Counting Rules and Binomial Probabilities

Counting Rules and Binomial Probabilities Stat 509 Lecture (E. Pena) February 12, 2001

Fundamental Theorem of Counting Suppose you have an experiment which could be performed in k steps, so the outcome of this experiment will be of form: (O1, O2, …, Ok) where Oi is the outcome from the ith step of the experiment. If on the ith step there are ni distinct possible outcomes, then the total number of possible (ordered) outcomes in the experiment is n1n2…nk. Counting Rules and Binomial Probabilities

Some Examples 1. How many possible outcomes are there when a coin is tossed 25 times? Solution: (2)(2)(2)…(2) = 225 2. Suppose that a man and a woman is to be chosen from a group of people where 10 are men and 15 are women. How many possible choices can be made? Solution: Step 1 is to choose the man, and there are 10 ways of doing this. Step 2 is to choose the woman and there are 15 ways of doing this. Therefore, the number of possible outcomes is (10)(15) = 150 ways. Counting Rules and Binomial Probabilities

3. How many possible subsets could be formed from a set with 10 elements, if we also include the empty subset and the whole set as a subset? Answer: 210 = 1024. Why? 4. Suppose that you have n distinct objects. In how many ways could these n objects be arranged in a sequence? For example, suppose that there are 20 students in this class, and there are available 20 chairs. In how many ways could you be seated? Solution: The first object can be chosen in n ways; the next in (n-1) ways; then (n-2) ways, …, until the last object has only one way to go. Therefore, the number of ways is: (n)(n-1)(n-2)…(2)(1) = n! with 0! = 1. Counting Rules and Binomial Probabilities

Factorial Notation For a positive integer, n, the factorial of n is defined via n! = (n)(n-1)(n-2)…(2)(1). Examples 4! = (4)(3)(2)(1) = 24 By convention, 0! = 1 7! = (7)(6!) = (7)(6)(5!) In general, n! = n(n-1)! = n(n-1)(n-2)! = etc. Counting Rules and Binomial Probabilities

Permutations 5. Suppose that you have n distinct elements. In how many ways could you choose r of these elements if sampling is without replacement, and the order in which the elements are chosen is important, that is, the samples are ordered? Solution:nPr = number of permutations of n objects taken r at a time. This number is given by Counting Rules and Binomial Probabilities

Combinations 6. Suppose again that you have n distinct objects. In how many ways could you form a sample of size r if sampling is without replacement and the order in which the objects are being included in the sample does not matter? Answer:nCr = number of combinations of n objects taken r at a time. This is defined by Counting Rules and Binomial Probabilities

Some Properties of Combinations 1. nC0 = 1 and nCn = 1 2. nC1 = n and nCn-1 = n 3. nCr = nCn-r 4. nCr are called the binomial coefficients since they appear as the coefficients in the binomial expansion Counting Rules and Binomial Probabilities

Applications of Counting Rules 1. A batch of 20 electronic items (e.g., computer chips) has 5 defective and 15 good items. Suppose that a sample of size 4 is taken without replacement from this batch, with the order in which they are taken being unimportant. A) How many possible samples are possible? Solution:20C4 = 20!/[4!(20-4)!] = 4845 samples. B) How many samples are possible which has exactly 2 defectives and 2 good items? Solution: (5C2)(15C2) = (10)(105) = 1050. Counting Rules and Binomial Probabilities

C) If sampling is performed randomly, what will be the probability of getting a sample with exactly 2 defectives and 2 good items? Solution: P(2D,2G) = N(A)/N(S) = 1050/4845 = 0.2167. D) What is the probability of getting a sample with at least 2 defective items? Solution: P(at least 2 D) = P(2D,2G) + P(3D,1G) + P(4D,0G) = [(5C2)(15C2) + (5C3)(15C1) + (5C4)(15C0)]/20C4 = [1050 + 150 + 5]/4845 = 1205/4845 = 0.2487. Counting Rules and Binomial Probabilities

2. Suppose that you have 8 (indistinguishable) objects of type I, and 12 (indistinguishable) objects of type II, and you have 20 positions on which to place these 20 objects. In how many ways could you place these 20 objects in these 20 positions? Solution: You may place these objects by first choosing the 8 positions out of the 20 positions on which to put the type I objects. The remaining 12 positions will then be occupied by the 12 type II objects. Therefore, the total number of ways equals the number of ways of choosing 8 positions out of 20, without regards to the order in which the positions are chosen. Therefore, the total number of ways is 20C8 = 20C12 = 20!/[8!(20-8)!] = 125,970 ways. Counting Rules and Binomial Probabilities

