn contains 25 white balls and 15 red balls. If 10 balls are selected at random and without replacement from the urn, what is the probability that k of them are white, k = 0, 1, 2,..., 10? There are ( ) ways of selecting the 10 balls from a total of 40 balls. There are ( ) 25 k ways of selecting k of the 25 white balls. There are ( 15 ) 10 k ways of selecting 10 k of the 15 red balls.

10 Example Thus, if X is the number of white balls in the sample, P(X = k) = k 10 k For example, if k = 7, then 25 7 P(X = 7) = In Mathematica, this probability can be computed as In[3]:= Out[3]= 25, 40D, 7D N

11 Sampling and Hypergeometric Probabilities Theorem Suppose the number of individuals in a population of size N with a certain trait is K. If a sample of size n is collected at random and without replacement from this population, then the probability that there are k individuals with the trait in the sample is ( K )( N K ) k n k P(X = k) = ( N. n) A random sample taken without replacement is called a simple random sample (SRS). The probabilities computed here are called hypergeometric distributions.

12 Example A SRS of size 20 is taken from a statistics class of 100 students. The statistics professor considers the class a success if at least 15 of the 20 students in the sample say that they have learned something useful in the class. If the actual number of students in the class who agree with this statement is K, then the probability that the class is labeled a success is P(X 15) = 20 k=15 ( K )( 100 K ) k 20 k ) = 1 ( If e.g. K = 50, the probability is ( 14 P(X 15) = 1 k=0 ( 50 )( 50 k 20 k ( 100 ) 20 ( 14 k=0 )) ( K )( 100 K )) k 20 k ). (

13 Example This probability can be computed in Mathematica as In[4]:= Out[4]= , 100D, 14D If we plot the probability of labelling the class as a success as a function of K, we obtain the following operating characteristic (OC) curve: PHsuccessL K

14 Homework Problems for Section 3.2 (Points) p : (1), (2), (2); p : (1), (2), (1), (2). Homework problems are due at the beginning of the class on Wednesday, January 30, 2013.

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