Презентация Probabilities. Week 5 (2) онлайн

Содержание слайда: BBA182 Applied Statistics Week 5 (2) Probabilities Dr Susanne Hansen Saral Email: susanne.saral@okan.edu.tr https://piazza.com/class/ixrj5mmox1u2t8?cid=4# www.khanacademy.org

Содержание слайда: Where do probabilities come from? Two different ways to determine probabilities: 1. Objective approach: a. Relative frequency approach, derived from historical data b. Classical or logical approach based on logical observations, ex. Tossing a fair coin 2. Subjective approach, based on personal experience

Содержание слайда: Types of Probability Relative frequency approach Objective Approach: a) Relative frequency We calculate the relative frequency (percent) of the event:

Содержание слайда: Objective probability – The Relative Frequency

Содержание слайда: Objective probability assessment – The Relative Frequency Approach

Содержание слайда: Types of Probability Classical approach Objective Approach:

Содержание слайда: Subjective approach to assign probabilities We use the subjective approach : No possibility to use the classical approach nor the relative frequency approach. No historic data available New situation that nobody has been in so far The probability will differ between two people, because it is subjective.

Содержание слайда: Types of Probability Subjective Approach: Based on the experience and judgment of the person making the estimate: Opinion polls (broad public) Judgement of experts (professional judgement) Personal judgement

Содержание слайда: Interpreting probability No matter what method is used to assign probabilities, we interpret the probability, using the relative frequency approach for an infinite number of experiments. The probability is only an estimate, because the relative frequency approach defines probability as the “long-run” relative frequency. The larger the number of observations the better the estimate will become. Ex.: Tossing a coin, birth of a baby, etc. Head and tail will only occur 50 % in the long run Girl and boy will only occur 50 % in the long run

Содержание слайда: Probability rules continued Rule 1 and 2 If A is any event in the sample space S, then a probability is a number between 0 and 1 The probability of the set of all possible outcomes must be 1 P(S) = 1 P(S) = Σ P(Oi ) = 1 , where S is the sample space

Содержание слайда: Objective probability assessment – The Relative Frequency Approach

Содержание слайда: Probability rules. Rule 3 Complement rule Suppose the probability that you win in the lottery is 0.1 or 10 %. What is the probability then that you don’t win in the lottery?

Содержание слайда: Probability rules. Rule 3 Complement rule The set of outcomes that are not in the event A, but are in the sample space is called the “complement” of event A and is denoted . The probability of an event occurring is 1 minus the probability that it does not occur: P (A) = 1 – P () Remember: Sample space , S = 1.0

Содержание слайда: Probability rule 4 Multiplication rule – calculating joint probabilities Independent events P(A B) = P(A) x P(B) A jar contains 5 red marbles and 5 blue marbles, a total of 10 marbles. Two marbles will be randomly chosen, one at a time. The first marble will be put back into the jar after the first trial (with replacement), then the second marble is picked. What is the probability that 2 red marbles will be randomly chosen in two consecutive trials?

Содержание слайда: Multiplication Rule for independent events (continued) The probability to chose two red marbles in a row is the same, because we put the red marble back into the jar. That is: First trial: P(R) = Second trial: P(RR) = Therefore: P(R RR) = P(R)P(RR) = ( = = .25 The probability that two red marbles are chosen consecutively is 25 %.

Содержание слайда: Independent events Events are independent from each other when the probability of occurrence of the first event does not affect the probability of occurrence of the second event. The probability of occurrence of the second event will be the same as for the first event.

Содержание слайда: Multiplication rule – calculating joint probabilities Dependent events P(A B) = P(A) x P() A jar contains 5 red marbles and 5 blue marbles a total of 10 marbles. Two marbles will be randomly chosen one at a time. The first marble will not be replaced after it has been removed from the jar. What is the probability that 2 red marbles will randomly be chosen?

