MATH 10: Elementary Statistics and Probability Chapter 3: Probability Topics

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1 MATH 10: Elementary Statistics and Probability Chapter 3: Probability Topics Tony Pourmohamad Department of Mathematics De Anza College Spring 2015

2 Objectives By the end of this set of slides, you should be able to: 1 Understand and use the terminology of probability 2 Determine whether two events are mutually exclusive and whether two events are independent 3 Calculate probabilities using Addition and Multiplication rules 4 Construct and interpret tree diagrams 2 / 53

3 Some Necessary Terminology Experiment: A planned operation carried out under controlled conditions Outcome: A result of an experiment Sample Space: Set of all possible outcomes of an experiment Event: Any combination of outcomes; A subset of the sample space Example: Flipping a fair coin is an example of an experiment Outcome: You either get a heads (H) or a tails (T) Sample Space: S = {H, T} Event: You get a heads Other examples? 3 / 53

4 So What Is Probability? This is like a debate on religion or philosophy in the field of statistics There is no one way to define probability So in this course how will we define probability? Probability The probability of any outcome is the long-term relative frequency of that outcome Probability is the likelihood or chance that something will happen A probability is a number between 0 and 1, inclusive Notation: P(A) is the probability that the event A occurs Example: A ="The event of getting a heads". So P(A) is the probability of getting a heads. 4 / 53

5 Equally Likely Events Equally likely means that each outcome of an experiment occurs with equal probability Example: If you toss a fair six sided die, each face (1,2,3,4,5 or 6) is as likely to occur as any other face Example: If you toss a fair coin, a Head (H) and a Tail (T) are equally likely to occur To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event A and divide by the total number of outcomes in the sample space P(A) = number of outcomes for event A total number of outcomes in S 5 / 53

6 Equally Likely Events (continued) Example: Tossing a Fair Coin Twice A fair coin has equal probability of landing on Head (H) or Tail (T). Find the probability of getting ONE HEAD in two tosses: Sample space of outcomes for tossing a coin TWICE: S = {HH, HT, TH, TT} A ="Getting ONE HEAD in two tosses" A = {HT, TH} number of outcomes for event A P(A) = total number of outcomes in S = 2 4 = / 53

7 Example #1 Imagine rolling 1 fair die The sample space S = {1, 2, 3, 4, 5, 6} Event Odd Even 2 or 4 number 4 Event A = {1, 3, 5} B = {2, 4, 6} D = {2, 4} T = {1, 2, 3, 4} Probability P(A) = 3 6 P(B) = 3 6 P(D) = 2 6 P(T) = 4 6 Are all the events equally likely? 7 / 53

8 Compound Events Compounds events are created using "AND" or "OR" "AND" event: "A and B" means BOTH events A and B occur The outcome that occurs satisfies both events A and B Event "A and B" includes items in common to both (intersection of) A and B Example: Let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}, then "A and B" = {4, 5} "OR" event: "A or B" means either event A occurs or event B occurs or both occur The outcome that occurs satisfies event A or event B or both Event "A or B" is the union of items from these events Example: Let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}, then "A or B" = {1, 2, 3, 4, 5, 6, 7, 8} 8 / 53

9 Complements The complement of an event A is denoted A (read "A prime") A means that the event A did NOT occur A consists of all outcomes that are NOT in A Example: Let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then A = {5, 6} Example: If A is the event that it rains tomorrow, then A is the event that it does NOT rain tomorrow Complement Rule: P(A) + P(A ) = 1 = P(A) = 1 P(A ) Example: Let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then A = {5, 6} and furthermore P(A) = 4 6 and P(A ) = 2 6 so P(A) + P(A ) = = 1 9 / 53

10 Conditional Probability Probability that event A occurs if we know that outcome B has occurred We say this as "given that": We write this using a line that means: given that P(A B) = "Probability that event A occurs given that outcome B has occurred" P(event condition) The outcome that we know has occurred is called the condition The condition is after the line The condition reduces the sample space to be smaller by eliminating outcomes that did not occur 10 / 53

11 Example #2: Tossing a Fair Coin Twice A fair coin has equal probability of landing on Head (H) or Tail (T) Sample space of outcomes for tossing a coin TWICE: S = {HH, HT, TH, TT} 1 Find the probability of getting TWO HEADS in two tosses of the coin: So what is A? P(A) = Find the probability of getting TWO HEADS in two tosses IF WE KNOW THAT (GIVEN THAT) the FIRST TOSS WAS A HEAD. A = the event of getting two heads in two tosses B = the first toss was a head S = {HH, HT}, so P(A B) = / 53

