Introduction to Probability and Random Variables and Discrete probability Distributions

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Chapter Introduction to Probability 6.1 Assigning probabilities to Events A random experiment is a process or course of action, whose outcome is uncertain Experiment • Flip a coin • The marks of a statistics test • The time to assemble a computer Outcomes Heads and Tails Numbers between and 10 Non-negative numbers • Repeated random experiments may result in different outcomes One can only refer to the experiment outcome in terms of the probability of each outcome to occur • To determine the probabilities we need to define the sample space, – which is exhaustive (list of all the possible outcomes) – in which the outcomes are mutually exclusive (outcome not overlap) Example 1: Build the sample space for the two random experiments described below Case 1: A ball is randomly selected from a bag containing blue and red balls An outcome is considered the ball color The sample space: {B, R} Case 2: Two balls are randomly selected from a bag that contains blue and red balls An outcome is considered the colors of the two balls drawn The sample space: {BB, BR, RB, RR} Example 2: The random experiment: randomly select two numbers from the set 1, 2, 3, 4, (a number cannot be selected twice) An outcome is defined by the sum of the two numbers 1+3 1+5, 2+4 3+5 The sample space: {3, 4, 5, 6, 7, 8, 9} 1+2 1+4, 2+3; 2+5, 3+4 4+5 Sample Space: S = {O1, O2, …,Ok} O1 O2 Simple events: The individual outcomes that cannot be further decomposed An event is a collection of simple events Example 3: A dice is rolled once Specify the sample points that belong to each event Event A = the number facing up is Event B = The number facing up is odd Event C = The number facing up is not greater than Solution: A = {6};B = {1, 3, 5}; C = {1, 2, 3) Example 4: A dice is rolled once Event A = the number facing up is Event B = The number facing up is odd Event C = The number facing up is less than The outcome is Which event takes place? Solution: Event B and event C take place because the outcome ‘3’ belongs to both events Sample Space: S = {O1, O2, …,Ok} Our objective objective isis to to Our determine P(A), P(A), the the determine probability that that event event AA probability will occur occur will 6.1 Assigning probabilities to Events Given a sample space S={O1,O2,…,Ok}, The probability of a simple event P(Oi): the following characteristics must hold: P(O ) 1 for each i i n  P(Oi ) 1 i 1 10 Checking the conditions • Only one out of two outcomes can occur (An answer can be either correct or incorrect) • There is a fixed finite number of trials (There are 10 questions in the test, n=10) • Each answer is independent of the others • The probability p of a correct answer does not change from question to question (20% chance that an answer is correct) 84 Let X = the number of correct answers 10! P(X 0)  (.20)0 (.80)10 .1074 0! (10  0)! 10! P(X 2)  (.20) (.80)10 .3020 2!(10  2)! Pat fails the test if she gets less than correct answers P(X4= p(0) + p(1) + p(2) + p(3) + p(4) = 1074 + 2684 + 3020 + 2013 +.0881 =.9672 This is called cumulative probability 85 Binomial Table… The probabilities listed in the tables are cumulative, i.e P(X ≤ k) – k is the row index; the columns of the table are organized by P(success) = p “What is the probability that Pat gets no answers correct?” i.e what is P(X = 0), given P(success) = 20 and n=10 ? P(X = 0) = P(X ≤ 0) = .1074 “What is the probability that Pat gets two answers correct?” i.e what is P(X = 2), given P(success) = 20 and n=10 ? P(X = 2) = P(X≤2) – P(X≤1) = .6778 – .3758 = .3020 remember, the table shows cumulative probabilities… What is the probability that Pat fails the quiz”? i.e what is P(X ≤ 4), given P(success) = 20 and n=10 ? P(X ≤ 4) = 9672 The binomial table gives cumulative probabilities for P(X ≤ k) P(X = k) = P(X ≤ k) – P(X ≤ [k–1]) P(X ≥ k) = – P(X ≤ [k–1]) Mean and Variance - Binomial Variable E(X) = m = np V(X) = s2 = np(1-p) 91 Example 11: If all the students in Pat’s class practice the same learning behavior like she does, what is the mean and the standard deviation of the quiz mark? Solution m = np = 10(.2) = s = [np(1-p)]1/2 = [10(.2)(.8)]1/2 = 1.26 92 Example 12 Records show that 30% of the customers in a shoe store make their payments using a credit card This morning 20 customers purchased shoes 93 • This is a binomial experiment: • There are two possible outcomes of which only one will take place (paid with the credit card or not) • There is a finite number of trials (20 customers are observed) • Customers pay independently • Each customer has the same probability to pay with a credit card ( 30) 94 Find the probability that at most 11 customers use a credit card Use the Binomial Table n = 20 p k 11 01……… 30 P(X  11) = 995 995 95 • What is the probability that at least but not more than customers used a credit card? Not more than p k 01……… 30 573 33 44 55 66 035 608 P(3X6)=P(X=3 or or or 6) =.608 - 035 = 573 P(X6) - P(X2) 96 Find the probability that exactly 14 customers did not use a credit card Let Y be the number of customers who did not use a credit card, while X (as before) the number of those who did use a credit card P(Y=14) = P(X=6) = P(X = 6) - P(X = 5) = 608 - 416 = 192 97 What is the expected number of customers who used a credit card? E(X) = np = 20(.30) = 98 ... event:C “A and with both C”, and “B andevent C” A and B A C B 16 A A and C B and C B 17 The probability of event C can be calculated as the sum of the two joint probabilities A C B P(C) = P(A and C)... Joint Probability The Probabilities of Joint Events The probability of the intersection of A and B is called also the joint probability of A and B = P(A and B) 14 Example 9: A potential investor... market (B1) the market (B2) P(Ai) Top 20 MBA program (A1) P(A1 and B1) + P(A1 and B2) = P(A1) Not top 20 MBA program (A2) P(A2 and B1) + P(A2 and B2) = P(A2) Marginal Probability P(Bj) 19 Mutual fund
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