The possible experimental outcomes: 1, 2, 3, 4, 5, 6 (Experimental outcomes are also known as sample points)

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1 Chapter 4 Introduction to Probability Experiments, Counting Rules and Assigning Probabilities Example Rolling a dice you can get the values: S = {1, 2, 3, 4, 5, 6} S is called the sample space. Experiment: Rolling a dice The possible experimental outcomes: 1, 2, 3, 4, 5, 6 (Experimental outcomes are also known as sample points) The number of experimental outcomes: 6 Defining the outcomes: E 1 = 1 E 2 = 2 E 3 = 3 E 4 = 4 E 5 = 5 E 6 = 6 Value on the dice is a one Value on the dice is a two Value on the dice is a three Value on the dice is a four Value on the dice is a five Value on the dice is a six Then: P(E 1 ) = P(E 2 ) = P(E 3 ) = P(E 4 ) = P(E 5 ) = P(E 6 ) = Note P(E 1 ) + P(E 2 ) + P(E 3 ) + P(E 4 ) + P(E 5 ) + P(E 6 ) = = =1 More examples of experiments: Coin: Select a part for inspection: Play a football game: Counting Rules, Combinations and Permutations Question: How many experimental outcomes are there? Copyright Reserved 1

2 2 Multiple-Step Experiments Example: Throwing 2 coins can be thought of as a two-step experiment in which step 1 is the throwing of the first coin and step 2 is the throwing of the second coin. Sample Space (S): The number of experimental outcomes: Tree Diagram: A graphical representation that is helpful in visualizing a multiple-step experiment is a tree diagram. A tree diagram for the experiment of tossing 2 coins: Copyright Reserved 2

3 Counting Rule for Multiple-Step Experiments: 3 If an experiment can be described as a sequence of k steps with n 1 = possible outcomes on step 1. n 2 = possible outcomes on step 2. n k = possible outcomes on step k. Then the total number of experimental outcomes is: (n 1 )(n 2 ) (n k ) The coin example: Example: How many different routes can a parcel follow if 2 starting venues 3 cars 4 delivery points. Copyright Reserved 3

4 Treediagram: 4 Copyright Reserved 4

5 5 Factorials! There are 3 people at a bus stop. How many different ways can the 3 people be arranged? Note: 0! = 1 Calculator: n! or x! Examples: 1. 6! = 3 factorial is: 3! = (3)(2)(1) = 6 5 factorial is: 5! = (5)(4)(3)(2)(1) = ! = Combinations Order of selection is NOT important The number of combinations of N objects taken n at a time is: = =!!! Example: Consider a quality control process in which an inspector selects 2 of 5 parts, to inspect for defects. In a group of 5 parts, how many combinations of 2 parts can be selected? Example: Lotto: Use numbers 1 to 49 Choose a group of 6 numbers Question: How many experimental outcomes are there? Calculator: or Examples: 1. = 2. = 3. = Copyright Reserved 5

6 Permutations 6 The counting rule for permutations allows one to compute the number of experimental outcomes when n objects are to be selected from a set of N objects where the order of selection is important. The number of permutations of N objects taken n at a time is given by: =! =!! Example: Note: Order of selection is important Consider a quality control process in which an inspector selects 2 of 5 parts to inspect for defects. How many permutations may be selected? Calculator: or Assigning Probabilities The 3 approaches most frequently used: 1. Classical Method 2. Relative Frequency Method 3. Subjective Method Classical Method The classical method of assigning probabilities is appropriate when all the experimental outcomes are equally likely. If n experimental outcomes are possible, a probability of is assigned to each experimental outcome. Examples: (i) Rolling a dice: P(E i ) = i = 1, 2, 3, 4, 5, 6 (ii) Throwing a coin: P(E i ) = i = 1, 2 Copyright Reserved 6

7 Relative Frequency Method 7 It s appropriate when data are available to estimate the proportion of the time the experimental outcome will occur if the experiment is repeated a large number of times. Example: Consider a study of waiting times in the X-ray department for a local hospital. The number of patients waiting for service at 9am was recorded for 20 successive days. The following results were obtained: Then: P(0) = P(1) = P(2) = P(3) = P(4) = Number waiting Frequency Note: The bigger your sample, the closer your probabilities will be to the true probabilities. Subjective Method i. It is most appropriate when it is unrealistic to assume that the experimental outcomes are equally likely (Can t use Classical Method) OR ii. when little relevant data are available (can t use Relative Frequency Method) Example: A ranger thinks that the probability of seeing a lion is 0.3. Then the probability of not seeing a lion is 0.7. Copyright Reserved 7

8 4.3 Some Basic Relationships of Probability 8 Figure 4.4 Complement of event A is shaded Figure 4.5 Union of events A and B is shaded P(A) + P(A C ) = 1 Figure 4.6 Intersection of events A and B is shaded Addition law: = + Copyright Reserved 8

9 Figure 4.7 Mutually exclusive events Conditional Probability Example 1: = Addition law for mutually exclusive events: = + with =0 Therefore, = + Gender Promoted Male Female Total Yes No Total Define the events: A = Promoted and M = Male. Then A C = Not promoted and M C = Female. 1. The probability that an employee is promoted? Note: This is a marginal probability. 2. The probability that an employee is not promoted? Note: This is a marginal probability. 3. The probability that an employee is male? Note: This is a marginal probability. Copyright Reserved 9

