Ch 14 Randomness and Probability We ll begin a new part: randomness and probability. This part contain 4 chapters: 14-17. Why we need to learn this part? Probability is not a portion of statistics. Instead it s a branch of mathematics Later when we are going to develop the advanced statistical techniques, we need to employ the knowledge of probability In chapter 14, we ll discuss random phenomenon Probabilities Rules for probability
Term 1: Random Phenomenon A phenomenon is random if we know what outcomes could happen, but not which particular outcomes will happen. Let s try the following example together: Questions: Can you identify which of the following are random? 1. Toss a coin 2. Is it going to rain on Jun 28 th,2013? 3. Date of the final exam of this course We are mainly interested in the random phenomenon when we are learning the probability part. The probability provides a numerical measure of the randomness.
Term 2: Several Definitions Term Definition Example Trial In general, each occasion upon which we observe a random phenomenon is called a trial. Rolling a fair dice once and record the number on top Outcome Event At each trial, we record the value of the random phenomenon, and call that the trial s outcome. An event is a combination of outcomes (denoted as A, B ) We say an event happens if any outcome of this event occurs. Could be any number among 1,2,3,4,5 and 6 Let event A={get a number greater than 3}={4, 5,6}, is the combination of the outcome {4},{5} and {6} Sample Space Sample space is the collection of all possible outcomes, denoted by S S={1, 2, 3, 4, 5, 6} as the collection of all possible outcomes
Term 2: Several Definitions Example: Suppose we are going to toss two fair coins (one 1 cent coin, one 1 quarter coin) at the same time and record the patterns on top. 1) Please identify the trial, outcome and the sample space Term Trial Outcome Sample Space Answer Toss two fair coins at the same time for once and record the top patterns. Any one result among the followings: Both heads, both tails, 1 cent head and 1 quarter tail, and 1 cent tail and 1 quarter head. S={HH,TT,HT,TH} 2) Try to express the event A={we get two different patterns} A={ HT, TH}
Term 2: Several Definitions Example: Please identify the sample space for the following trials 1) Toss a coin repeatedly; record the number of the tosses you have done until you get the first head S= 2) Toss a coin 10 times; record the number of heads. S= 3) Roll two dice at the same time; record the sum of two numbers. S=
Term 3: Modeling Probability The probability, like the chance, is a numerical measurement of the likelihood of the random phenomenon Unlike the chance which is mostly expressed by percentages, the probability is mostly expressed in decimal-place form. Fro example, we write 50% chance in the probability form as 0.5 We denote P(event) for the probability of a particular event.
Term 3: Modeling Probability Let s start with something simple. Consider the following trials: 1) Flip a fair coin, record the pattern on top S={H, T} 2) Toss a fair die, record the number on top S={1, 2, 3, 4, 5, 6} The common place between the two events is: All possible outcomes are equally likely to happen. Example: Are the outcomes equally likely to happen in the following trial? 1) Choose a card from a complete deck with 52 cards and record its color. 2) Toss two fair coins at the same time, and record the number of heads we get.
Term 3: Modeling Probability When outcomes are equally likely to happen, the probability that any single outcome happens is easy to compute-it is just one divided by the total number of possible outcomes. For example, 1) The probability of getting a head with a fair coin is ½=0.5 2) The probability of rolling a 3 with a fair die is 1/6. For the event A that are made up of several equally likely outcomes, P( A) = # # all outcomes possible in A outcomes For example, The probability of rolling an odd number with a fair die is 3/6=0.5 since # of outcomes in rolling an odd number is 3 (i.e. {1,3,5}) and # of all possible outcomes is 6 (i.e. {1,2,3,4,5,6})
Term 3:Modeling Probability Example: Drawing one card from a well-shuffled deck with 52 cards. Q1:What is the probability of picking a face card (JQK)? Q2: What is the probability of picking a heart? Q3: What is the probability of picking a seven? Q4: What is the probability of picking a black card?
Term 3:Modeling Probability Caution: The formula P( A) = # all possible outcomes can only be used when the outcomes are equally likely to happen. Example: Suppose we toss two fair coins together and record the number of heads we can get on top. What is the probability that we get at least one head? Here is a student s work: Sample space={0 heads, 1 head, 2 heads} # of all possible outcomes=3 Event={at least 1 head} ={1 head, 2 heads} # of outcomes in event =2 So P(at least 1 head)=2/3. # outcomes in A Question: Is this correct? Actually, it s wrong! Because the outcomes in the sample space are not equally likely to happen. P(0 heads)= ¼=0.25, P(1 head)=2/4=0.5, P(2 heads)=1/4=0.25 So P(at least one head)= ¾=0.75
Term 3:Modeling Probability Venn diagram is the most commonly used diagram to describe the events. The way we connect the probabilities with the Venn diagram is to treat the area as the corresponding probability. For example, in the Venn diagram above, The area of the Event A circle corresponds to P(A) When we solve the probability problems, it s always a good idea to draw a Venn diagram first.
