Textbook: pp Chapter 2: Probability Concepts and Applications


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1 1 Textbook: pp Chapter 2: Probability Concepts and Applications
2 2 Learning Objectives After completing this chapter, students will be able to: Understand the basic foundations of probability analysis. Describe statistically dependent and independent events. Use Bayes theorem to establish posterior probabilities. Describe and provide examples of both discrete and continuous random variables. Explain the difference between discrete and continuous probability distributions. Calculate expected values and variances and use the normal table.
3 3 Learning Objectives After completing this chapter, students will be able to: Understand the binomial distribution. Understand the normal distribution and use the normal table. Understand the F distribution. Understand the exponential distribution and its relation to queuing theory. Understand the Poisson distribution and its relation to queuing theory.
4 4 Introduction Life is uncertain; we are not sure what the future will bring Probability is a numerical statement about the likelihood that an event will occur For example, the market for the new Iphone X might be good with a chance of 60% (a probability of 0.6) or not good with a chance of 40% (a probability of 0.4).
5 5 Two Basic Rules of Probability The probability, P, of any event or state of nature occurring is greater than or equal to 0 and less than or equal to 1. That is, 0 P(event) 1 A probability of 0 indicates that an event is never expected to occur. A probability of 1 means that an event is always expected to occur. The sum of the simple probabilities for all possible outcomes of an activity must equal 1. Regardless of how probabilities are determined, they must adhere to these two rules.
6 6 Types of Probability (1 of 3) Objective Approach Relative frequency approach P(event) = Number of occurrences of the event Total numbers of trials or outcomes
7 7 Diversey Paint Example (1 of 2) Historical demand for white latex paint at = 0, 1, 2, 3, or 4 gallons per day Observed frequencies over the past 200 days
8 8 Diversey Paint Example (2 of 2) Individual probabilities Historical demand are all for white latex paint at = 0, 1, 2, 3, or between 4 gallons 0 and per day 1 0 P (event) 1 Observed frequencies over the past 200 days Total of all event probabilities equals 1 P (event) = 1.00
9 9 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace Types of Probability (2 of 3) Objective Approach o Classical or logical method Perform a series of trials P P 1 Number of ways of getting a head head = 2 Number of possible outcomes( head or tail) 13 Number of chances of drawing a spade spade = 52 Number of possible outcomes 1 = = 0.25 = 25% 4
10 Example: What is the probability that the Chinese economy will be in a severe depression in 2018? 10 Types of Probability (3 of 3) Subjective Approach o When logic and past history are not appropriate, probability values can be assessed subjectively! o Based on the experience person making the estimate and judgment of the Opinion polls (e.g. to determine election returns) Judgment of experts Delphi method (a panel of experts is assembled to make their predictions of the future)
11 11 Mutually Exclusive and Collectively Exhaustive Events (1 of 2) Events are said to be mutually exclusive if only one of the events can occur on any one trial o Tossing a coin will result in either a head or a tail o Rolling a die will result in only one of six possible outcomes
12 12 Mutually Exclusive and Collectively Exhaustive Events (2 of 2) Events are said to be collectively exhaustive if the list of outcomes includes every possible outcome Both heads and tails as possible outcomes of coin flips All six possible outcomes of the roll of a die:
13 13 Venn Diagrams A B A B Events that are mutually exclusive Events that are not mutually exclusive
14 14 Drawing a Card Draw one card from a deck of 52 playing cards A = event that a 7 is drawn B = event that a heart is drawn P (a 7 is drawn) = P(A)= 4 / 52 = 1 / 13 P (a heart is drawn) = P(B) = 13 / 52 = 1 / 4 These two events are not mutually exclusive since a 7 of hearts can be drawn These two events are not collectively exhaustive since there are other cards in the deck besides 7s and hearts
15 15 Unions and Intersections of Events (1 of 3) Intersection the set of all outcomes that are common to both events Intersection of event A and event B = A and B Probability notation = A B = AB P(Intersection of event A and event B) = P(A and B) Sometimes called joint probability = P(A B) = P(AB)
16 16 Unions and Intersections of Events (2 of 3) Union the set of all outcomes that are contained in either of two events Union of event A and event B = A or B Probability notation P(Union of event A and event B) = P(A or B) = P(A B)
17 17 Unions and Intersections of Events (3 of 3) In the previous example Intersection of event A and event B (A and B) = the 7 of hearts is drawn P(A and B) = P(7 of hearts is drawn) = 1 /52 Union of event A and event B (A or B) = either a 7 or a heart is drawn P(A or B) = P(any 7 or any heart is drawn) = 16 /52
