Probability of Compound Event

1. Probability of Compound Event Probability & Statistics 1.0 Students know the definition of the notion of independent events and can use the rules for addition, multiplication, and complementation to solve for probabilities of particular events in finite sample spaces. 2.0 Students know the definition of conditional probability and use it to solve for probabilities in finite sample spaces.

2. Probability of Compound Event Objectives Key Words Compound Event The union of intersection of two events Overlapping Events Two events that have outcomes in common Disjoint or Mutually Exclusive Events Two events that have no outcomes in common Complement of an Event All outcomes that are not in the event • Solve for the probabilities of Unions of events • Solve for the probabilities of Intersections of events

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4. Probability of Compound Events • Overlapping Events • If A and B are overlapping events, then P(A and B)≠0, and the probability of A or B is: • Disjoint Events • If A and B are disjoint events, then P(A and B)=0, and the probability of A or B is:

5. Example 1 Cards A card is randomly selected from a standard deck of 52 cards. What is the probability that the card is a 10or a face card (a jack, queen, or king)? 0.308 ~ ~ SOLUTION Let event A be selecting a 10 and let event B be selecting a face card. Event A has 4 outcomes and event B has 12 outcomes. Events A and B are disjoint events. 4 12 16 4 P ( P ( P ( = = = A orB ( A ( B ( + + + 52 52 52 13 Find P(A or B) for Disjoint Events

6. Example 2 A card is randomly selected from a standard deck of 52 cards. What is the probability that the card is a face card or a spade? 12 13 3 Substitute probabilities. – = + 52 52 52 22 , or about 0.42 Simplify. = 52 SOLUTION Let event A be selecting a face card. Let event B be selecting a spade. Event A has 12 outcomes, and event B has 13 outcomes. Three of the outcomes are common to both A and B. P ( P ( P ( P ( – = Union of overlapping events A orB ( A ( B ( A andB ( + Find P(A or B) for Overlapping Events

7. Checkpoint In Exercises 1 and 2, a card is randomly selected from a standard deck of 52 cards. Find the probability of the given event. 1. The card is an ace or a jack. 2 0.15 ANSWER 13 ~ ~ ~ ~ 2. The card is a heart or a 7. 4 0.31 ANSWER 13 Find the Probability of a Compound Event

8. Example 3 SOLUTION Let A represent the event “has applied to a state university.” Let B represent event “has applied to a private university.” You need to find P(A and B). Find P(A and B) for Overlapping Events College In a class of 20 seniors, 17 have applied to a state university and 11 have applied to a private university. In all, 19 of the 20 students have applied to a university. What is the probability that a class member chosen at random has applied to a state university and a private university?

9. Example 3 17 11 19 You are given: P ( P ( P ( A orB A ( = B ( = ( = 20 20 20 P ( P ( P ( P ( – = Write formula for A ( B ( A andB ( A orB ( + P . ( ) AorB 19 17 11 Substitute probabilities. P ( – A andB = ( + 20 20 20 17 11 19 Solve for P ( – A andB = ( + P . ( ) AandB 20 20 20 9 , or 0.45 Simplify. = 20 Find P(A and B) for Overlapping Events

10. Checkpoint 1 0.2 ANSWER = 5 Find the Probability of a Compound Event 3. Pets Of 25 students, 13 have a dog and 8 have a cat. In all, 16 of the 25 have a dog or a cat. What is the probability that any one of the 25 has a dog and a cat?

11. Probability of the Complement of an Event The sum of the probabilities of an event and its complement is 1. So, Recall: Complement of an Event All outcomes that are not in the event

12. Example 4 3 33 11 Dice When two six-sided dice are rolled, 36 outcomes are possible, as shown. Find the probability of The given event. b. P(sum 10) 1 P(sum 10) ≤ 1 > – – = = = = 36 36 12 a. The sum is not 2. b. The sum is less than or equal to 10. ~ ~ ~ ~ SOLUTION 1 35 a. P(sum is not 2) 1P(sum is 2) 1 0.972 – – = = = 36 36 0.917 Find Probabilities of Complements

13. Example 5 365P23 Express the probability. 1 – = 36523 SOLUTION Use a calculator. The complement of “at least 2 of the students share a birthday” is “each student has a different birthday.” Because each student has a birthday on one of 365 days (ignoring leap year), there are 36523 possibilities for all the birthdays. The number of possibilities in which each of the 23 birthdays is different is: ~ ~ . . . 365 364 363 345 344 343 This is 365P23. • • • • • • P(at least 2 are the same) 1 P(all are different) – Write the complement. = 0.507 Find Probabilities of Complements Birthdays A class has 23 students. What is the probability that at least 2 of the students share a birthday?

14. Conclusions Summary Assignment Probability of Compound Event Page 565 #(18,20,22,30,34,42) • A problem such as “find the probability of getting 0,1,2,3 or 4 red cards when you draw 5 cards” can require many computations. How can you simplify the solution? • If you know P(not A), the probability of the complement of A, then P(A)=1-P(not A)