1. Holt CA Course 1 8-5 Disjoint Events Warm Up Warm Up California Standards Lesson Presentation Preview

2. Holt CA Course 1 8-5 Disjoint Events Warm Up A bag contains 18 marbles; 4 are green, 8 are red, and 6 are yellow. Find the probability of each event when a marble is chosen at random. 1. yellow2. red 3. green4. blue 5. not red 1313 0 4949 2929 5959

3. Holt CA Course 1 8-5 Disjoint Events SDAP3.4 Understand that the probability of either of two disjoint events occurring is the sum of the two individual probabilities and that the probability of one event following another, in independent trials, is the product of the two probabilities. Also covered: SDAP3.1 California Standards

4. Holt CA Course 1 8-5 Disjoint Events On a game show, the letters in the word Hollywood are printed on cards and shuffled. A contestant will win a trip to Hollywood if the first card she chooses is printed with an O or an L. Choosing an O or an L on the first card is an example of a set of disjoint events. Disjoint events cannot occur in the same trial of an experiment.

5. Holt CA Course 1 8-5 Disjoint Events Vocabulary disjoint events

6. Holt CA Course 1 8-5 Disjoint Events Probability of Two Disjoint Events P(A or B) = P(A) + P(B) Probability of either event Probability of one event Probability of other event Disjoint events are sometimes called mutually exclusive events. Reading Math

7. Holt CA Course 1 8-5 Disjoint Events Determine whether each set of events is disjoint. Explain. Additional Example 1: Identifying Disjoint Events A. choosing a dog or a poodle from the animals at an animal shelter The event is not disjoint. A poodle is a type of dog, so it is possible to choose an animal that is both a dog and a poodle. B. choosing a fish or a snake from the animals at a pet store The event is disjoint. Fish and snakes are different types of animals, so you cannot choose an animal that is both a fish and a snake.

8. Holt CA Course 1 8-5 Disjoint Events Find the probability of each set of disjoint events. Additional Example 2: Finding the Probability of Disjoint Events A. choosing an A or an E from the letters in the word mathematics P(A) = 2 11 P(E) = 1 11 P(A or E) = P(A) + P(E) Add the probabilities of the individual events. = + = 2 11 1 11 3 11 3 11 The probability of choosing an A or an E is.

9. Holt CA Course 1 8-5 Disjoint Events P(sum of 2) = 1 36 P(sum of 8) = 5 36 P(sum of 2 or sum of 8) = P(sum of 2) + P(sum of 8) = + = = 1 36 5 36 6 36 1616 The probability that the sum of the cubes is 2 or 8 is. 1616 Sharon rolls two number cubes. What is the probability that the sum of the numbers shown on the cubes is 2 or 8? Additional Example 3 Continued Step 2 Find the probability of the set of disjoint events.

10. Holt CA Course 1 8-5 Disjoint Events Sharon rolls two number cubes. What is the probability that the sum of the numbers shown on the cubes is 2 or 8? Additional Example 3: Recreation Application Step 1 Use a grid to find the sample space. 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 First Roll Second Roll The grid shows all possible sums. There are 36 equally likely outcomes in the sample space.

11. Holt CA Course 1 8-5 Disjoint Events Determine whether each set of events is disjoint. Explain. Check It Out! Example 1 A. choosing a bowl of soup or a bowl of chicken noodle soup from the cafeteria The event is not disjoint. Chicken noodle is a type of soup, so it is possible to choose a bowl of chicken noodle soup and soup. B. choosing a bowl of chicken noodle soup or broccoli cheese soup The event is disjoint. Chicken noodle and broccoli cheese are different types of soups, so you cannot choose a soup that is both chicken noodle and broccoli cheese.

12. Holt CA Course 1 8-5 Disjoint Events Find the probability of each set of disjoint events. Check It Out! Example 2 A. choosing an I or an E from the letters in the word centimeter P(I) = 1 10 P(E) = 3 10 P(I or E) = P(I) + P(E) Add the probabilities of the individual events. = + = = 1 10 3 10 4 10 2525 The probability of choosing an I or an E is. 2525

13. Holt CA Course 1 8-5 Disjoint Events Find the probability of each set of disjoint events. Check It Out! Example 2 B. spinning a 2 or a 4 on a spinner with eight equal sectors numbered 1-6 P(2) = 1616 P(4) = 1616 P(2 or 4) = P(2) + P(4) Add the probabilities of the individual events. = + = = 1616 1616 2626 1313 The probability of choosing a 2 or a 4 is. 1313

14. Holt CA Course 1 8-5 Disjoint Events Sun Li rolls two number cubes. What is the probability that the sum of the numbers shown on the cubes is 3 or 4? Check It Out! Example 3 Step 1 Use a grid to find the sample space. 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 First Roll Second Roll The grid shows all possible sums. There are 36 equally likely outcomes in the sample space.

15. Holt CA Course 1 8-5 Disjoint Events P(sum of 3) = 2 36 P(sum of 4) = 3 36 P(sum of 3 or sum of 4) = P(sum of 3) + P(sum of 4) = + = 2 36 3 36 5 36 5 36 The probability that the sum of the cubes is 3 or 4 is. Check It Out! Example 3 Continued Step 2 Find the probability of the set of disjoint events. Sun Li rolls two number cubes. What is the probability that the sum of the numbers shown on the cubes is 3 or 4?

16. Holt CA Course 1 8-5 Disjoint Events Lesson Quiz: Part I 1. Determine whether choosing a fiction book or a nonfiction book is a set of disjoint events. Explain. Find the probability of each set of disjoint events. 2. Choosing an O or an E from the letters in outcome 3. Rolling an odd number or a 6 on a number cube Disjoint; you cannot choose a book that is both fiction and nonfiction. 3737 2323

17. Holt CA Course 1 8-5 Disjoint Events Lesson Quiz: Part II 4. David rolls two number cubes at the same time. If the product of the numbers rolled is 25 or 30, he will win the game. Use a grid to find the sample space. The find the probability that David will win the game. 1 12

18. Holt CA Course 1 8-5 Disjoint Events Find the probability of each set of disjoint events. B. spinning a 3 or a 4 on a spinner with eight equal sectors numbered 1-8 P(3) = 1818 P(4) = 1818 P(3 or 4) = P(3) + P(4) Add the probabilities of the individual events. = + = = 1818 1818 2828 1414 The probability of choosing a 3 or a 4 is. 1414 Additional Example 2: Finding the Probability of Disjoint Events