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Click to edit Master title style,*,Weather forecast,Psychology,Games,Sports,3,Elementary Statistics,Business,Medicine,Probability,1,Section 3.1,Basic Concepts of Probability,2, 1 2 3 4 5 6 , Die is even = 2 4 6 ,4,Roll a die,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:,Important Terms,The result of a single trial,3,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, , ,Another Experiment,4,Classical,(equally probable outcomes),Probability blood pressure will decrease after medication,Probability the line will be busy,Empirical,Intuition,Types of Probability,5,Two dice are rolled.,Describe the,sample space.,1st roll,36 outcomes,2nd roll,Start,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,Tree Diagrams,6,1,1,1,2,1,3,1,4,1,5,1,6,2,1,2,2,2,3,2,4,2,5,2,6,3,1,3,2,3,3,3,4,3,5,3,6,4,1,4,2,4,3,4,4,4,5,4,6,5,1,5,2,5,3,5,4,5,5,5,6,6,1,6,2,6,3,6,4,6,5,6,6,Find the probability the sum is 4,Find the probability the sum is 11,Find the probability the sum is 4 or 11,Two dice are rolled and the sum is noted.,Sample Space and Probabilities,3/36 = 1/12 = 0.083,2/36 = 1/18 = 0.056,5/36 = 0.139,7,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.,The days production consists of 12 cars, 5 of which are defective. If one car is selected at random, find the probability it is not defective.,E,E,Solution:,P(defective) =,5/12,P(not defective) = 1 - 5/12 =,7/12,=,0.583,P(E) = 1 - P(E),8,Section 3.2,Conditional Probability and the,Multiplication Rule,9,The probability an event B will occur, given (on the condition) that another event A has occurred,.,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?,We write this as P(B|A) and say “probability of B, given A”.,Given a defective car has been selected, the conditional sample space has 4 defective out of 11. P(B|A) =,4/11,Conditional Probability, , , ,10,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,The conditional probability, P(B|A),=,1/6,Independent Events,11,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= taking an aspirin each day,B= having a heart attack,A= being a female,B= being under 64” tall,Two events that are not independent are,dependent.,A= Being female,B=Having type O blood,A= 1st child is a boy,B= 2nd child is a boy,Independent Events,12,If events A and B are independent, then P(B|A) = P(B),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.,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.,Independent Events,Conditional Probability,Probability,13,The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is:,1. P(Yes),2. P(Seattle),3. P(Miami),4. P(No, given Miami),Omaha,Seattle,Miami,Total,Yes,100,150,150,400,No,125,130,95,350,Undecided,75,170,5,250,Total,300,450,250,1000,One of the responses is selected at random. Find:,Contingency Table,14,1. P(Yes),2. P(Seattle),3. P(Miami),4. P(No, given Miami),100,150,150,125,130,95,350,75,170,5,250,Omaha,Seattle,Miami,Total,Yes,No,Und,Total,300,450,250,400,1000,=,400,/ 1000 = 0.4,=,95,/ 250 = 0.38,=,250,/ 1000 = 0.25,Answers: 1) 0.4 2) 0.45 3) 0.25 4) 0.38,=,450,/ 1000 = 0.45,Solutions,15,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,Multiplication Rule, , , ,16,Two dice are rolled. Find the probability both are 4s.,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.,Multiplication Rule,17,Section 3.3,The Addition Rule,18,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,B,A,or,B,A,and,B,A,B,19,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,B,Mutually exclusive,P(A and B) = 0,When event A occurs it excludes event B in the same trial.,20,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.,A,B,Non-mutually exclusive,P(A and B),0,A and B,21,The Addition Rule,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 = 0.538,22,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),23,The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is:,Contingency Table,3. P(Miami or Yes),4. P(Miami or Seattle),Omaha,Seattle,Miami,Total,Yes,100,150,150,400,No,125,130,95,350,Undecided,75,170,5,250,Total,300,450,250,1000,One of the responses is selected at random. Find:,1. P(Miami and Yes),2. P(Miami and Seattle),24,Contingency Table,1. P(Miami and Yes),2. P(Miami and Seattle),= 250/1000 * 150/250 = 150/1000 =,0.15,=,0,Omaha,Seattle,Miami,Total,Yes,100,150,150,400,No,125,130,95,350,Undecided,75,170,5,250,Total,300,450,250,1000,One of the responses is selected at random. Find:,25,Contingency Table,3 P(Miami or Yes),4. P(Miami or Seattle),250/1000 + 450/1000 - 0/1000,=700/1000 = 0.7,Omaha,Seattle,Miami,Total,Yes,100,150,150,400,No,125,130,95,350,Undecided,75,170,5,250,Total,300,450,250,1000,250/1000 + 400/1000 - 150/1000,=500/1000 = 0.5,26,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, dont forget to subtract the probability of both occurring.,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.,Summary,27,Section 3.4,Counting Principles,28,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?,=,12 meals,Start,2,Soup,*,3,Main,2,*,Dessert,Fundamental Counting Principle,29,Factorials,Suppose you want to arrange,n,objects in order.,There are,n,choices for 1,st,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,!,30,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?,There are 6720 permutations of 8 books reading 5.,Permutations,31,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.,There are 56 combinations of 8 objects taking 5.,Combinations,32,2,3,Combinations of 4 objects choosing 2,1,4,Each of the 6 groups represents a combination,1,2,1,3,1,4,3,4,2,4,2,3,33,2,3,Permutations of 4 objects choosing 2,1,4,Each of the 12 groups represents a permutation.,1,2,1,2,1,3,1,3,2,3,2,3,1,4,3,4,2,4,1,4,3,4,2,4,34,
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