Probability Rules

260 Views

Download Presentation
## Probability Rules

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Probability Rules**Chapter 6**Review of Approaches to Probability**1.) What is the probability that the Dow Jones Industrial Average will exceed 12,000? Which approach to probability would you use to answer this question? 2.) The National Center for Health Statistics reports that of 883 deaths, 24 resulted from an automobile accident, 182 from cancer, and 333 from heart disease. What is the probability that a particular death is due to an automobile accident? Which approach to probability did you use to answer this question? 3.)One card will be randomly selected from a standard 52-card deck. What is the probability the card will be an Ace? Which approach to probability did you use to answer this question?**Review of Basic Probability Rules**Complement Rule P(A) = 1 – P(Ac) Addition Rule for Mutually Exclusive Events P(A or B) = P(A) + P(B) Multiplication Rule for Independent Events P(A and B) = P(A)*P(B) “At Least One” Rule P(At least one) = 1 – P(none)**Venn Diagram: Mutually Exclusive Events**One card is drawn from a standard deck of cards. What is the probability that it is an ace or a nine? A B Ace nine Events A and B are mutually exclusive. A card can be either an Ace or a nine, but can not be both**Venn Diagram:Events that are Not Mutually Exclusive**One card is drawn from a standard deck of cards. What is the probability that it is red or an ace? Red Ace Both red and an ace These events are not mutually exclusive as it is possible for a card to be both red and an ace (ace of hearts, ace of diamonds)**Probabilities of Events that Are Not Mutually Exclusive**Find P(Red or Ace): 0.5 0.077 Red Ace 0.038 = P(Red) + P(Ace) – P(Both Red and Ace) = 0.5 + 0.077 -0.038 =0.539**General Addition Rule**Used when events are not mutually exclusive. P(A or B) = P(A) + P(B) – P(A and B) Example: The Illinois Tourist Commission selected a sample of 200 tourists who visited Chicago during the past year. The survey revealed that 120 tourists went to the Sears Tower, 100 went to Wrigley Field and 60 visited both sites. What is the probability of selecting a person at random who visited both the Sears Tower and Wrigley Field?**General Addition Rule (continued)**P(Sears Tower) = 120/200 = 0.6 P(Wrigley Field) = 100/200 = 0.5 P(Both) = 60/200 = 0.3 0.3 0.6 0.5 S W P(S or W) = P(S) + P(W) – P(S and W) = 0.6 + 0.5 - 0.3 = 0.8**General Addition Rule (continued)**What is the probability that a randomly selected person visited either the Sears Tower or Wrigley Field but NOT both? P(S or W but NOT both) = P(S or W) – P(S and W) = 0.8 – 0.3 = 0.5 Second approach: P(S and Wc) = 0.6 – 0.3 = 0.3 P(W and SC) = 0.5 – 0.3 = 0.2 P(S or W but NOT both) = P(S and Wc) + P(W and SC) = 0.3 + 0.2 = 0.5**General Addition Rule (continued)**What is the probability that a randomly selected tourist went to neither location? P(neither location) = 1 – P(either location) = 1 – P(S or W) = 1 – 0.8 = 0.2**Conditional Probabilities**Here is a contingency table that gives the counts of students by their gender and political views. (Data are from Fall 2005 Class Survey) P(Female) = 77/137 = 0.562 P(Female and Liberal) = 30/137 = 0.219 What is the probability that a selected student has moderate political views given that we have selected a female?**Conditional Probabilities (continued)**What is the probability that a selected student has moderate political views given that we have selected a female? P(Moderate | Female) = 24/77 = 0.311 Conditional probability, P (B|A) – the probability of event B given event A.**Conditional Probabilities (continued)**Formal Definition: P(B|A) = P(A and B) P(A) Example: P(Moderate and Female) P(Female) =(24/137) / (77/137) = 0.175 / 0.562 = 0.311**Multiplication Rule**Multiplication Rule for Independent events: P(A and B) = P(A) * P(B) Independent – the occurrence of one event has no effect on the probability of the occurrence of another event. Example: A survey by the American Automobile Association (AAA) revealed that 60 percent of its members made airline reservations last year. Two members are selected at random. What is the probability both made airline reservations last year? P(R1 and R2) = P(R1)*P(R2) = (0.6)*(0.6) = .36**General Multiplication Rule**Use when events are Dependent. P(A and B) = P(A) * P(B|A) For two events A and B, the joint probability that both events will happen is found by multiplying the probability event A will happen by the conditional probability of event B occurring.**General Multiplication Rule (continued)**Example: A county welfare agency employs 10 welfare workers who interview prospective food stamp recipients. Periodically the supervisor selects, at random, the forms completed by two workers to audit for illegal deductions. Unknown to the supervisor, three of the workers have regularly been giving illegal deductions to applicants. What is the probability that both of the two workers chosen have been giving illegal deductions?