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Chapter 8 Counting , Probability Distributions, and Further Probability Topics 1. 2. 3. 8.5 Probability Distributions & Expected Value 8.1, 8.2 Multiplication Principle, Permutations, & Combinations 8.4 Binomial Probability 8.5 Probability Distributions & Expected .......Value Random Variables A random variable is a numerical description of the outcome of an experiment. A random variable can be classified as being either discrete or continuous depending on the numerical values it assumes. A discrete random variable may assume either a finite number of values or an infinite sequence of values. A continuous random variable may assume any numerical value in an interval or collection of intervals. Discrete Random Variables Discrete random variable with a finite number of values Let x = number of passengers on a flight to New York where x can take on the values 0, 1, 2,…, 450. Discrete random variable with an infinite sequence of values Let x = number of customers visiting Disney World in one day where x can take on the values 0, 1, 2, . . . We can count the customers arriving, but there is no finite upper limit on the number that might arrive. Discrete Probability Distributions The probability distribution for a random variable describes how probabilities are distributed over the values of the random variable. The probability distribution is defined by a probability function, denoted by f(x), which provides the probability for each value of the random variable. The required conditions for a discrete probability function are: 0 ≤ f(xi) ≤ 1 f(x1) + f(x2) + f(x3) + ∙ ∙ ∙ + f(xn) = 1 We can describe a discrete probability distribution with a table, graph, or equation. Example: JSL Appliances Using past data on TV sales (below left), a tabular representation of the probability distribution for TV sales (below right) was developed. Number Units Sold (x) of Days 0 80 1 50 2 40 3 10 4 20 200 x 0 1 2 3 4 P(x) .40 .25 .20 .05 .10 1.00 Example: JSL Appliances Graphical Representation of the Probability Distribution (Histogram) Probability, f(x) .50 .40 .30 .20 .10 0 1 2 3 4 Values of Random Variable x (TV sales) Example: JSL Appliances Graphical Representation of the Probability Distribution (Histogram) Probability, f(x) P x 1 P 1 P 2 P 3 P 4 .25 .20 .05 .10 .60 .50 .40 .30 .20 .10 0 1 2 3 4 Values of Random Variable x (TV sales) Discrete Uniform Probability Distribution The discrete uniform probability distribution is the simplest example of a discrete probability distribution given by an equation. The discrete uniform probability function is P(x) = 1/n where: n = the number of values the random variable may assume Note that the values of the random variable are equally likely. Expected Value The expected value E(x), or average, of a random variable is a measure of its central location. If a random variable x can take on n values (x1, x2, x3, …, xn) with corresponding probabilities P(x1), P(x2), P(x3),…, P(xn), then the expected value of the random variable is E x x1P x1 x2 P x2 x3P x3 xn P xn or E x xi P xi Example: JSL Appliances Expected Value of a Discrete Random Variable x 0 1 2 3 4 P(x) xP(x) .40 .00 .25 .25 .20 .40 .05 .15 .10 .40 xP(x) = 1.20 = E(x) The expected value (or average) of the distribution is 1.2 TVs. 8.1, 8.2 The Multiplication Principle, Permutations, Combinations Multiplication Principle: For experiments involving multiple steps. Combinations: For experiments in which r elements are to be selected from a larger set of n objects. Permutations: For experiments in which r elements are to be selected from a larger set of n objects, in a specific order of selection. A Counting Rule for Multiple-Step Experiments THE MULTIPLICATION PRINCIPLE If an experiment consists of a sequence of k steps in which there are n1 possible results for the first step, n2 possible results for the second step, and so on, then the total number of experimental outcomes is given by (n1)(n2)...(nk). Example: Roll a 6-sided die, and toss a coin. How many experimental outcomes are possible? Example: Multiplication Principle 2-Step Experiment n1 = 6 1 2 3 4 5 n2 = 2 H H H H T (1,H) (1,T) (2,H) (2,T) T (3,H) (3,T) T T H T 6 H T (4,H) (4,T) (5,H) (5,T) (6,H) (6,T) 6 2 = 12 Counting Rule for Combinations Another useful counting rule enables us to count the number of experimental outcomes when r elements are to be selected from a larger set of n objects. The number of combinations of n elements taken r at a time is n n! n Cr C r r !(n r )! n r FACTORIAL NOTATION For any natural number n n! = n(n - 1)(n - 2) . . . (3)(2)(1) Also, 0! = 1 Combination - Example In the Florida Lottery contestants choose 6 numbers from 1 to 53. How many possible combinations of 6 numbers are there? n = 53 r=6 53 53! 22,957, 480 53 C6 6 6! 53 6 ! Using the TI-30XA Factorial 10! 