12 Example #2 Continued 3 Find the probability of getting ONE HEAD in two tosses. A = the event of getting ONE HEAD in two tosses P(A) = 2 4 = Find the probability of getting ONE HEAD in two tosses GIVEN THAT the FIRST TOSS WAS A HEAD. A = the event of getting ONE HEAD in two tosses B = the first toss was a head S = {HH, HT}, so P(A B) = Find the probability of getting ONE HEAD in two tosses, GIVEN THAT AT LEAST ONE HEAD was obtained. 12 / 53

13 Example #2 Continued 5 Find the probability of getting ONE HEAD in two tosses, GIVEN THAT AT LEAST ONE HEAD was obtained. A = the event of getting ONE HEAD in two tosses B = at least one head was obtained S = {HH, HT, TH}, so P(A B) = / 53

14 Example #3: Rolling a Fair Die Imagine rolling 1 fair die The sample space S = {1, 2, 3, 4, 5, 6} Event Odd Even 2 or 4 number 4 Event A = {1, 3, 5} B = {2, 4, 6} D = {2, 4} T = {1, 2, 3, 4} Probability P(A) = 3 6 P(B) = 3 6 P(D) = 2 6 P(T) = Find the probability of rolling a number 4 GIVEN THAT the outcome is even P(T B) = P(1 or 2 or 3 or 4 2 or 4 or 6) = 2 3 The "reduced" (smaller) sample space is S = {2, 4, 6}; odd numbers were removed from the sample space 14 / 53

15 Example #3 Continued: Rolling a Fair Die 2 Use the reduced sample space to find the probability of rolling an odd number given that the outcome is 4. P(odd 4) = P(A T) = P(1 or 3 or 5 1 or 2 or 3 or 4) = 2 4 = Find the probability of rolling an odd number P(odd) = P(A) = 3 6 = 1 2 Is the event in 2) more likely to occur than the event in 3)? 15 / 53

16 Important to Note For AND and OR events, the order of listing the events does not matter and can be switched, i.e., P(A and B) = P(B and A) and P(A or B) = P(B or A) For CONDITIONAL PROBABILITY the order is important, i.e., in most situations P(A B) P(B A) 16 / 53

17 Mutually Exclusive Events Mutually Exclusive Events: Two events A and B that cannot occur at the same time are said to be mutually exclusive (also called disjoint) If two events are mutually exclusive, then P(A and B) = 0 Examples: List a pair of events above that are mutually exclusive and are not complements List a pair of events above that are mutually exclusive and are complements Are being a part-time and a full-time student at De Anza mutually exclusive? Are being a day student and a night student at De Anza mutually exclusive? 17 / 53

18 Independent Events Independent Events: Two events, A and B, are independent if and only if the probability of one event occurring is not affected by whether the other event occurs or not If A and B are independent, then P(A B) = P(A) and P(B A) = P(B). Why? If two events are NOT independent, then we say that they are dependent Example: Two events that are independent A = the event of getting a HEAD on a coin flip B = the event of getting a 4 when rolling a die Other examples of independent or dependent events? 18 / 53

19 With and Without Replacement With Replacement: If each member of the population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, the events are independent Without Replacement: When sampling is done without replacement, each member of a population may be chosen only once When sampling is done without replacement, the events are dependent Examples? 19 / 53

20 With and Without Replacement Example: I have a bag with 4 blue marbles and 6 red marbles. What is the probability of pulling out a blue marble? A = pulling out a blue marble S = {B 1, B 2, B 3, B 4, R 1, R 2, R 3, R 4, R 5, R 6 } P(A) = 4 10 Now imagine I pull marbles out of the bag without replacement, and on the first draw I pulled out a blue marble, what is the probability of pulling out a blue marble now? P(A) = 3 9 Why? What would have happened if I had done it with replacement? 20 / 53

21 Mutually Exclusive Versus Independence Note that the concepts of mutually exclusive and independent are not equivalent For example, two roles of a fair die are independent events, however, they are not mutually exclusive Two events that are not mutually exclusive may or may not be independent Two events that are mutually exclusive must be dependent 21 / 53

22 Example #4 Are the events independent? Show justification using probabilities and state your conclusion Event Odd Even 2 or 4 number 4 Event A = {1, 3, 5} B = {2, 4, 6} D = {2, 4} T = {1, 2, 3, 4} Probability P(A) = 3 6 P(B) = 3 6 P(D) = 2 6 P(T) = Are events A = "odd number" and T = "number 4" independent? and P(A T) = 2 4 = 1 2 P(A) = 3 6 = 1 2 Conclusion: The events A and T are independent since P(A T) = P(A) 22 / 53

23 Example #5 Which of the following describe independent events? Repeated tosses of a coin Selecting 2 cards consecutively from a deck of 52 cards, without replacement Selecting 2 cards from a deck of cards, with replacement The numbers that show on each of two dice when tossed The color of two marbles selected consecutively from a jar of colored marbles, without replacement The color of two marbles selected from a jar of colored marbles, with replacement 23 / 53