**General Multiplication Rule (continued)**Solution: Define the following two events: A = First worker selected gives illegal deductions B = Second worker selected gives illegal deductions We want to find the probability that both A and B occur. To find the P(A) consider the following Venn Diagram. I = worker with illegal deductions N = worker not giving illegal D Each observation in the sample space is equally likely. P (A) = P(I1) + P(I2) + P(I3) = 1/10 + 1/10 + 1/10 = 3/10 or 0.30 N1 N2 N3 N4 N5 N6 N7 I1 I2 I3 A**General Multiplication Rule (continued)**To find the conditional probability, P(B|A), we need to make changes to the sample space. Remember our assumption is that the first worker selected is giving illegal deductions. P (B|A) = P(I1) + P(I2) = 1/9 + 1/9 = 2/9 Substituting P(A) and P(B|A) into the formula for the general multiplication rule, we find P(A and B) = P(A)P(B|A) = (3/10) * (2/9) = 6/90 = 1/15 or 0.067 N1 N2 N3 N4 N5 N6 N7 I1 I2 B|A**Tree Diagram**A tree diagram is a display of conditional events or probabilities that is helpful in thinking through conditioning. N (6/9) N and N = (7/10)(6/9) = 42/90 N (7/10) N and I = (7/10)(3/9) = 21/90 I = (3/9) (3/10) N =(7/9) I and N = (3/10)(7/9) = 21/90 I I = (2/9) I and I = (3/10)(2/9) = 6/90**Independent Events?**Again, events are independent when the outcome of one event does not influence the probability of the other. Events A and B are independent whenever P(B|A) = P(B) In the case of independent events the general multiplication rule reduces to the simple multiplication rule. P(A and B) = P(A) * P(B|A) = P(A) * P(B)**Exploring Independence**Is the probability of being liberal independent of gender for students? In other words, does P(Liberal | Female) = P(Liberal)? P(Liberal|Female) = 30/77 = 0.39 P(Liberal) = 47/137 = 0.343 Because these probabilities are not equal, we can be pretty sure that liberal political views are not independent of the student’s gender**Let’s Try Some Examples**1.) Two cards are drawn without replacement. What is the probability they are both aces? 2.) What is the probability of getting 5 hearts in a row? 3.) I draw one card and look at it. I tell you that it is red. What is the probability it is a heart? And what is the probability it is red, given that it is a heart?**Examples**4.) Are “red card” and “spade” independent? Mutually exclusive? 5.) Are “face card” and “king” independent? Mutually exclusive?**Example - Travel**Suppose the probability that a U.S. resident has traveled to Canada is 0.18, to Mexico is 0.09, and to both countries is 0.04. What’s the probability that an American chosen at random has: A.) traveled to Canada but not Mexico? B.) traveled to either Canada or Mexico? C.) not traveled to either country? D.) Are travel to Mexico and Canada mutually exclusive events? E.) Are travel to Mexico and Canada independent events? Explain.**Example - Sick Cars**Twenty percent of cars that are inspected have faulty pollution control systems. The cost of repairing a pollution control system exceeds $100 about 40% of the time. When a driver takes her car in for inspection, what’s the probability that she will end up paying more than $100 to repair the pollution control system?**Example - Health**The probabilities that an adult American man has high blood pressure and/or high cholesterol are shown in the table: A.) What is the probability that a man has both conditions? B.) What’s the probability that he has high blood pressure? C.) What’s the probability that a man with high blood pressure has high cholesterol? D.)Are high blood pressure and high cholesterol independent?**Example - Absenteeism**A company’s records indicate that on any given day about 1% of their day shift employees and 2% of the night shift employees will miss work. Sixty percent of the employees work the day shift. A.) Is absenteeism independent of shift worked? Explain. B.) What percent of employees are absent on any given day?**Example - Blood Type**The American Red Cross says that about 45% of the U.S. population has Type O blood, 40% Type A, 11% Type B, and the rest Type AB. Among four potential donors, what is the probability that: A.) All are Type O? B.) No one is Type AB? C.) They are not all Type A? D.) At least one person is Type B?**Graduation**• A private college report contains these statistics: • 70% of incoming freshmen attended public schools. • 75% of public school students who enroll as freshmen eventually graduate • 90% of other freshmen eventually graduate. • A.) Is there any evidence that a freshman’s chances to graduate may depend upon what kind of high school the student attended? Explain. • B.) What percent of freshman eventually graduate?