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C.2 Probability Computations A7 C.2 Probability Computations ■ Find the Probability of Mutually Exclusive Events

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C.2 Probability Computations A7 C.2 Probability Computations ■ Find the Probability of Mutually Exclusive Events APPENDIX C.2 Probability Computations A7 C.2 Probability Computations ■ Find the probability of mutually exclusive events. ■ Find the probability of the complement of an event. ■ Find the odds of an event. Mutually Exclusive Events In Appendix C.1, you looked at a formula for calculating simple probabilities. In this section, you will study formulas for calculating probabilities of more involved events and combinations of events. First, you will study a formula for the probability of the union of two disjoint (or mutually exclusive) events. Two events A and B (from the same sample space) are mutually exclusive if A and B have no outcomes in common. That is,A ʝ B ϭл. For instance, if two dice are tossed, the event A of rolling a total of 6 and the event B of rolling a total of 9 are mutually exclusive because the following sets have no elements in com- mon. (The ordered pairs represent the numbers on the dice.) A ϭ ͭ͑1, 5͒, ͑2, 4͒, ͑3, 3͒, ͑4, 2͒, ͑5, 1͒ͮ B ϭ ͭ͑3, 6͒, ͑4, 5͒, ͑5, 4͒, ͑6, 3͒ͮ To find the probability that one or the other of two mutually exclusive events will occur, you can add their individual probabilities, as indicated in Figure C.4. B A P(A ∪ B) = P(A) + P(B) FIGURE C.4 Probability of Mutually Exclusive Events If A and B are mutually exclusive events, then for a given experiment the probability that A or B will occur is P͑A ʜ B͒ ϭ P͑A͒ ϩ P͑B͒. A8 APPENDIX C Probability and Probability Distributions Example 1 Finding the Probability of Mutually Exclusive Events A card is drawn from a standard deck of 52 cards. What is the probability that it is a face card (event A) or a 4 (event B)? SOLUTION The two events are mutually exclusive because no one card can be both a face card and a 4. Because n͑A͒ 12 3 P͑A͒ ϭ ϭ ϭ n͑S͒ 52 13 and n͑B͒ 4 1 P͑B͒ ϭ ϭ ϭ n͑S͒ 52 13 it follows that P͑A ʜ B͒ ϭ P͑A͒ ϩ P͑B͒ 3 1 4 ϭ ϩ ϭ . 13 13 13 So, you can conclude that the probability of selecting a face card or a 4 is 4 Ϸ 13 0.308. ✓CHECKPOINT 1 A card is drawn from a standard deck of 52 cards. What is the probability that the card is a jack or an ace? ■ Three events A, B, and C in the same sample space are mutually exclusive if A ʝ B ϭл, A ʝ C ϭл, B ʝ C ϭл. The probability that A, B, or C will occur is given by P͑A ʜ B ʜ C͒ ϭ P͑A͒ ϩ P͑B͒ ϩ P͑C͒. This pattern can be applied to four or more events that are mutually exclu- sive. That is, if A1, A2, A3, . , An are mutually exclusive events in the same sample space then ͑ ʜ ʜ ʜ . ʜ ͒ ϭ ͑ ͒ ϩ ͑ ͒ ϩ ͑ ͒ ϩ . ϩ ͑ ͒ P A1 A2 A3 An P A1 P A2 P A3 P An . APPENDIX C.2 Probability Computations A9 Example 2 Finding the Probability of Mutually Exclusive Events The weights of adult Lhasa Apsos at a breeding facility are shown in the table. Weight (in pounds) 6–8 9–11 12–14 15–17 18–20 21–23 Number of dogs 0 2 7 5 3 1 What is the probability that an adult Lhasa Apso chosen at random will weigh between 12 and 20 pounds? SOLUTION Let A, B,and C represent the following events. A ϭ weight between 12 and 14 pounds B ϭ weight between 15 and 17 pounds C ϭ weight between 18 and 20 pounds Because these three events are mutually exclusive, the probability that the adult Lhasa Apso’s weight will fall between 12 and 20 pounds is P͑A ʜ B ʜ C͒ ϭ P͑A͒ ϩ P͑B͒ ϩ P͑C͒ 7 5 3 ϭ ϩ ϩ 18 18 18 15 ϭ 18 Ϸ 0.833 ✓CHECKPOINT 2 In Example 2, find the probability that an adult Lhasa Apso chosen at random weighs between 9 and 17 pounds. ■ If a sample space S is subdivided into a collection of mutually exclusive events A1, A2, A3, . , An such that ʜ ʜ ʜ . ʜ ϭ A1 A2 A3 An S then the sum of the probabilities of the events must be 1. This Addition Principle is summarized as