Содержание слайда: Multiplication rule – Dependent events (continued) Let, R represent the event that the first marble chosen is red RR the event that the second marble chosen is also red. We want to calculate the joint probability: P(R RR) = P(R)P() There are 5 red marbles in the jar out of ten, the probability that the first marble chosen is red: P(R) =

Содержание слайда: Multiplication Rule - Dependent events (continued) After the first marble is chosen, there are only nine marbles left. Given that the first marble chosen is red, there are now only 4 red marbles left in the jar. It follows that: P(RR|R) = Thus the joint probability P(R RR) is: P(R and RR) = P(R)P(RR|R) = () (= = .22 or 22% The probability that two red marbles are chosen consecutively is 22 %

Содержание слайда: Multiple choice quiz: 1 correct 3 false You are going to take a multiple choice exam. You did not have time to study and will therefore guess. The questions are independent from each other. There are 5 multiples choice questions with 4 alternative answers. Only one answer is correct. What is the probability that you will pick the right answer out of the 4 alternatives? What is the probability that you will pick the wrong answer out of the 4 alternatives? What is the probability that you will pick two answers correctly? What is the probability of picking two wrong answers? What is the probability that you will pick all the correct answers out of the 5 questions? What is the probability that you will pick all wrong answers out of the 5 questions?

Содержание слайда: Multiple choice quiz: 1 correct 3 false You are going to take a multiple choice exam. You did not have time to study and will therefore guess. The questions are independent from each other. There are 5 multiples choice questions with 4 alternative answers. Only one answer is correct. What is the probability that you will pick the right answer out of the 4 alternatives? P = 0.25 What is the probability that you will pick the wrong answer out of the 4 alternatives? P = = 0.75 What is the probability that you will pick two answers correctly? P = ( = 0.06 What is the probability of picking two wrong answers? P= ( = .56 What is the probability that you will pick all the correct answers out of the 5 questions? P=( = 0.001 What is the probability that you will pick all wrong answers out of the 5 questions? P = ( = .24

Содержание слайда: Probability Rule 5: Addition rule for mutually exclusive events A and B are mutually exclusive events in a sample space. Then the probability of the union A B is the sum of their individual probabilities: P(A B) = P(A) + P(B)

Содержание слайда: Probability rule 5: Addition rule for mutually exclusive events Example An online shop, Netpoint, a clothing retailer, receives 1’000 visits on a particular day. From past experience it has been determined that every 1’000 hits, result in 10 large sales of at least $ 500 and 100 sales of less than $ 500. Assuming that all hits have the same probability of a sale = (1) What is the probability of a large sale from a particular hit? (2) What is the probability of a small sale? (3) What is the probability of any sale?

Содержание слайда: Addition rule of mutually exclusive events: Example – Definition of events Our single hit is selected over a total hit of 1’000 on a particular day. Let, Event A: «selected hit results in a large sale» = 10 Event B: «selected hit results in a small sale» = 100 Event A B : Hit results in any sale = 110 What is P(A) ? What is P(B) ? What is the probability of any sale, A B (A and B are mutually exclusive events)?

Содержание слайда: Addition rule of mutually exclusive events: Example - Solution P(A) = = = 0.01 or 1 % P(B) = = = 0.10 or 10 % P(A B) = = = 0.11 or 11 % (mutually exclusive)

Содержание слайда: Addition rule of mutually exclusive events: Class exercise A corporation receives a shipment of 100 units of computer chips from a manufacturer. Research indicates the probabilities of defective parts per shipment shown in the following table: What is the probability that there will be fewer than three defective parts in a shipment? P(x < 3) What is the probability that there will be more than one defective part in a shipment? P(x > 1) The five probabilities in the table sum up to 1. Why must this be so?

Содержание слайда: Probability rule 6: Addition rule for non- mutually exclusive events

Содержание слайда:

Содержание слайда: Addition rule of mutually non-exclusive events rolling a dice

Содержание слайда: Addition rule of mutually non-exclusive events: Example: P (A U B) = P(A) + P(B) – P(A ∩ B) A video store owner finds that 30 % of the customers entering the store ask an assistant for help, and that 20 % of the customers buy a video before leaving the store. It is also found that 15 % of all customers both ask for assistance and make a purchase. What is the probability that a customer does at least one of these two things?

Содержание слайда: Addition rule of non-mutually exclusive events: Example: A video store owner finds that 30 % of the customers entering the store ask an assistant for help, and that 20 % of the customers buy a video before leaving the store. It is also found that 15 % of all customers both ask for assistance and make a purchase. What is the probability that a customer does at least one of these two things?

Содержание слайда: Addition rule of non-mutually exclusive events: P(A U B) = P(A) + P(B) – P(A ∩ B) Class exercise It was estimated that 30 % of all students in their 4th year at a university campus were concerned about employment future. 25 % were seriously concerned about grades, and 20 % were seriously concerned about both. What is the probability that a randomly chosen 4th year student from this campus is seriously concerned with at least one of these two concerns?