10 2nd x! 3,628,800 Using the TI-30XA Combination 53 C6 53 2nd nCr 6 Counting Rule for Permutations A third useful counting rule enables us to count the number of experimental outcomes when r elements are to be selected from a larger set of n objects, in a specific order of selection. The number of permutations of n objects taken r at a time is n! n Pr P n, r (n r )! Example - Permutation If the Florida Lottery required contestants to pick 6 numbers out of 53 in the correct order, how many experimental outcomes would there be? 53! P 53,6 16,529,385,600 53 6 ! Using the TI-30XA Permutation P 53 2nd 53 6 nPr 6 1.65293856 10 10 1.65293856 10 16,529,385, 600 Now You Try 1. 2. 3. An automobile manufacturer produces 8 models, each available in 7 different exterior colors, with 4 different upholstery fabrics, and 5 interior colors. How many varieties of automobiles are available? Hamburger Hut sells hamburgers with a selection of cheese, relish, lettuce, tomato, mustard, or ketchup. How many different hamburgers can be ordered with exactly 3 extras? A horse race consists of 12 horses. What is the probability of selecting the 1st, 2nd, and 3rd place horses in the correct order? 8.4 Binomial Probability Distribution Properties of a Binomial Experiment The experiment consists of a sequence of n identical trials. Two outcomes, success and failure, are possible on each trial. The probability of a success, denoted by p, does not change from trial to trial. The trials are independent. Notation for Binomial Probability Distributions P(S) = p P(F) = 1 – p = q n x p p = probability of a success q = probability of a failure denotes the fixed number of trials denotes a specific number of successes in n trials, so x can be any whole number between 0 and n, inclusive. denotes the probability of success in one of the n trials. q denotes the probability of failure in one of the n trials. P(x) denotes the probability of getting exactly x successes among the n trials. INT 221 – B. Potter 23 Example: Evans Electronics Binomial Probability Distribution Evans is concerned about a low retention rate for employees. On the basis of past experience, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year. Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year? Let: p = .10, n = 3, x = 1 Example: Evans Electronics Success = leave within 1 year; Fail = not leave 1. Determine the number of experimental outcomes involving 1 success in 3 trials (x = 1 success in n = 3 trials). 2. Determine the probability of each of the above outcomes. Example: Evans Electronics 1. Determine the number of experimental outcomes involving 1 success in 3 trials (x = 1 success in n = 3 trials). n n! 3! 3 x x ! n x ! 1! 3 1! Example: Evans Electronics 2. Determine the probability of each of the above outcomes. P S p, P F 1 p P SFF p 1 p 1 p 0.1 0.9 0.9 0.1 0.9 0.081 2 Probability of a particular sequence of trial outcomes = p 1 p x n x With x successes in n trials Binomial Probability Distribution Binomial Probability Function n x (n x ) P ( x) p (1 p ) x where: f(x) = the probability of x successes in n trials (x = 1) n = the number of trials (3) p = the probability of success on any one trial (.10) Binomial Probability Distribution Binomial Probability Function n x (n x ) P ( x) p (1 p ) x Step 1 Step 2 Example: Evans Electronics Using the Binomial Probability Function n x P( x) p (1 p)( n x ) x 3! P(1) (0.1)1 (0.9)2 1!(3 1)! = (3)(0.081) = .243 Example: Evans Electronics Using a Tree Diagram First Worker Second Worker Leaves (.1) Leaves (.1) Third Worker 3 .0010 S (.9) 2 .0090 2 .0090 1 .0810 2 .0090 1 .0810 L (.1) 1 .0810 S (.9) 0 .7290 L (.1) S (.9) L (.1) .081 3S(.9).243 Leaves (.1) Stays (.9) Probab. L (.1) Stays (.9) Stays (.9) Value of x Binomial Probability Distribution Expected Value E(x) = = np Evan’s Electronics: E(x) = = 3(.1) = .3 Binomial Probability Distribution Seventy percent of the students applying to a university are accepted. 1. What is the probability that among the next 18 applicants exactly 10 will be accepted? n 18, x 10, p .70 Solution #1 n! n x x P x p 1 p x ! n x ! n 18, x 10, p .7 18! 1810 10 P 10 .7 1 .7 10!18 10 ! 43, 758.028247525.00006561 .0811 Binomial Probability Distribution Seventy percent of the students applying to a university are accepted. 1. What is the probability that among the next 18 applicants exactly 10 will be accepted? 2. Determine the expected number of acceptances if n = 18. Solution #2 E x np 18 .70 12.6 Binomial Probability Distribution Seventy percent of the students applying to a university are accepted. 1. What is the probability that among the next 18 applicants exactly 10 will be accepted? 2. Determine the expected number of acceptances if n = 18. 3. What is the probability that among the next 5 applicants no more than 3 are accepted? P 0 x 3 P 3 P 2 P 1 P 0 Or , 1 P 4 P 5 Solution #3 5! 53 3 P 3 0.7 1 0.7 .3087 3! 5 3! P 2 .1323 P 1 .0284 P 0 .0024 P 0 x 3 P 3 P 2 P 1 P 0 .3087 .1323 .0284 .0024 .4718 Now You Try 1. 2. In 2003, the percentage of children under 18 years of age who lived with both parents was approximately 70%. Find the probabilities that the following number of persons selected at random from 10 children under 18 years of age in 2003 lived with both parents. Exactly 6 (x = 6) At most 4 (0 x 4) End of Chapter 8