1. The Practice of Statistics Third Edition Chapter 6: Probability and Simulation: The Study of Randomness 6.3 General Probability Rules Copyright © 2008 by W. H. Freeman & Company Daniel S. Yates

2. Essential Questions What is the addition rule for disjoint events? What is the general addition rule for union of two events? How do you compute P(A U B)? What is a joint event and joint probability? What is the general multiplication rule for any two events? What is meant by the conditional probability P(A| B)? How do you define independent events in terms of conditional probability?

3. Probability Rules from 6.2

4. Venn Diagrams: Disjoint Events A B S

5. Union P(A U B U C) = P(A) + P(B) + P(C)

7. Venn Diagrams: Non-disjoint Events A B S A and B

8. Venn Diagram, Non-Disjoint, in set notation Suppose set A consists of the following and set B consists of: Notice that 6 is included in both sets. So, in order to find the set (A U B) you must subtract one 6.

9. Venn Diagrams: Non-disjoint Events A B S A and B

10. Joint Event Joint Event is the simultaneous occurrence of two events. For example, the outcome in set of A is six and the outcome in set of B is also six. The joint probability of a joint event is P(A and B). P(A and B) is the same as P(A ∩ B).

11. General Rule for Union of Two Events

12. Example 1 In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes. A person selected at random from the town. What is the probability that the person will have brown hair or brown eyes? P( BH or BE) P( BH or BE) = P(BH) + P(BE) – P(BH ∩ BE) =.40 +.25 -.15 = 0.5 What is the probability that person selected does not have brown hair or brown eyes? 1 – P( BH or BE) = 1 -.5 = 0.5

13. Two Methods for Picturing Probabilities In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes. Venn Diagram Brown Hair Brown Eyes.25.10.15.5 Table BrownHair YesNoTotal BrownYes0.150.25 EyesNo Total0.401.00

14. Independent events The outcome of one trial does not influence or change the outcome of another trial.

15. Multiplication Rule For two independent events A and B, the probability that both A and B occur is the product of the probabilities of the two events. P(A and B) = P(A) x P(B) P(A ∩ B) = P(A) x P(B) A B A∩B

16. Independence Please note, that we can use the multiplication rule for independent events, to verify if two events are independent. If P(A and B) ≠ P(A)·P(B), then the events are not independent.

17. Example 1 Continued In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes. BrownHair YesNoTotal BrownYes0.150.25 EyesNo Total0.401.00 What is the probability of selecting a person with brown hair? P(A) What is the probability of selecting a person with brown eyes? P(B) P(A) = 0.40 P(B) = 0.25 What is the probability of selecting a person with brown hair and brown eyes? P( A and B) P(A and B) = 0.15

18. Conditional Probability Lets consider the situation of finding the probability of one event under the condition that we know the results of the other event. The notation for conditional probability is: P(A| B) The bar means “given the information that…” So, P(A| B) reads, “The probability of event A given event B occurs.”

19. Lets Consider Two Events That May or May Not Be Independent If we take the equation P(A ∩ B) = P(A)P(B| A) and solve for P(B| A) we get the equation for calculating P(B| A).

21. Example 1 – Independence? Are Events A and B independent? P(A)●P(B) = (0.40)(0.25) = 0.10 Since P(A and B) ≠ P(A)●P(B), Events A and B are not independent. Second Method – Test P(B| A) = P(B) In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes. BrownHair YesNoTotal BrownYes0.150.100.25 EyesNo0.250.500.75 Total0.400.601.00 Event A = Brown Hair Event B = Brown Eyes Since P(A| B) ≠ 0.40, then Events are not independent.

22. Example 1 – Conditional Probability In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes. BrownHair YesNoTotal BrownYes0.150.100.25 EyesNo0.250.500.75 Total0.400.601.00 Event A = Brown Hair Event B = Brown Eyes What is the P(A ∩ B) ? Ans: P(A ∩ B) = 0.15 What is the probability that the randomly selected person has brown hair given that he has brown eyes? Ans: P(A| B) = P(A ∩ B) / P(B) = 0.15/0.25 = 0.6

23. Example 1 – Conditional Probability Continued BrownHair YesNoTotal BrownYes0.150.100.25 EyesNo0.250.500.75 Total0.400.601.00 Event A = Brown Hair Event B = Brown Eyes What is the probability of the randomly selected person has brown eyes given the person has brown hair? Ans: P(B| A) = P(A ∩ B) / P(A) = 0.15/0.40 = 0.375 What is the probability that the randomly selected person has neither brown hair nor brown eyes? Ans: P( A c ∩ B c ) = 0.50

24. Example 2 Age 18-2930-6465 and overTotal Married7,84243,8088,27059,920 Never Married 13,9307,18475121,865 Widowed362,5238,38510,944 Divorced7049,1741,26311,141 Total22,51262,68918,669103,870

25. Example – Question 1 Age 18-2930-6465 and overTotal Married7,84243,8088,27059,920 Never Married 13,9307,18475121,865 Widowed362,5238,38510,944 Divorced7049,1741,26311,141 Total22,51262,68918,669103,870 A= young (between 18 and 29) P(A)=? –22512/103870