24 Example #6 In India, adult (15+ years) literacy rates are 82.1% for men and 65.5% for women The overall literacy rate is estimated as approximately 74%. Is the literacy rate in India independent of gender? Justify your answer using appropriate probabilities. Consider the population of residents of India age 15 and over: Events: F = female, M = male, L = literate Now, P(L M) = 82.1% yet P(L) = 74% so P(L M) P(L) and so they are not independent. Source: 24 / 53

25 Two Basic Rules of Probability When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not The rules 1 The Addition Rule and 2 The Multiplication Rule 25 / 53

26 The Addition Rule The Addition Rule: Let A and B be events, then P(A or B) = P(A) + P(B) P(A and B) If A and B are mutually exclusive events, then P(A or B) = P(A) + P(B) Example: Imagine flipping a fair coin twice A = the event of getting a HEADS (H) on the first flip B = the event of getting a HEADS (H) on the second flip A and B are not mutually exclusive, why? So what is the P(A or B)? Does it make sense? P(A or B) = P(A) + P(B) P(A and B) = = / 53

27 Example #7: Rolling Two Dice The sample space for when rolling two dice is (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) What is the probability of a rolling a sum of 6? A = rolling a sum of 6 P(A) = 5 36 What is the probability of rolling a double (i.e., the same numbers)? B = rolling a double P(B) = / 53

28 Example #7 Continued: Rolling Two Dice (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) What is the probability of a rolling a sum of 6 or a double? A = rolling a sum of 6 B = rolling a double P(A or B) = P(A) + P(B) P(A and B) Are A and B mutually exclusive? = = / 53

29 The Multiplication Rule The Multiplication Rule: Let A and B be events, then P(A and B) = P(A B)P(B) If A and B are independent events, then P(A and B) = P(A)P(B) Example: Imagine flipping a fair coin twice A = the event of getting a HEADS (H) on the first flip B = the event of getting a HEADS (H) on the second flip A and B are independent, why? So what is the P(A and B)? Does it make sense? P(A and B) = P(A)P(B) = = / 53

30 Example #8: Bag of Marbles Imagine you are picking marbles out of a bag without replacement There are 4 blue marbles and 6 red marbles What is the probability of selecting a blue marble on the 1st draw and a red marble on the 2nd? B = selecting a blue marble R = selecting a red marble Are these events independent? P(B and R) = P(B)P(R B) = = What about the same problem but with replacement? P(B and R) = P(B)P(R) = = / 53

31 Example #8 Continued: Bag of Marbles Imagine picking marbles out of a bag without replacement There are 4 blue marbles and 6 red marbles What is the probability of selecting a blue marble and a red marble (in any order)? B = selecting a blue marble R = selecting a red marble P(B and R) = P(B on 1st and R on 2nd OR R on 1st and B on 2nd) = P(B on 1st and R on 2nd) + P(R on 1st and B on 2nd) = = Why didn t I subtract P(B on 1st and R on 2nd AND R on 1st and B on 2nd) 31 / 53

32 The Conditional Probability Rule The Conditional Probability Rule: If P(B) 0 then P(A B) = P(A and B) P(B) Why is this true? Recall the multiplication rule: P(A and B) = P(A B)P(B) If I multiply both sides by 1/P(B) then I obtain 1 1 P(A and B) = P(A B)P(B) P(B) P(B) P(A and B) = = P(A B) P(B) P(B) P(B) P(A and B) = = P(A B) P(B) 32 / 53

33 Example #9 In a certain neighborhood: 65% of residents subscribe to the Mercury News 30% of residents subscribe to the SF Chronicle 20% of residents subscribe to both newspapers Let M = person subscribes to the Mercury News and C = person subscribes to the Chronicle Find the probability that a person subscribes to the Mercury News given that he/she subscribes to the SF Chronicle What does this mean in the language of probability? We want to find P(M C), so using the conditional probability rule P(M C) = P(M and C) p(c) = = / 53

34 Example #10: Bag of Marbles Imagine you are picking marbles out of a bag without replacement There are 4 blue marbles and 6 red marbles What is the probability of selecting a blue marble on the 2nd draw given that a red marble was selected on the 1st draw? B = selecting a blue marble R = selecting a red marble P(B R) = = Could I have solved this faster? P(B and R) P(R) = = / 53

35 Contingency Tables A contingency table displays data for two variables The contingency table shows the number of individuals or items in each category Example of a contingency table: Lunger Cancer No Lung Cancer Total Smoker Non-Smoker Total / 53