Term 5 Probability rules (Part I) Rule#1 A probability is a number between 0 and 1 For any event A, 0<=P(A)<=1 Rule#2 Probability Assignment Rule: The set of all possible outcomes of a trial must have probability 1: P( S) = 1 Venn diagram
Term 5 Probability rules (Part I) Example: For a given kind of lottery, there are 3 levels of awards: 100 dollars, 10 dollars and 1 dollar. Which of the following probability assignments is possible? Award $100 $10 $1 (a) 0.15 0.50 0.10 (b) -0.1 0.5 0.6 (c) 0.5 0.5 0.5 (d) 0.1 0.2 0.7
Term 5 Probability rules (Part I) Rule#3 Complement Rule: The set of outcomes that are not in the event A is called the compliment of A, and is denoted by A C For example, 1) when we toss a coin, let event A={ H }, then A C ={T} 2) when we toss a die, let event A= { getting a number bigger than 3 }, then A C ={ getting a number smaller than or equal to 3} = {1,2,3} The probability of an event occurring is 1 minus the probability that it doesn t occur. P( A) = 1 P( C A )
Term 5 Probability rules (Part I) Rule#3 Complement Rule: Example: During the product testing, the inspector observed that 10% of the products are defective. Question: What is the probability that a randomly chosen product is not defective? Solution: P(not defective)=p(defective C ) By the complement rule, P(defective C )=1-P(defective)=1-0.10=0.90
Term 5 Probability rules (Part I) The operations for two events A and B Operation Interpretation Venn Diagram A or B A and B 1) For event A and B, at least one of these two events happens. 2) On the Venn diagram, it s the union of Circle A and Circle B 3) The set of its outcomes is the union of two sets of outcomes of event A and B 1) For event A and B, both of them happen at the same time. 2) On the Venn diagram, it s the intersection of Circle A and Circle B 3) The set of its outcomes is the intersection of two sets of outcomes of event A and B
Term 5 Probability rules (Part I) The operations for two events A and B Example: Suppose we toss a fair dice and record the number on its top. Denote two events: event A={get an odd number} and event B={get a number bigger than 3} Question: Please identify the set of the outcomes of a) A and B Answer: A and B= { 5 } b) A or B Answer: A or B={ 1, 3, 4, 5, 6}
Term 5 Probability rules (Part I) Rule#4 Addition Rule: We say two events are disjoint (or mutually exclusive) if they have no outcomes in common. In other words, two disjoint events can t happen together at the same time. B Provided that A and B are disjoint, A Two disjoint sets, A and B Q: When we toss a fair dice, which of the following events are disjoint? 1) A={ get an odd number} B={get an even number} 2) A= {get an odd number} B={get a number bigger than 3} P ( AorB) = P( A) + P( B)
Term 5 Probability rules (Part I) Rule#4 Addition Rule: Example: Suppose at an intersection, we will meet a green light with a chance of 35% and a yellow light with a chance of 50%. Question: What is the probability that we will meet a greed light or a yellow light? Solution: P(green)=0.35 P(yellow)=0.50 and {meet a green light} and {meet a yellow light} are disjoint. By the addition rule, P(green or yellow)= P(green)+P(yellow)=0.35+0.50=0.85
Review of Chapter 14 In this chapter, we learned Random Phenomenon (Definition of trail, outcome, event, sample space) Modeling Probability What is probability? How to calculate probability when outcomes are equally likely to happen? Venn diagram P( A) = # # all outcomes possible in A outcomes Some Probability rules (To be reviewed in detail)
Review of Probability Rules in Chapter 14 Rule #1: A probability is a number between 0 and 1. Rule #2(Probability Assignment Rule): The set of all possible outcomes of a trail must have probability 1. P( S) = 1 Rule #3(Complement Rule):The probability of an event occurring is 1 minus the probability that it doesn t occur. P( A) = 1 P( A The operations for two events A and B. A or B: Union of two sets of outcomes of event A and B: Intersection of two sets C ) Rule #4: Additional Rule. If A and B are disjoint, P(A or B)=P(A)+P(B) Rule #5: Multiplication Rule. (next)
Term 5 Probability rules (Part I) Rule#5 Multiplication Rule: We say two events are independent intuitively if the occurrence of one event doesn t affect the probability that the other event happens. Disjoint Independent because: 1) If two events are independent with each other, then they still can happen together at the same time. So two independent events can never be disjoint! 2) If two events are disjoint with each other, then one event happens will directly eliminate the possibility that the other event can happen. So two disjoint events can never be independent!
Term 5 Probability rules (Part I) Rule#5 Multiplication Rule: Disjoint Independent Example: Suppose we toss two fairs coins one-by-one. For the following two pairs of events, please identify which pair is disjoint and which pair is independent. a) A={first coin is a head} B={First coin is a tail} Answer: A and B are disjoint since they can t happen together at the same time. b) A={first coin is a head} B={Second coin is a tail} Answer: A and B are independent since the occurrence of event A doesn t affect the probability that B happens. We toss the two coins one-by-one (independent tosses).