follows. Addition Principle If A1, A2, A3, . , An are mutually exclusive events in a sample space S such that ʜ ʜ ʜ . ʜ ϭ A1 A2 A3 An S, then ͑ ͒ ϩ ͑ ͒ ϩ ͑ ͒ ϩ . ϩ ͑ ͒ ϭ P A1 P A2 P A3 P An 1. A10 APPENDIX C Probability and Probability Distributions Example 3 Applying the Addition Principle Two six-sided dice are tossed. Construct a table showing the probabilities for the various totals of the two dice. Then show that the sum of these probabilities is 1. SOLUTION The outcomes representing various totals are shown in the following table, which is called a frequency distribution. Number of Total Outcomes Outcomes 21͑1, 1͒ 32͑1, 2͒, ͑2, 1͒ 43͑1, 3͒, ͑2, 2͒, ͑3, 1͒ 54͑1, 4͒, ͑2, 3͒, ͑3, 2͒, ͑4, 1͒ 65͑1, 5͒, ͑2, 4͒, ͑3, 3͒, ͑4, 2͒, ͑5, 1͒ 76͑1, 6͒, ͑2, 5͒, ͑3, 4͒, ͑4, 3͒, ͑5, 2͒, ͑6, 1͒ 85͑2, 6͒, ͑3, 5͒, ͑4, 4͒, ͑5, 3͒, ͑6, 2͒ 94͑3, 6͒, ͑4, 5͒, ͑5, 4͒, ͑6, 3͒ 10͑4, 6͒, ͑5, 5͒, ͑6, 4͒ 3 11͑5, 6͒, ͑6, 5͒ 2 12͑6, 6͒ 1 From this frequency distribution, you can construct the following table showing the probability of each of the various totals. This table is called a probability distribution. Total 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 5 4 3 2 1 Probability 36 36 36 36 36 36 36 36 36 36 36 ✓ CHECKPOINT 3 Note that the events in this probability distribution are mutually exclusive and Five coins are tossed. Construct a represent every possible total (or outcome). From the Addition Principle, you can table showing the probabilities for see that the sum of the given probabilities is 1. the number of heads that turn up. 1 2 3 4 5 6 5 4 3 2 1 ϩ ϩ ϩ ϩ ϩ ϩ ϩ ϩ ϩ ϩ ϭ 1 Then show that the sum of the 36 36 36 36 36 36 36 36 36 36 36 probabilities is 1. ■ Be sure you see that the probability of A ʜ B is equal to the sum of the prob- abilities of A and B only if A and B are mutually exclusive. For instance,the probability of drawing a heart (event A) or a face card (event B) from a standard deck of 52 cards is not equal to P͑A͒ ϩ P͑B͒ because A and B are not mutually A ∩ B exclusive. To find the probability of A ʜ B, where A and B are not necessarily mutually A B exclusive events, use the fact that n͑A ʜ B͒ ϭ n͑A͒ ϩ n͑B͒ Ϫ n͑A ʝ B͒. A From this, it follows that ͑ ʜ ͒ ϭ ͑ ϩ ͑ ͒ Ϫ ͑ ʝ ͒ P(A ∪ B) = P(A) + P(B) − P(A ∩ B) P A B P A) P B P A B FIGURE C.5 as summarized in the following box. (See Figure C.5) APPENDIX C.2 Probability Computations A11 Probability of the Union of Two Events If A and B are events in the same sample space S, then the probability of A ʜ B is P͑A ʜ B͒ ϭ P͑A͒ ϩ P͑B͒ Ϫ P͑A ʝ B͒. Example 4 Finding the Probability of the Union of Two Events One card is selected from a standard deck of 52 playing cards. What is the probability that the card is either a heart or a face card. SOLUTION Because the deck has 13 hearts, the probability of selecting a heart Hearts (event A) is ♥ ∩ 13 1 A 2♥ n(A B) = 3 P͑A͒ ϭ ϭ . Selecting a heart 3♥ 52 4 4♥ 5♥ 6♥ ♥ 7♥ J J♠ Because the deck has 12 face cards, the probability of selecting a face card (event ♥ J♦ 8♥ 9♥ Q B) is K♥ Q♠ ♣ 10♥ ♦ J Q 12 3 K♠ K♦ P͑B͒ ϭ ϭ . Selecting a face card Q♣ 52 13 K♣ Face cards Because three of the cards are hearts and face cards (See Figure C.6) it follows FIGURE C.6 that 3 P͑A ʝ B͒ ϭ . 52 ✓CHECKPOINT 4 Finally, applying the formula for the probability of the union of two events, you can conclude that the probability of selecting a heart or a face card is One card is selected from a stan- dard deck of 52 playing cards. P͑A ʜ B͒ ϭ P͑A͒ ϩ P͑〉͒ Ϫ P͑A ʝ B͒ What is the probability that the 13 12 3 22 card is either a king or a face ϭ ϩ Ϫ ϭ Ϸ 0.423 52 52 52 52 card? ■ The Complement of an Event The complement of an event A is the collection of all outcomes in the sample space that are not in A. The complement of event A is denoted by AЈ. Because P͑A ʜ AЈ͒ ϭ 1 and because A and AЈ are mutually exclusive, it follows that P͑A͒ ϩ P͑AЈ͒ ϭ 1. So, the probability of AЈ is P͑AЈ͒ ϭ 1 Ϫ P͑A͒.
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