Survey

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3.6 Binomial probabilities 3.6.1 Binomial probabilities and random variables. Binomial probabilities occur when we repeatedly do something that has two possible outcomes and we count the number of times that one or the other outcome occurs. Example 3.6.1. A bank customer has two options. 1. Go into the bank = g 2. Use the drive through service = d Suppose the probability that a customer goes into the bank is Pr{g} = 0.6 and the probability that the customer uses the drive through service is Pr{d} = 0.4. Furthermore, suppose whether one customer drives goes in is independent of what all the other customers do. Suppose we observe the next four customers. a. What is the probability that exactly 2 of them go into the bank? b. More generally, what is the probability that exactly k of them go into the bank? c. What is the probability that at least 2 of them go into the bank? d. More generally, what is the probability that at least k of them go into the bank? These are typical questions involving binomial probabilities. In order to answer these, let X1, ..., Xn = a sequence of independent random variables each with two possible outcomes that we take to be the numbers 1 and 0. Suppose that each of the random variables has the same probability distribution. We shall call 1 a success and 0 a failure and the observations of the value of any of the random variables a trial. The probability of a success is the same for any trial. In Example 3.6.1 we have n = 4 and a success is the customer going in and a failure is any customer driving through. Thus g corresponds to 1 and d corresponds to 0 and Xk = 1 if the kth custormer goes in and Xk = 0 if the kth customer drives through. (x1, ..., xn) = { X1 = x1, ..., Xn = xn } = a sequence of n values each of which could be 1 or 0. This is a possible outcome of observing the outcomes of the n trials X1, ..., Xn. There are 2n possible sequences. In Example 3.6.1 where n = 4 there are 16 possible sequences, namely 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 p = Pr{Xk = 1} = the probability of success in any trial, i.e. the probability that any of the random variables Xk takes on the value 1. In Example 3.6.1 one has p = 0.6. q = 1 – p = Pr{Xk = 0} = the probability of failure in any trial, i.e. the probability that any of the random variables Xk takes on the value 0. In Example 3.6.1 one has q = 0.6. 3.6 - 1 Pr{ x1, ..., xn } Pr{ X1 = x1, ..., Xn = xn } = Pr{X1 = x1} ... Pr{Xn = xn} = the probability = of an outcome (x1, ..., xn) k n-k pq = p (1 – p )n-k = the probability Pr{ x1, ..., xn } of any particular outcome (x1, ..., xn) k which has k successes and n – k failures. For example, in Example 3.6.1 one has Pr{ 1011 } = (0.6)3(0.4) (nk ) = n! = Cn,k = the number of possible outcomes (x1, ..., xn) that has k of the k! (n-k)! values 1 and n-k of the values 0. This is also called the number of possible combinations of n things taken k at a time or the binomial coefficient of n things taken k at a time. See below for a derivation of this formula. In the context of Example 3.6.1 the different outcomes where 2 or the 4 customers 4 4! go in are 0011, 0101, 0110, 1001, 1010 and 1100. So 2 = = 6. 2! (4-2)! N = X1 + ... + Xn = the number of successes in n trials, i.e. the number of times the outcome is 1 in the n observations X1, ..., Xn. N is a binomial random variable. n k ( )p q k n-k = n k ( )p (1 – p ) k n-k = b(k; n, p) = Pr{ N = k } = the probability of k successes in n trials. These are called binomial probabilities. This is the probability mass function of N. k Fk = F(k; n, p) = Pr{ N k } = b0 + b1 + ... + bk = ( )p (1 – p ) n j j n-j = the j=0 probability of no more than k successes in n trials. These is the cumulative distribution function of N. In question a in Example 3.6.1 we want 1 - F54 , in question b we want 1 - Fk-1. The answer to question a in Example 3.6.1 is 4 b2 = 2 (0.6)2(0.4)2 = 6(0.36)(0.16) = 0.3024). 4 The answer to b is bk = k (0.6)k(0.4)4-k. The answer to c is 4 4 4 b2 + b3 +b4 = 2 (0.6)2(0.4)2 + 3 (0.6)3(0.4)1 + 4 (0.6)4(0.4)0 = 6(0.36)(0.16) + 4(0.216)(0.4) + 0.1296 = 0.3024 + 0.3456 + 0.1296 = 0.777 Note that the mean of the binomial random variable N is E(N) = E(X1 + ... + Xn) = E(X1) + ... + E(Xn) = np since for each k we have E(Xk) = 0 * Pr{Xk = 0} + 1 Pr{Xk = 1} = p 3.6 - 2 Problem 3.6.2. Suppose that customers that go into the bank require 2 minutes of time to service and customers that drive through require 1 minute of time to service. What is the average time to serve 100 customers? Example . A bakery is deciding how many loaves of its special raisin bread to bake each day. It costs them 50 cents a loaf to make and they sell it retail for $2.00 a loaf. Any loaves that are unsold at the end of the day can be sold to a wholesaler for 20 cents a loaf. The bakery estimates that they have 50 customers each day and the probability that any particular customer will buy the raisin bread is 10%. Furthermore, whether one customer buys the raisin bread is independent of whether any of the other customers buy it. Treat this as a single period inventory problem to determine the optimal number of loaves to bake each day. The retail price r = 2, the wholesale price w = 0.5 and the salvage price s = 0.2. Let N be the number of loaves they sell. N is a binomial random variable with n = 50 and p = 0.01. So the probability mass function of N is fk = (50k )(0.1) (0.9) k n-k and the cumulative distribution function is k Fk = fj. We should bake y loaves where y is the first value j=0 such that Fy r-w = 5/6 0.833. From the table at the right r-s k 0 1 2 3 4 5 6 7 8 9 10 fk 0.00515378 0.0286321 0.0779429 0.138565 0.180905 0.184925 0.154104 0.107628 0.0642779 0.0333293 0.0151833 Fk 0.00515378 0.0337859 0.111729 0.250294 0.431198 0.616123 0.770227 0.877855 0.942133 0.975462 0.990645 the first y such that Fy exceeds 5/6 is y = 7. So the bakery should bake 7 loaves each day. (n ) In order to derive the formula for k , we first need a brief discussion of permutations. 