3. If a coin is tossed 25 times, in how many ways could you get an outcome with exactly 15 heads and 10 tails? Solution: The number of ways would be the number of ways of choosing the 15 positions out of the 25 tosses in which the heads should appear. The other 10 positions will have the tails. Thus, the total number of ways of getting exactly 15 heads and 10 tails is 25C15 = 3,268,760. Question: What will be the probability of getting exactly 15 heads in 25 tosses of a fair coin? Answer: P(15H,10T) = 25C15/225 = 25C15 (1/2)15 (1/2)10 = 3268760/33554432 = 0.097, or about 10%. Counting Rules and Binomial Probabilities

4. Consider the experiment of rolling a fair die 25 times. How many possible outcomes are there in this experiment? Solution: By the Fundamental Theorem of Counting, there will be (6)(6)…(6) = 625 possible outcomes. B) How many possible outcomes are there with exactly 15 “6”s and 10 “non-6”s? Solution: Any configuration with 15 6’s and 10 non-6’s could occur in (115)(510) possible ways. There are 25C15 configurations with 15 6’s and 10 non-6’s. Therefore, the total number of ways of getting 15 6’s and 10 non-6’s is 25C15 (1)15(5)10. Counting Rules and Binomial Probabilities

Binomial Probabilities In the preceding example of rolling a fair die 25 times, the probability of obtaining exactly 15 6’s and 10 non-6’s is therefore P(15 6’s and 10 non-6’s) = 25C15 (1/6)15 (5/6)10. This is an example of a binomial probability. Definition: A discrete random variable X taking values in {0,1,2,…,n} is said to have a binomial probability function with parameters n and p (0 < p < 1) if its probability function is given by: Counting Rules and Binomial Probabilities

Binomial Probability Models A random experiment that gives rise to binomial random variables typically has the following characteristics: 1. The experiment consists of n trials or steps. 2. Each trial or step has only two possible outcomes: labeled a success (S) or a failure (F). 3. The probability of getting a success (S) at a given trial is equal to p. 4. The outcome on any given trial is unaffected by the outcomes of the other trials (this is the assumption of independent trials). If X is the random variable denoting the number of successes out of the n trials, then it has a binomial probability distribution with parameters n and p. Counting Rules and Binomial Probabilities

Binomial Probability Shapes Counting Rules and Binomial Probabilities

Computing Binomial Probabilities • Use the binomial tables in the textbook. Provides probabilities of form P(X < k), so P(X = k) = P(X < k) - P(X < k-1). • Use the binomial function in your calculator. • Use Minitab using the “Probability Distributions” option in the “Calc” Menu. • Calculate it by hand and calculator. Example: Assume that the probability of a newly manufactured computer of a certain brand having a defect is 0.10. Suppose that 20 of these newly manufactured computers are tested for defects. Let X denote the number out of these 20 computers which has defects. Then X has a binomial distribution with n = 20 and p = .10. Counting Rules and Binomial Probabilities

Some Questions Pertaining to the Preceding Situation • What is the probability that none of the 20 computers will have defects? Answer: P(X=0) = 20C0 (.1)0(.9)20-0 = (0.9)20 = 0.1216. • What is the probability that at least one of these 20 computers will be found to have a defect? Answer: P(X > 1) = 1 - P(X=0) = 1 - 0.1216 = 0.8784. • What is the probability that the number of defective computers will be at most 2? Answer: P(X < 2) = P(X=0) + P(X=1) + P(X=2) = .1216 + .2702 + .2852 = .6769 using Minitab. Counting Rules and Binomial Probabilities

Mean and Standard Deviation of a Binomial Random Variable If X is a random variable whose distribution is binomial with parameters n and p, then Mean = m = np; Variance = s2 = np(1-p); Standard Deviation = s = [np(1-p)]1/2. Example: In the preceding example, m = (20)(.1) = 2 and s = [20(.1)(1-.1)]1/2 = (1.8)1/2 = 1.34. Counting Rules and Binomial Probabilities

Computational Limitations when Dealing with Binomial Distributions Situation: Suppose that you toss a fair coin 6 million times, and you denote by X the number of heads that comes up. Then X will be binomial with n = 6,000,000 and p = 0.5. Consequently, On average, we should get m = (6,000,000)(.5) = 3,000,000 heads in this experiment; and The standard deviation will be s = [(6000000)(.5)(.5)]1/2 = 1224.74. But, what will be the probability that X will differ from 3 million by at most 500? This is tough computationally, so later we will simply approximate this probability! [Relate this situation to the Florida election!!] Counting Rules and Binomial Probabilities

Counting Rules and Binomial Probabilities