26. Example 2 – Question 2 Age 18-2930-6465 and overTotal Married7,84243,8088,27059,920 Never Married 13,9307,18475121,865 Widowed362,5238,38510,944 Divorced7049,1741,26311,141 Total22,51262,68918,669103,870 B=married P(B)=? –59920/103870

27. Example 2 – Question 3 Age 18-2930-6465 and overTotal Married7,84243,8088,27059,920 Never Married 13,9307,18475121,865 Widowed362,5238,38510,944 Divorced7049,1741,26311,141 Total22,51262,68918,669103,870 A=is young (between 18 and 29) B=married P(A and B)=? –7842/103870

28. Example 2 – Question 4 Age 18-2930-6465 and overTotal Married7,84243,8088,27059,920 Never Married 13,9307,18475121,865 Widowed362,5238,38510,944 Divorced7049,1741,26311,141 Total22,51262,68918,669103,870 A=is young (between 18 and 29) B=married P(A | B)= (Read as “the probability of A given B”) –7842/59920 This is known as a “conditional probability”

29. Example 2 – Question 5 Age 18-2930-6465 and overTotal Married7,84243,8088,27059,920 Never Married 13,9307,18475121,865 Widowed362,5238,38510,944 Divorced7049,1741,26311,141 Total22,51262,68918,669103,870 A=is young (between 18 and 29) B=married P(B | A)= (Read as “the probability of B given A”) –7842/22512

30. Example 2 Question 6 Age 18-2930-6465 and overTotal Married7,84243,8088,27059,920 Never Married 13,9307,18475121,865 Widowed362,5238,38510,944 Divorced7049,1741,26311,141 Total22,51262,68918,669103,870 P(A and B)= 7842/103870 P(A and B)= P(A)*P(B|A)

31. Example 2 – Question 7 Age 18-2930-6465 and overTotal Married7,84243,8088,27059,920 Never Married 13,9307,18475121,865 Widowed362,5238,38510,944 Divorced7049,1741,26311,141 Total22,51262,68918,669103,870 P(A and B)= 7842/103870 P(A and B)= P(B)*P(A|B)

32. Multiplication Rule Extended to Several Events Recall the general multiplication rule for the intersection of two events. Now, consider the intersection of three events. Remember

33. Example 6.29 page 448 Only 5% of male high school basketball, and football players go on to play at the college level. Of these, only 1.7% enter major league professional sports. About 40% of the athletes who compete in college and then reach the pros have a career of more than 3 years. Define – A = { competes in college} P(A) = 0.05 – B = { competes professionally} P(B|A) = 0.017 – C = { pro career longer than 3 years} P(C|A and B) = 0.40 What is the probability that a high school athlete competes in college and then goes on to have a pro career of more than 3 years? P(A and B and C) P(A and B and C) = P(A)●P(B|A)●P(C|A and B) = (0.05)(0.017)(0.40) = 0.00034 SO, only 3 out of 10,000 high school athletes will have a pro-career of more than 3 years.

34. Example 3 Using Tree Diagram Using the same probabilities from the previous problem: – A = { competes in college} P(A) = 0.05 – B = { competes professionally} P(B|A) = 0.017 Suppose the probability of a few high school athlete enter the pro’s directly from high school (does not compete in college) is 0.0001. P(B| A c ) = 0.0001 What is the probability that a high school athlete will go on to the professional sport?

35. Example 3 Using the same probabilities from the previous problem: A = { competes in college} P(A) = 0.05 B = { competes professionally} P(B|A) = 0.017 P(B| A c ) = 0.0001 High School Athletes A B BcBc AcAc B BcBc 0.05 P(B|A) = 0.017 P(B|A c ) = 0.0001 0.95 0.983 0.9999 What is the probability that a high school athlete will go on to the professional sport? P(B) = P(A ∩ B|A) + P(A c ∩ B|A c ) = 0.00085 + 0.000095 = 0.000945 P(A ∩ B|A) = 0.00085 P(A ∩ B c |A) = 0.04915 P(A c ∩ B|A c ) = 0.000095 P(A c ∩ B c |A c ) = 0.94991

36. Tree Diagrams Tree diagrams combine the addition and multiplication rules. The multiplication rule is use to reach the end of any complete branch The probability of any outcome is found by adding the probabilities of all branches that are part of the event.

37. A Simulation Problem Run a simulation for throwing two dice and finding the probability of rolling a sum of six. –Step 1: State the Problem or describe the random event. –Step 2: State the assumptions. –Step 3: Assign digits to represent outcomes. –Step 4: Simulate many repetitions. –Step 5: State your conclusions.

38. Assigning Probability to Events 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 A23456789101112 P(A) 1/362/363/364/365/366/365/361/91/121/181/36

39. Solution –Step 1: State the Problem or describe the random event. Find the probability of having a sum of six after a toss of two die. –Step 2: State the assumptions. The toss of the dice are independent. The outcome (the sum) are not equally likely. –Step 3: Assign digits to represent outcomes. Use 1-36 for the 36 combinations Number correspondence: 1 = sum of 2; 2-3 = sum of 3; 4-6 = sum of 4; 7-10 = sum of 5; 11-15 = sum of 6 –Step 4: Simulate many repetitions. Randint(1,36,20)→L1 L1≥11 and L1≤15→L2 Sum(L2) –Step 5: State your conclusions The simulation resulted in approximately 30% of the rolls having a sum of six.