18 18 Probability Rules (1 of 5) General rule for union of two events, additive rule P(A or B) = P(A) + P(B) P(A and B) o Union of two events, a 7 or a heart P(A or B) = P(A) + P(B) P(A and B) = 4 / / 52 1 / 52 = 16 / 52
19 19 Probability Rules (2 of 5) Conditional probability probability that an event occurs given another event has already happened P(A B) = P(AB) P(B) P(AB) = P(A B) P(B) o Probability of a 7 given a heart has been drawn P(AB) 1 / 52 P(A B) = = = 1 / P(B) / 52
20 20 Probability Rules (5 of 5) Independent one event has no effect on the other event P(A B) = P(A) P(A and B) = P(A)P(B) For a fair coin tossed twice A = event that a head is the result of the first toss B = event that a head is the result of the second toss P(A) = 0.5 and P(B) = 0.5 P(AB) = P(A)P(B) = 0.5(0.5) = 0.25
21 21 Independent Events (1 of 2) A bucket contains 3 black balls and 7 green balls Draw a ball from the bucket, replace it, and draw a second ball! 1. The probability of a black ball drawn on first draw is: P(B) = The probability of two green balls drawn is: P(GG) = P(G) x P(G) = 0.7 x 0.7 = 0.49
22 4. The probability of a green ball drawn on the second draw if the first draw is green is: P(G G) = P(G) = Independent Events (2 of 2) A bucket contains 3 black balls and 7 green balls Draw a ball from the bucket, replace it, and draw a second ball! 3. The probability of a black ball drawn on the second draw if the first draw is green is: P(B G) = P(B) = 0.30
23 23 Dependent Events (1 of 3) An urn contains the following 10 balls: o 4 are white (W) and lettered (L) o 2 are white (W) and numbered (N) o 3 are yellow (Y) and lettered (L) o 1 is yellow (Y) and numbered (N) P(WL) = 4 / 10 = 0.4 P(YL) = 3 / 10 = 0.3 P(WN) = 2 / 10 = 0.2 P(YN) = 1 / 10 = 0.1 P(W) = 6 / 10 = 0.6 P(L) = 7 / 10 = 0.7 P(Y) = 4 / 10 = 0.4 P(N) = 3 / 10 = 0.3
24 Dependent Events (2 of 3) 24
25 25 Dependent Events (3 of 3) The conditional probability that the ball drawn is lettered, given that it is yellow We can verify P(YL) using the joint probability formula:
26 26 Probability Rules (3 of 5) Which sets are independent? 1. (a) Your education (b) Your income level 2. (a) Draw a jack of hearts from a full 52card deck (b) Draw a jack of clubs from a full 52card deck 3. (a) Snow in Zhengzhou (b) Rain in Bangor / Wales
27 27 Probability Rules (4 of 5) Which sets are independent? 1. (a) Your education (b) Your income level 2. (a) Draw a jack of hearts from a full 52card deck (b) Draw a jack of clubs from a full 52card deck 3. (a) Snow in Zhengzhou (b) Rain in Bangor / Wales Dependent events Independent events Independent events
28 28 Revising Probabilities with Bayes Theorem (1 of 7) Bayes theorem is used to incorporate additional information and help create posterior probabilities from original or prior probabilities This means that we can take new or recent data and then revise and improve upon our old probability estimates for an event!
29 29 Revising Probabilities with Bayes Theorem (2 of 7) A cup contains two dice identical in appearance but one is fair (unbiased) and the other is loaded (biased) o The probability of rolling a 3 on the fair die is 1/6 or o The probability of tossing the same number on the loaded die is 0.60 o We select one by chance, toss it, and get a 3 o What is the probability that the die rolled was fair? o What is the probability that the loaded die was rolled?
30 We can answer these questions by using the formula for joint probability under statistical dependence and Bayes theorem. 30 Revising Probabilities with Bayes Theorem (3 of 7) The probability of the die being fair or loaded is P(fair) = 0.50 P(loaded) = 0.50 and that P(3 fair) = P(3 loaded) = 0.60 The probabilities of P(3 and fair) and P(3 and loaded) are P(3 and fair) = P(3 fair) x P(fair) = (0.166)(0.50) = P(3 and loaded) = P(3 loaded) x P(loaded) = (0.60)(0.50) = 0.300
31 31 Revising Probabilities with Bayes Theorem (4 of 7) A 3 can occur in combination with the state fair die or in combination with the state loaded die. The sum of their probabilities gives the unconditional (marginal probability) of a 3 on the toss: P(3) = = 0.383
32 32 Revising Probabilities with Bayes Theorem (5 of 7) If a 3 does occur, the probability that the die rolled was the fair one is The probability that the die was loaded is
33 33 Revising Probabilities with Bayes Theorem (6 of 7) If a 3 does occur, the probability that the die rolled was the fair one is These are the revised or posterior probabilities for the next roll of the die The We probability use these to that revise the die ourwas prior loaded probability is estimates
34 35 General Form of Bayes Theorem (1 of 2) We can compute revised probabilities more directly by using P(A B) = P(B A)P(A) P(B A)P(A)+ P(B A )P( A ) where A = the complement of the event A; for example, if A is the event fair die, then A is loaded die
35 36 General Form of Bayes Theorem (2 of 2) Conditional probability From the previous example Replace A with fair die, A with loaded die, B with 3 rolled
36 37 Homework  Chapter 2 Please read Chapter 3!
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