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Basic Concepts of Probability Section 3.1 Basic Concepts of Probability

Probability experiment: Important Terms Probability experiment: Roll a die An action through which counts, measurements or responses are obtained Sample space: { 1 2 3 4 5 6 } The set of all possible outcomes Event: { Die is even }={ 2 4 6 } Basic Definitions are given first. Each concept is then illustrated. A subset of the sample space. Outcome: {4} The result of a single trial

                 Another Experiment Probability Experiment: An action through which counts, measurements, or responses are obtained Sample Space: The set of all possible outcomes Event: A subset of the sample space. Outcome: The result of a single trial Choose a car from production line            A quality control example utilizing the newly defined terms      

Classical (equally probable outcomes) Types of Probability Classical (equally probable outcomes) Empirical Probability blood pressure will decrease after medication The three types of probability and an example for each. Formulas for classical and empirical probabilities are given. In classical probability all simple outcomes are equally likely. Intuition Probability the line will be busy

Tree Diagrams Start 36 outcomes Two dice are rolled. Describe the sample space. Start 1st roll 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 Illustration of a tree diagram using the roll of two dice. 2nd roll 36 outcomes

Sample Space and Probabilities Two dice are rolled and the sum is noted. 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 The sum is 4 can be successful in 3 ways. P(4) = 3/36 = 1/12 = 0.083 The sum is 11 can be successful in 2 ways. P11) = 2/36 = 118 =0.056 Find the probability the sum is 4 3/36 = 1/12 = 0.083 Find the probability the sum is 11 2/36 = 1/18 = 0.056 Find the probability the sum is 4 or 11 5/36 = 0.139

 Complementary Events The complement of event E is event E¢. E¢ consists of all the events in the sample space that are not in event E. E E¢ P(E´) = 1 - P(E)  The day’s production consists of 12 cars, 5 of which are defective. If one car is selected at random, find the probability it is not defective. Sometimes it is easier to calculate the probability that an event won’t happen. Then subtract from 1 to find the probability that it will Solution: P(defective) = 5/12 P(not defective) = 1 - 5/12 = 7/12 = 0.583 

Conditional Probability and the Section 3.2 Conditional Probability and the Multiplication Rule

Conditional Probability The probability an event B will occur, given (on the condition) that another event A has occurred.  We write this as P(B|A) and say “probability of B, given A”.  Two cars are selected from a production line of 12 cars where 5 are defective. What is the probability the 2nd car is defective, given the first car was defective?             2 illustrations for conditional probability, one with dependent events the other with independent events.            Given a defective car has been selected, the conditional sample space has 4 defective out of 11. P(B|A) = 4/11

Independent Events the second die is a 4, given the first was a 4.  Two dice are rolled, find the probability the second die is a 4, given the first was a 4. Original sample space: {1, 2, 3, 4, 5, 6 } Given the first die was a 4, the conditional sample space is : {1, 2, 3, 4, 5, 6} 2 illustrations for conditional probability, one with dependent events the other with independent events. Note that knowing what happens on the first die, does not affect the probability of rolling a 4 on the second. The conditional probability, P(B|A) = 1/6

Independent Events Two events A and B are independent if the probability of the occurrence of event B is not affected by the occurrence (or non-occurrence) of event A. A= Being female B=Having type O blood A= 1st child is a boy B= 2nd child is a boy Two events that are not independent are dependent. Intuitive examples for pairs of independent events and pairs of dependent events A= taking an aspirin each day B= having a heart attack A= being a female B= being under 64” tall

Independent Events If events A and B are independent, then P(B|A) = P(B) Conditional Probability Probability  12 cars are on a production line where 5 are defective and 2 cars are selected at random. A= first car is defective B= second car is defective. The probability of getting a defective car for the second car depends on whether the first was defective. The events are dependent. Distinguish between 2 events that are independent and 2 events that are dependent.  Two dice are rolled. A= first is a 4 and B = second is a 4 P(B)= 1/6 and P(B|A) = 1/6. The events are independent.

Contingency Table The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is: Omaha Seattle Miami Total Yes 100 150 400 No 125 130 95 350 Undecided 75 170 5 250 300 450 1000 One of the responses is selected at random. Find: 1. P(Yes) 2. P(Seattle) 3. P(Miami) 4. P(No, given Miami) A 3 by 3 contingency table

Solutions 100 150 125 130 95 350 75 170 5 250 Omaha Seattle Miami Total Yes No Und 300 450 400 1000 = 400 / 1000 = 0.4 = 450 / 1000 = 0.45 1. P(Yes) 2. P(Seattle) 3. P(Miami) 4. P(No, given Miami) Notice how the sample space changes when an event is “given”. =250 / 1000 = 0.25 = 95 / 250 = 0.38 Answers: 1) 0.4 2) 0.45 3) 0.25 4) 0.38

                     Multiplication Rule To find the probability that two events, A and B will occur in sequence, multiply the probability A occurs by the conditional probability B occurs, given A has occurred. P( A and B) = P(A) × P(B|A)  Two cars are selected from a production line of 12 where 5 are defective. Find the probability both cars are defective.                      A = first car is defective B = second car is defective. P(A) = 5/12 P(B|A) = 4/11 P(A and B) = 5/12 × 4/11 = 5/33= 0.1515

When two events A and B are independent, then Multiplication Rule  Two dice are rolled. Find the probability both are 4’s. A= first die is a 4 and B= second die is a 4. P(A) = 1/6 P(B|A) = 1/6 P(A and B) = 1/6 × 1/6 = 1/36 = 0.028 When two events A and B are independent, then P (A and B) = P(A) × P(B) Note for independent events P(B) and P(B|A) are the same.