Term 5 Probability rules (Part I) Rule#5 Multiplication Rule: Come back to the Multiplication Rule Provided that A and B are independent, P ( A and B) = P( A) P( B) Example: (Continue with the light problem) Suppose at an intersection, we will meet a green light with a chance of 35% and a yellow light with a chance of 50%. Question: What is the probability of meeting a green light on Monday and Tuesday? Solution: Denote A={green on Mon} B={green on Tue} Then A and B are independent with each other. By Multiplication rule, P(green on Mon and Tue)= P(A and B)=P(A) P(B)=0.35 2 =0.1225
Term 5 Probability rules (Part I) Finally, let s learn how to apply all the rules under a complex situation Example: The Masterfoods company said yellow candies made up 20% of their plain M&M, red another 20%, and orange, blue, and green each made up 10%. The rest were brown a)if you pick an M&M at random, what is the probability that 1) it is brown? 2) It is yellow or orange? 3) It is not green? 4) It is red and green? b)if you pick two M&M s in a row independently, what is the probability that 1) They are all brown? 2) The second one is the first one that s red? 3) None are yellow? 4) at least one green?
Ch15 Probability Rules In this chapter, we are going to expand the discussion in Ch14 General Addition Rule Conditional Probability General Multiplication Rule Redefine the Independence Tree diagram
Term 1 : General Addition Rule Review of addition rule P(A or B)= P(A) + P(B), provided that A and B are disjoint. A B
Term 1 : General Addition Rule Let s look at the following example. A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program. Q: What s the probability that a randomly selected student either lives or eats on campus? Is it safe to say : P(live or eat)=p(live)+p(eat)=0.56+0.62=1.18? No, we are missing the condition disjoint!
Term 1 : General Addition Rule If the two events are not disjoint, how to find P(A or B)? Q: Is it still true that P( A or B) =P(A) +P(B)? The Answer is : NO!. If you simply add P(A) and P(B) together, we add the intersection part P(A and B) twice!
Term 1 : General Addition Rule To fix this problem, we just need to subtract one P(A and B) from the sum. This gives us the General Addition Rule For any two events A and B, P( A or B) = P(A) +P(B)-P(A and B)
Term 1 : General Addition Rule Continue with the previous example. A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program, and 42% do both. Q: What s the probability that a randomly selected student either lives or eats on campus? Notation: L={live on campus} M={has a meal plan} P(L)=0.56 P(M)=0.62 P(L and M)=0.42
Term 1 : General Addition Rule Solution: Draw a Venn diagram first. With the info: P(L)=0.56 P(M)=0.62 and P(L and M)=0.42 By the General Addition Rule: P(a student either live or eat on campus) =P(L or M)= P(L)+P(M)-P(L and M) =0.56+0.62-0.42=0.76
Term 1 : General Addition Rule Comparison between the addition rule and the general addition rule Addition Rule P(A or B)= P(A) +P(B), provided events A and B are disjoint General Addition Rule P(A or B)= P(A) +P(B)-P(A and B), for any two events A and B 1) Just like their names, the general addition rule is a generalization of the addition rule i.e. when A and B are disjoint, we have P(A and B) = 0 P(A or B)= P(A)+P(B)-P(A and B)=P(A)+P(B) 1) To use the addition rule, we always have to check the disjoint condition first. But for the general addition rule, we don t need to check
Term 2: Advanced Use of Venn Diagram Suppose we have two events A and B now, and their relationship is displayed by using the following Venn Diagram So far, we have learned the following interpretations:
Term 2: Advanced Use of Venn Diagram Next, let s look at some new interpretations from the Venn Diagrams Example (Continue) : A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program, and 42% do both. P(L)=0.56 P(M)=0.62 P(L and M)=0.42 Q1: What is the probability that a randomly selected student lives on campus but has no meal plan? Solution: P(L and M C ) =P(L) P(L and M) =0.56-0.42=0.14
Term 2: Advanced Use of Venn Diagram Example (Continue) : A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program, and 42% do both. P(L)=0.56 P(M)=0.62 P(L and M)=0.42 Q2: What is the probability that a randomly selected student has a meal plan but lives off campus? Solution: P(M and L C ) =P(M) P(L and M) =0.62-0.42=0.20
Term 2: Advanced Use of Venn Diagram Example (Continue) : A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program, and 42% do both. P(L)=0.56 P(M)=0.62 P(L and M)=0.42 Q3: What is the probability that a randomly selected student neither has a meal plan nor lives on campus? Solution: P(M C and L C ) =1-P(M or L) =1-(0.56+0.62-0.42) =1-0.76=0.24
Term 2: Advanced Use of Venn Diagram Example (Continue) : A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program, and 42% do both. P(L)=0.56 P(M)=0.62 P(L and M)=0.42 Q4: What is the probability that a randomly selected student either has a meal plan or lives on campus, but doesn t own both? Solution: P(L or M)-P(L and M) =P(L)+P(M)-P(L and M)-P(L and M) =P(L)+P(M)-2*P(L and M) =0.56+0.62-2*0.42=0.34 Or P(L and M C ) + P(M and L C )=0.14+0.20=0.34 Same!