36 Contingency Tables We can use the data in the table to find probabilities All probabilities, EXCEPT conditional probabilities, have the grand total in the denominator Lunger Cancer No Lung Cancer Total Smoker Non-Smoker Total Example: What is the probability of not having lung cancer? P(No Lung Cancer) = = 1 4 Example: What is the probability of being a smoker? P(Being a smoker) = = / 53

37 Contingency Tables The condition limits you to a particular row or column in the table The denominator will be the total for the row or column in the table that corresponds to the condition Lunger Cancer No Lung Cancer Total Smoker Non-Smoker Total Example: What is the probability of not having lung cancer given you were a smoker? P(No Lung Cancer Smoker) = = / 53

38 Contingency Tables Lunger Cancer No Lung Cancer Total Smoker Non-Smoker Total Example: What is the probability of not having lung cancer and being a smoker? P(No Lung Cancer and Smoker) = = 2 10 alternatively, P(No Lung Cancer and Smoker) = P(No Lung Cancer Smoker)P(Smoker) = = / 53

39 Contingency Tables Lunger Cancer No Lung Cancer Total Smoker Non-Smoker Total Example: What is the probability of not having lung cancer or being a smoker? P(No Lung Cancer or Smoker) = P(No Lung Cancer) + P(Smoker) P(No Lung Cancer and Smoker) = = / 53

40 Example #11 A large car dealership examined a sample of vehicles sold or leased in the past year. Car (C) SUV (S) Van (V) Truck (T) Total New Vehicle (N) Used Vehicle (U) Leased Vehicle (L) Total Example: What is the probability that the vehicle was leased? P(L) = Example: What is the probability that the vehicle was a SUV? P(S) = / 53

41 Example #11 Continued A large car dealership examined a sample of vehicles sold or leased in the past year. Car (C) SUV (S) Van (V) Truck (T) Total New Vehicle (N) Used Vehicle (U) Leased Vehicle (L) Total Example: What is the probability that the vehicle was a truck? P(T) = Example: What is the probability that the vehicle was not a truck? P(T ) = 1 P(T) = = / 53

42 Example #11 Continued Test yourself: 1 Find the probability that the vehicle was a car AND was leased 2 Find the probability that the vehicle was used GIVEN THAT it was a van 3 Find the probability that the vehicle was used OR was a van Solutions: 1 34/ / /300 Hint: Use the addition rule 42 / 53

43 Example #11 Continued Car (C) SUV (S) Van (V) Truck (T) Total New Vehicle (N) Used Vehicle (U) Leased Vehicle (L) Total Example: Are the events N and V independent? and P(N V) = P(N) = P(N V) P(N) so not independent 43 / 53

44 Tree Diagrams A Tree Diagram is a special type of graph used to determine the outcomes of an experiment It consists of "branches" that are labeled with either frequencies or probabilities Each "branch" represents a mutually exclusive outcome Tree diagrams can make some probability problems easier to visualize and solve 44 / 53

45 Example #12 An urn contains 11 marbles, 3 Red and 8 Blue We are selecting 2 marbles randomly from the urn with replacement Draw the tree diagram using frequencies. 45 / 53

46 Example #12 Continued Why 9RR? Imagine we label all the marbles {B 1, B 2, B 3, B 4, B 5, B 6, B 7, B 8, R 1, R 2, R 3 } There are 9 unique ways I can draw two red marbles with replacement, i.e., {(R 1, R 1 )(R 1, R 2 ), (R 2, R 1 ), (R 1, R 3 ), (R 3, R 1 ), (R 2, R 2 ), (R 2, R 3 ), (R 3, R 2 ), (R 3, R 3 )} 46 / 53

47 Example #12 Continued Could also write the tree using probabilities 47 / 53

48 Example #12 Continued What do each of those values in the previous slide mean? 48 / 53

49 Example #12 Continued What is the probability of getting a blue marble on the 1st draw and a blue marble on the 2nd? What is the probability of getting a red marble on the 1st draw and a blue marble on the 2nd? 49 / 53

50 Example #12 Continued What is the probability of getting a blue marble on the 1st draw OR a red marble on the 1st draw? 50 / 53

51 Example #12 Continued What is the probability of getting a red on the 1st and blue on the 2nd OR a blue on the 1st or red on the 2nd? P(RB or BR) = = / 53

52 Example #12 Continued What is the probability of getting a red on the 2nd given that you picked a blue on the 1st? 52 / 53

53 Tree Diagrams: Some Things to Remember Each complete path through the tree represents a separate mutually exclusive outcome in the sample space Some steps 1 Draw a tree representing the possible mutually exclusive outcomes 2 Assign conditional probabilities along the branches of the tree 3 Multiply probabilities along each complete path through the tree to find probabilities of each "AND" outcome in the sample space 4 Add probabilities for the appropriate paths of a tree to find the probability of a compound OR event 53 / 53

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