10 4. The probability that an employee is female? Note: This is a marginal probability The probability that a randomly selected employee is male and promoted? Note: This is a joint probability. 6. The probability that a randomly selected employee is female and promoted? Note: This is a joint probability. 7. The probability of getting a promotion, given the employee is male. Note: This is a conditional probability. 8. The probability of getting a promotion, given the employee is female. Note: This is a conditional probability. 9. The probability of being male, given that the employee was promoted. Note: This is a conditional probability. 10. The probability of being female, given that the employee was not promoted. Note: This is a conditional probability. Copyright Reserved 10

11 Independent events: Two events A and B are independent if P(A B) = P(A) or P(B A) = P(B) 11 Use this definition to test for dependence / independence: We have already calculated all the appropriate marginal and conditional probabilities. In order to test for dependence / independence between gender and whether of not an employee gets promoted, any one of the following marginal and conditional probabilities may be compared to one another: Option 1: P(Male) = P(Male Promoted) = The two events are dependent Option 2: P(Female) = P(Female Not promoted) = The two events are dependent Option 3: P(Promoted) = P(Promoted Female) = The two events are dependent There are other options, for example, comparing P(Not promoted) to P(Not promoted Female) ect. Here we ve only shown 3 possible ways to test for dependence / independence between gender and whether of not an employee gets promoted. Copyright Reserved 11

12 12 Alternative definition for independence: Two events A and B are independent if = Use this definition to test for dependence / independence: We have already calculated all the appropriate marginal and joint probabilities. In order to test for dependence / independence between gender and whether of not an employee gets promoted, any one of the following marginal and joint probabilities may be compared to one another: Option 1: P(Male) = P(Promoted) = 0 P(Male Promoted) = =.. =. = The two events are dependent Option 2: P(Female) = P(Promoted) = P(Female Promoted) = =.. =. = The two events are dependent There are other options. Here we ve only shown 2 possible ways to test for dependence / independence between gender and whether of not an employee gets promoted. Example: Independent events Given: P( ) = P(Seeing a lion) = 0.3 and P( ) = P(Seeing a cheetah) = Suppose these two events are independent. Question: Calculate : Answer: It is given that events and are independent, therefore = = + = + = =. Addition law Because of independence Copyright Reserved 12

13 Typical exam questions 13 Questions 1 to 3 are based on the following information: A political analyst knows from previous experience that the probability that political party A will win the election is 0.4 and the probability that political party B will win the election is Define the events: = Political party A wins the election. = Political party B wins the election. = Political party C wins the election. Assume: There are only 3 political parties running for election. Note: Only one party can win the election. Question 1 The probability that political party C will win the election is: Answer 1 P(Party C will win) = 1 P(Party A will win) P(Party B will win) = = Question 2 Events A and B are: (A) Mutually exclusive since 0. (B) Mutually exclusive since =. (C) Not mutually exclusive since =0. (D) Not mutually exclusive since =0. (E) Mutually exclusive since = Question 3 The probability that political party A or political party B wins the election is: (A) 0.14 (B) 0.05 (C) 0.75 (D) 0.67 (E) 0.61 Answer 3 = + = =0.75. Copyright Reserved 13

14 14 Questions 4 to 7 are based on the following information: Suppose we have a sample space with five equally likely experimental outcomes, namely: E 1, E 2, E 3, E 4 and E 5. Let A = {E 1, E 2 } B = {E 3, E 4 } C = {E 2, E 3, E 5 } Question 4 Calculate P(A), P(B), P(C), and P(A C ), P(B C ), P(C C ): Answer 4 P(A) = = 0.4 and using the fact that P(A) + P(AC ) = 1 we find P(A C ) = 1 - P(A) = = 0.6. P(B) = = 0.4 and using the fact that P(B) + P(BC ) = 1 we find P(B C ) = 1 - P(B) = = 0.6. P(C) = = 0.6 and using the fact that P(C) + P(CC ) = 1 we find P(C C ) = 1 - P(C) = = 0.4. Another way to calculate P(A C ), P(B C ) and P(C C ): A C = {E 3, E 4, E 5 } and consequently P(A C ) = = 0.6. B C = {E 1, E 2, E 5 } and consequently P(B C ) = = 0.6. C C = {E 1, E 4 } and consequently P(C C ) = = 0.4. Question 5 Calculate : Answer 5 = + = + =0.8 Question 6 Calculate : Answer 6 = Question 7 Calculate : Answer 7 = =0.2 = =0 Copyright Reserved 14

15 Questions 8 to 10 are based on the following information: A survey of undergraduate students in Commerce at the XYZ University revealed the following regarding the gender and majors of the students. Major Gender Accounting Economics Total Male Female Total A student is selected at random from the 320 students. Define the events as follows: M: Male and E: Economics 15 Question 8 The probability of selecting a female who majors in Economics is: Answer 8 = =0.22. Question 9 The probability of selecting an Accounting major, given that the person selected is a male, is: Answer 9 = = =0.65 Question 10 Events M and E are not mutually exclusive because: (A) P(M E) = 0 (B) P(E M) 0 (C) P(M E) = 0 (D) P(M E) 0 (E) P(M) P(E) 0 Questions 11 and 12 are based on the following information: Suppose that we have two events, A and B, with P(A) = 0.5, P(B) = 0.6 and =0.4. Question 11 = Answer 11 = =. =0.6.. Question 12 The events A and B are: (A) dependent because (B) dependent because = (C) independent because = (D) dependent because (E) independent because = Answer 12 =0.6 and =0.5. The events are dependent, because. Answer = D. Copyright Reserved 15