1. Section 1.1, Slide 1 Copyright © 2014, 2010, 2007 Pearson Education, Inc. Section 13.1, Slide 1 13 Probability What Are the Chances?

2. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Section 13.1, Slide 2 The Basics of Probability Theory 13.1 Calculate probabilities by counting outcomes in a sample space. Use counting formulas to compute probabilities.

3. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Section 13.1, Slide 3 The Basics of Probability Theory 13.1 Understand how probability theory is used in genetics. Understand the relationship between probability and odds.

4. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 4 Sample Space and Events Random phenomena are occurrences that vary from day-to-day and case-to-case. Although we never know exactly how a random phenomenon will turn out, we can often calculate a number called a probability that it will occur in a certain way.

5. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 5 Example: Determine a sample space for the experiment of selecting an iPhone from a production line and determining whether it is defective. Solution: This sample space is {defective, nondefective}. Sample Space and Events

6. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 6 Example: Determine the sample space for the experiment. Three children are born to a family and we note the birth order with respect to gender. Sample Space and Events (continued on next slide)

7. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 7 Solution: We use the tree diagram to find the sample space: {bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg}. Sample Space and Events

8. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 8 Example: Determine the sample space for the experiment. We roll two dice and observe the pair of numbers showing on the top faces. Sample Space and Events Solution: The solution space consists of the 36 listed pairs.

9. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 9 Sample Space and Events

10. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 10 Solution: Sample Space and Events a) A head occurs when we flip a single coin. b) Two girls and one boy are born to a family. c) A sum of five occurs on a pair of dice. Example: Write each event as a subset of the sample space. a) The set {head} is the event. b) The event is {bgg, gbg, ggb}. c) The event is {(1, 4), (2, 3), (3, 2), (4, 1)}.

11. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 11 Sample Space and Events

12. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 12 Example: A pharmaceutical company is testing a new drug. The company injected 100 patients and obtained the information shown. Based on the table, if a person is injected with this drug, what is the probability that the patient will develop severe side effects? Sample Space and Events (continued on next slide)

13. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 13 Solution: We obtain the following probability based on previous observations. Sample Space and Events

14. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 14 Example: The table summarizes the marital status of men and women (in thousands) in the United States in 2006. If we randomly pick a male, what is the probability that he is divorced? Sample Space and Events (continued on next slide)

15. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 15 Solution: We are only interested in males, so we consider our sample space to be the 60,955 + 2,908 + 10,818 + 2,210 + 39,435 = 116,326 males. The event, call it D, is the set of 10,818 men who are divorced. Therefore, the probability that we would select a divorced male is Sample Space and Events

16. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 16 Counting and Probability Probabilities may based on empirical information. For example, the result of experimental data. Probabilities may based on theoretical information, namely, combination formulas.

17. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 17 Example: We flip three fair coins. What is the probability of each outcome in this sample space? Solution: Eight equally likely outcomes are shown below. Each has a probability of. Counting and Probability

18. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 18 Example: We draw a 5-card hand randomly from a standard 52-card deck. What is the probability that we draw one particular hand? Solution: In Chapter 13, we found that there are C(52, 5) = 2,598,960 different ways to choose 5 cards from a deck of 52. Each hand has the same chance of being drawn, so the probability of any particular hand is Counting and Probability

19. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 19 Counting and Probability

20. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 20 Counting and Probability

21. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 21 Example: What is the probability in a family with three children that two of the children are girls? Solution: We saw earlier that there are eight outcomes in this sample space. We denote the event that two of the children are girls by the set G = {bgg, gbg, ggb}. Counting and Probability

22. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 22 Example: What is the probability that a total of four shows when we roll two fair dice? Solution: The sample space for rolling two dice has 36 ordered pairs of numbers. We will represent the event “rolling a four” by F. Then F = {(1, 3), (2, 2), (3, 1)}. Counting and Probability

23. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 23 Example: If we draw a 5-card hand from a standard 52-card deck, what is the probability that all 5 cards are hearts? Solution: We know that there are C(52, 5) ways to select a 5-card hand from a 52-card deck. If we want to draw only hearts, then we are selecting 5 hearts from the 13 available, which can be done in C(13, 5) ways. Counting and Probability

24. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 24 Example: Four friends belong to a 10-member club. Two members of the club will be chosen to attend a conference. What is the probability that two of the four friends will be selected? Solution : We can choose 2 of the 10 members in Event E, choosing 2 of the 4 friends, can be done in C(4, 2) = 6 ways. Counting and Probability

25. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 25 Probability and Genetics Y – produces yellow seeds (dominant gene) g – produces green seeds (recessive gene)

26. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 26 Probability and Genetics Crossing two first generation plants: Punnett Square

27. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 27 Example: Sickle-cell anemia is a serious inherited disease. A person with two sickle-cell genes will have the disease, but a person with only one sickle-cell gene will be a carrier of the disease. If two parents who are carriers of sickle- cell anemia have a child, what is the probability of each of the following: Probability and Genetics (continued on next slide) a) The child has sickle-cell anemia? b) The child is a carrier? c) The child is disease free?

28. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 28 Solution: Use a Punnett square: s denotes sickle cell n denotes normal cell. Probability and Genetics

29. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 29 Odds If a family has 3 children, what are the odds against all 3 children being of the same gender? 6:2 or 3:1 What are the odds in favor? 1:3

30. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 30 Odds Example: A roulette wheel has 38 equal-size compartments. Thirty-six of the compartments are numbered 1 to 36 with half of them colored red and the other half black. The remaining 2 compartments are green and numbered 0 and 00. A small ball is placed on the spinning wheel and when the wheel stops, the ball rests in one of the compartments. What are the odds against the ball landing on red? (continued on next slide)

31. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 31 Odds Solution: There are 38 equally likely outcomes. 18 are in favor of the event “the ball lands on red” and 20 are against the event. The odds against red are 20 to 18 or 20:18, which we reduce to 10:9.

32. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 32 Odds If the probability of E is 0.3, then the odds against E are We may write this as 70:30 or 7:3.

33. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.1, Slide 33 Odds Example: If the probability of Green Bay winning the Super Bowl is 0.35. What are the odds against Green Bay winning the Super Bowl? Solution: From the diagram we compute That is, the odds against are 13 to 7.