Содержание слайда: Class exercise - solution Let P(A) be the probability of the event A: The student is concerned about employment prospect: 30 % Let P(B) be the probability of the event B: The student is concerned about the final grade, 25 % P(AB) the probability of both: 20 % P(A) = 30 %; P(B) = 25 %; P(AB) = 20% Addition rule: P(A∪B) = P(A) + P(B) – P(A∩B) = .30 + .25 - .20 = .35

Содержание слайда: Calculating probabilities of complex events Now we will look at how to calculate the probability of more complex events from the probability of related events. Example: Probability of tossing a 3 with two dices is 2/36. This probability is derived by combining two possible events: tossing a 1 (1/36) and tossing a 2 (1/36)

Содержание слайда: How to calculate probabilities of intersecting events Intersection of Events A and B - Events that are not mutually exclusive The intersection of events A and B is the event that occurs when both A and B occur. It is denoted as A and B (A The probability of the intersection is called the joint probability: P(AB)

Содержание слайда: Drawing a Card – not mutually exclusive Draw one card from a deck of 52 playing cards A = event that a 7 is drawn B = event that a heart is drawn

Содержание слайда: Joint probabilities - A business application A manufacturer of computer hardware buys microprocessors chips to use in the assembly process from two different manufacturers A and B. Concern has been expressed from the assembly department about the reliability of the supplies from the different manufacturers, and a rigorous examination of last month’s supplies has recently been completed with the results shown:

Содержание слайда: Manufacturer of computer hardware- Contingency table - joint probabilities

Содержание слайда: Manufacturer of computer hardware Contingency table joint probabilities It looks as if the assembly department is correct in expressing concern. Manufacturer B is supplying a smaller quantity of chips in total but more are found to be defective compared with Manufacturer A. However, let us consider this in the context of the probability principles we have developed: Relative frequency method (based on available data)

Содержание слайда: Manufacturer of computer hardware Marginal probabilities Let us consider the total of 9897 as a sample. Suppose we had chosen one chip at random from this sample. The following events and their probabilities can then be obtained: Find the probability of the following – marginal probabilities : Event A: the chip was supplied by Manufacturer A Event B: the chip was supplied by Manufacturer B Event C: the chip was satisfactory Event D: the chip was defective

Содержание слайда: Manufacturer of computer hardware Joint probabilities (Continued) Let us consider the total of 9897 as a sample. Suppose we had chosen one chip at random from this sample. The following joint events and their probabilities can be obtained: And the joint probabilities: P(A and C) supplied by A and satisfactory Joint probabilities P(B and C) supplied by B and satisfactory P(A and D) Supplied by A and defective P(B and D) supplied by B and defective

Содержание слайда: Interpretation of the joint probabilities in the example The joint probability that a chip is defective and that it is delivered from Manufacturer A is 0.012 The joint probability that a chip is satisfactory and it is delivered by Manufacturer A is 0.589 The probability that a chip is satisfactory and it is delivered by Manufacturer B is 0.379 The probability that a chip is defective and it is delivered by Manufacturer B is 0.020

Содержание слайда: Notations for the marginal and joint events Let, A1 = The chip is delivered from Manufacturer A A2 = The chip is delivered from Manufacturer B1 = The chip is satisfactory B2 = The chip is defective Therefore: P(A1 B1) = .589 P(A2 B1) = .379 P(A1 B2) = .012 P(A2 B2) = .029

Содержание слайда: Marginal probabilities The joint probabilities in the table allow us to calculate marginal probabilities: Marginal probabilities, computed by adding across rows or down columns. They are called so, because they are calculated in the margin of the table: Example: The chip was delivered by manufacturer A: Marginal probability P(A1) : P(A1 B1) + P (A1 B2) = .589 + .012 = .601

Содержание слайда: The following contingency table shows opinion about global warming among U.S. adults, broken down by political party affiliation.

Содержание слайда: A) What is the probability that a U.S. adult selected at random believes that global warming is a serious problem? B) What type of probability did you find in part A? (marginal or joint probability) C) What is the probability that a U.S. adult selected at random is a Republican and believes that global warming is a serious issue? D) What type of probability did you find in part C?

Содержание слайда: A) What is the probability that a U.S. adult selected at random believes that global warming is a serious problem? 63 % B) What type of probability did you find in part A? (marginal or joint probability) Marginal probability C) What is the probability that a U.S. adult selected at random is a Republican and believes that global warming is a serious issue? 18 % D) What type of probability did you find in part C? Joint probability

Содержание слайда: A Probability Table