3.6.1 Permutations. Sometimes when we make k observations (i1, ..., ik) we are only interested in the case where i1, ..., ik are all distinct. Example 3.6.2. How many 3 letter words can we make where all 3 letters are distinct? We can count them as follows. We can choose the first letter arbitrarily. There are 26 possible choices. For each choice of the first letter, there are 25 choices for the second letter if it is to be different from the first. So there are (26) (25) two letter words where the 2 letters are different. For each of these we can choose a third letter in 24 different ways so it is different from the first two. This gives (26) (25) (24) = 15,600 three letter words which have all 3 letters distinct. This example illustrates the following counting formula. Suppose we have n objects x1, ..., xn and we form a vector (i1, ..., ik) using k of these where k n. We are interested in the number of such vectors where i1, ...ik are all distinct. Arguing as above, we see that the number of such vectors is n (n-1)...(n-k+1). This number can be written as n!/(n-k)! and is called the number of permutations of n things taken k at a time. Sometimes it is denoted by P(n,k). Thus 3.6 - 3 (1) P(n,k) = n! (n-k)! = number of ways of selecting k different items in order from n objects = number of permutations of n things taken k at a time An important special case is where k = n. A vector (i1, ..., in) which uses all of x1, ..., xn exactly once is just a rearrangement of x1, ..., xn and is often called a permuatation of x1, ..., xn. The number of different permuatations of x1, ..., xn is n!. Example 3.6.3. Suppose a room has 17 people. What is the probability that the people have birthdays which are all distinct? We shall simplify the problem by assuming that none of them have birthdays which fall on February 29. If we number the people in some fashion from 1 to 17, then the birthdays of the 17 people can be represented by a vector (b1, ..., b17), where each bi is the birthday of the i-th person and can have any of 365 values. So there are 36517 possible outcomes and we assume they are all equally likely. The birthdays being all distinct corresponds to b1, ..., b17 being all distinct. By the paragraph above, the number of vectors (b1, ..., b17) with this property is (365) (364) ... (349). So the probability that they (365) (364) ... (349) are all distinct is = 0.685. 36517 Problem 3.6.2. A license plate has 3 letters followed by 3 digits. What is the probability that two or more of the letters or digits are the same? 3.6.3 Combinations. Here we are interested in the number of ways we can select k different objects from a set of n objects without regard to order. Here is an example that illustrates this. Example 3.6.4. How many ways can you choose 3 different letters from the 5 letters A, B, C, D, E? In this case we can list the different ways as follows ABC ABE ACE BCD BDE ABD ACD ADE BCE CDE In contrast, there are (5)(4)(3) = 60 three letter words which we can form using 3 different letters chosen from A, B, C, D, E. There is a relationship between the 60 three letter words and the 10 ways to choose 3 letters from 5. Suppose we choose 3 letters from A, B, C, D, E. For example we might choose B, D, and E. Then there are 3! = 6 different ways we can rearrange these to form 3 letter words with all letters different (namely BDE, BED, DBE, DEB, EBD, EDB). Similarly, each choice of 3 letters from A, B, C, D, E gives rise to 6 different 3 letter words. So # different 3 letter words with all letters different = 3.6 - 4 (# ways to choose 3 letters) (# permutations of 3 letters) i.e. 60 = (10) (6). More generally suppose we want to count the number of different ways of selecting a set of k different objects {i1, ..., ik} from a set of n items {x1, ..., xn}. In the last section this quantity was called the number (n ) or C(n,k). It is also called the binomial of combinations of n things taken k at a time and denoted by k coefficient of n things taken k at a time. This is related to the number of different ways of selecting a vector (i1, ..., ik) where i1, ..., ik are different elements of {x1, ..., xn}. Arguing as above, we have # different vectors = (# ways to choose k objects) (# permutations of k objects) i.e. n! = (# ways to choose k objects) = k! (n-k)! So, (2) (nk ) = n! = number of ways of selecting k objects from n k! (n-k)! = number of combinations of n things taken k at a time Example 3.6.5. The Michigan State Lottery has a game called Super Lotto where six different balls are drawn at random from a set of 47 balls numbered 1, 2, ..., 47. By (7.5), there are 47 = 47! = 47*46*45*44*43*42 = 10,737,573 6 6! 41! 6*5*4*3*2 ways to do this. Prior to the drawing, people may select their own set of six different numbers from the numbers 1, 2, ..., 47. If their set of six numbers matches the ones drawn by the state, they win a big prize. If we assume that each group of six different numbers is equally likely, then the probability of winning is 1/10,737,573. 3.6 - 5