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1 The Binomial Distribution The Binomial Distribution is a probability experiment that meets the following criteria. 1. Each trial has only 2 possible outcomes that are considered success or failure. 2. There must be a fixed number of trials. For example, flip a coin 8 times, take a multiple choice test that has 20 questions, or randomly choose 10 people from the population. 3. The outcomes are independent. This means that past events do not affect the probability of future events. These are events with replacement or when the population is assumed to be sufficiently large, like the population of the United States. 4. The probability of a success must remain the same for each trial. This might be best explained by example. Consider a multiple choice test that has 20 questions. If each question has four possible answers, then the probability of guessing the correct answer is one-fourth; however, if some of the questions have five possible answers instead of four, then the probability of guessing correctly is changed to one-fifth. In this second scenario, the probability of success did not remain the same for each trial. Following is an example problem that will be used to illustrate how to calculate problems regarding the binomial distribution. About one in 10 people in the United States are left-handed (L or lefties). Randomly draw 5 people from the population. Construct a probability distribution and use it to find the following probabilities. 1. three people are left-handed, 2. at most three people are left-handed, 3. at least three people are left-handed. Constructing the probability distribution will allow us to see the sample space and to better understand what the binomial distribution is describing. The sample space for drawing five lefties is that it is possible that zero lefties are picked from the crowd, one of the five people could be lefties, two of the five could be lefties, three of the five could be lefties, four of the five could be lefties, and finally, all of the people selected could be lefties. The sample space has six different outcomes. Let x be the probability of selecting a desired number of lefties, then part of the probability distribution can be constructed as follows. 1 x 0 1 2 3 4 5 P (x) For the sake of illustration, begin by finding P (2), the probability that 2 of the five people selected are lefties. One approach to the problem is to write an example of picking two lefties, which also means there will be 3 righties in the outcome. Let lefties be denoted L and righties be R. Then one example of how this could happen is L L R R R. This is not the only way this could happen, but it is an example that quickly comes to mind. From here, applying what has been learned about counting and probability will uncover the answer. The probability of selecting a person’s handedness is independent of the people already selected. The probability of 1 and the probability of a rightie is the complement, or selecting a leftie is 10 1 9 1 − 10 = 10 . This gives the following solution. L L R R R 1 10 1 10 9 10 9 10 9 10 From what has been learned, multiply the probabilities to find the answers; however, this would not be the correct because because this event can happen more ways than this one example. It can happen in other ways, for example, LRLRR, or RRLRL, and there are other arrangements. Instead of trying to find all other possibilities, notice the LLRRR is a word, or arrangement of letters. The number of words that can found is 5!/(2!3!). Multiply the above probabilities by the number of ways of selecting 2 lefties, or P (2) = 5! 2!3! · L L 1 10 · 1 10 · R 9 10 · R 9 10 · R 9 10 This process can be continued to find all probabilities of the sample space.1 1 P (0) = 5! · 5! R R R R R 9 · 10 9 · 10 9 · 10 9 · 10 9 10 L R R R R P (1) = 5! · 1!4! 1 · 10 9 · 10 9 · 10 9 · 10 9 10 L L L R R P (3) = 5! · 3!2! 1 · 10 1 · 10 1 · 10 9 · 10 9 10 2 The example of P (2) has more to offer. The problem has been stated in terms of selecting a leftie as a success. Consider the following analysis of P (2). 5! 1 1 9 9 9 · · · · · 3!2! 10 10 10 10 10 2 3 5! 9 1 = · · (5 − 2)!2! 10 10 2 5−2 1 9 =5 C2 · 10 10 P (2) = The problem was that lefties are success. The equation above translates as, “5 choose 2 successes times the probability of a success raised to the number of successes times the probability of failures raised to the number of failures.” In general, any individual outcome of a binomial distribution can be found using this technique. The formal equation for a binomial experiment with x successes and n trials is P (x) =n Cx px q n−x , where p is the probability of success, and q = 1 − p is the probability of failure. Now the original probability distribution can be completed. 0 1 2 3 4 5 P (x) P (0) P (1) P (2) P (3) P (4) P (5) P (x) 59049 100000 32805 100000 7290 100000 810 100000 45 100000 1 100000 x Notice that the sum of the probabilities is one, so the probability distribution seems correct. Return to the initial problem and the three question to address. 1. Find the probability that exactly three people are left-handed. This probability can be taken directly from the probability distribution or can be P (4) = P (5) = 5! · 4!1! 5! · 5! L L L L R 1 · 10 1 · 10 1 · 10 1 · 10 9 10 L L L L L 1 · 10 1 · 10 1 · 10 1 · 10 1 10 3 calculated directly as P (3) =5 C3 · 1 10 3 9 10 5−3 . Further, the TI-84 has the capability of solving this as the function: binompdf(5,1/10,3), where the syntax means that 5 is the number of trials, 1/10 is the probability of a success, and 3 is the number of successes. Access binompdf by pressing 2nd , vars to access “distr”, then scroll down to binompdf and press enter to select it. On the older model calculators, type binompdf(5,1/10,3), and on the new models answer the indicated questions then select “calculate”, then enter . 2. Find the probability that at most three people are left-handed. This means the probability of zero lefties, or the probability of 1 leftie, or the probability of 2 lefties, or the probability of 3 lefties: P (0) + P (1) + P (2)+P (3). Since the probability distribution has been constructed, these probabilities can be added directly as 32805 7290 81 99954 59049 + + + = = 0.99954 100000 100000 100000 100000 100000 The concept of adding from zero success to a certain number of successes is often helpful, so the TI-84 has the function binomcdf. The binomcdf function always begins adding from the probability of zero successes to, and including, the last denoted success. Try the calculation again using binomcdf. Access binomcdf by pressing 2nd , vars to access the “distr”, then scroll down to binomcdf and press enter to select it. On the older model calculators, type binomcdf(5,1/10,3). On the newer model TI84s, answer the indicated questions, then select “calculate”, then enter . Notice that the syntax of binompdf and binomcdf are the same, but the functions are different. 3. Find the probability that at least three people are left-handed. At least three means the probability of three lefties, or the probability of four lefties, or the probability of five lefties, i.e., P (3) + P (4) + P (5). Add each probability directly from the probability distribution, 45 1 856 810 + + = = 0.0856. 100000 100000 100000 100000 Another approach is to utilize binomcdf. To see this recall that the sum 4 of the probabilities of a probability distribution is one. Then, P (0) + P (1) + P (2) + P (3) + P (4) + P (5) = 1 −P (0) − P (1) − P (2)) = − P (0) − P (1) − P (2)) P (3) + P (4) + P (5) = 1 − (P (0) + P (1) + P (2)) | {z } | {z } P (at least 3) = 1 − (binomcdf(5,1/10,2)) P (at least 3) = 0.0856 5