Section 3.3 The Addition Rule

Compare “A and B” to “A or B” The compound event “A and B” means that A and B both occur in the same trial. Use the multiplication rule to find P(A and B). The compound event “A or B” means either A can occur without B, B can occur without A or both A and B can occur. Use the addition rule to find P(A or B). A and B A B A B A or B

Mutually Exclusive Events Two events, A and B are mutually exclusive, if they cannot occur in the same trial. A = A person is under 21years old B = A person is running for the U.S. Senate A = A person was born in Philadelphia B = A person was born in Houston A Mutually exclusive P(A and B) = 0 B When event A occurs it excludes event B in the same trial.

Non-Mutually Exclusive Events If two events can occur in the same trial, they are non-mutually exclusive. A = A person is under 25 B = A person is a lawyer A = A person was born in Philadelphia B = A person watches West Wing on TV. Non-mutually exclusive P(A and B) ¹ 0 A and B A B

The Addition Rule A= the card is a king B = the card is red. The probability that one or the other of two events will occur is: P(A) + P(B) - P(A and B)  A card is drawn from a deck. Find the probability it is a king or it is red. A= the card is a king B = the card is red. ¤ P(A) = 4/52 and P(B) = 26/52 but P( A and B) = 2/52 P(A or B) = 4/52 + 26/52 - 2/52 = 28/52 = 0.538

When events are mutually exclusive, The Addition Rule  A card is drawn from a deck. Find the probability the card is a king or a 10. A = the card is a king and B = the card is a 10. P(A) = 4/52 and P(B) = 4/52 and P( A and B) = 0/52 P(A or B) = 4/52 + 4/52 - 0/52 = 8/52 = 0.054 When events are mutually exclusive, P(A or B) = P(A) + P(B)

Contingency Table One of the responses is selected at random. Find: The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is: Omaha Seattle Miami Total Yes 100 150 400 No 125 130 95 350 Undecided 75 170 5 250 300 450 1000 One of the responses is selected at random. Find: 1. P(Miami and Yes) 2. P(Miami and Seattle) 3. P(Miami or Yes) 4. P(Miami or Seattle)

Contingency Table One of the responses is selected at random. Find: Omaha Seattle Miami Total Yes 100 150 400 No 125 130 95 350 Undecided 75 170 5 250 300 450 1000 One of the responses is selected at random. Find: 1. P(Miami and Yes) 2. P(Miami and Seattle) = 250/1000 * 150/250 = 150/1000 = 0.15 = 0

Contingency Table 3 P(Miami or Yes) 4. P(Miami or Seattle) Omaha Total Yes 100 150 400 No 125 130 95 350 Undecided 75 170 5 250 300 450 1000 250/1000 + 400/1000 - 150/1000 =500/1000 = 0.5 3 P(Miami or Yes) 4. P(Miami or Seattle) 250/1000 + 450/1000 - 0/1000 =700/1000 = 0.7

P(A and B) = P(A) *P(B|A) P(A or B) = P(A) + P(B) - P(A and B) Summary For complementary events P(E') = 1 - P(E) Subtract the probability of the event from one. The probability both of two events occur P(A and B) = P(A) *P(B|A) Multiply the probability of the first event by the conditional probability the second event occurs, given the first occurred. Probability at least one of two events occur P(A or B) = P(A) + P(B) - P(A and B) Add the simple probabilities but to prevent double counting, don’t forget to subtract the probability of both occurring.

Section 3.4 Counting Principles

Fundamental Counting Principle If one event can occur m ways and a second event can occur n ways, the number of ways the two events can occur in sequence is m*n. This rule can be extended for any number of events occurring in a sequence. If a meal consists of 2 choices of soup, 3 main dishes and 2 desserts, how many different meals can be selected? 2 * Dessert 2 Soup * 3 Main The probability of ordering a particular soup and particular main dish and a particular desert is 1/12 or 0.083 Start = 12 meals

Factorials This is called n factorial and written as n! Suppose you want to arrange n objects in order. There are n choices for 1st place Leaving n-1 choices for second, then n –2 choices for third place and so on until there is one choice of last place. Using the Fundamental Counting Principle, the number of ways of arranging n objects is: This is called n factorial and written as n!

The number of permutations for n objects is n! A permutation is an ordered arrangement The number of permutations for n objects is n! n! = n (n - 1) (n -2)…..3 * 2 * 1 The number of permutations of n objects taken r at a time (where r  n) is:  You are required to read 5 books from a list of 8. In how many different orders can you do so? Emphasize that in counting permutations, order is important. There are 6720 permutations of 8 books reading 5.

Combinations A combination is an selection of r objects from a group of n objects. The number of combinations of n objects taken r at a time is You are required to read 5 books from a list of 8. In how many different ways can you choose the books if order does not matter. In counting combinations, order does not matter. An object is either selected or it is not. There are 56 combinations of 8 objects taking 5.

2 4 1 3 2 1 3 1 4 1 2 3 3 4 2 4 Combinations of 4 objects choosing 2 The next two slides show the difference between permutations and combinations. With 4 objects, there are 6 ways to chose 2 without regard to order. 3 4 2 4 Each of the 6 groups represents a combination

2 4 1 3 Permutations of 4 objects choosing 2 1 2 1 2 1 3 1 3 1 4 1 4 2 3 2 3 If order is important, there are 12 ways of arranging 4 objects, choosing 2. If there are r members in each subset, then each subset can be arranged in r! ways. In this case there are 2 members in each subset so each subset (combination) can be arranged in 2! = 2*1 = 2 ways 3 4 3 4 2 2 4 Each of the 12 groups represents a permutation.