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Chapter 8 Counting Principles: Further Probability Topics

Chapter 8 Counting Principles: Further Probability Topics. Section 8.3 Probability Applications of Counting Principles. Recall that tree diagrams can be used to solve probability problems involving dependent events, but tree diagrams sometimes require a large number of branches.

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Chapter 8 Counting Principles: Further Probability Topics

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  1. Chapter 8Counting Principles: Further Probability Topics Section 8.3 Probability Applications of Counting Principles

  2. Recall that tree diagrams can be used to solve probability problems involving dependent events, but tree diagrams sometimes require a large number of branches. • Permutations and combinations are especially helpful when the numbers involved are large.

  3. The use of combinations to solve probability problems depends on the basic probability principle: Let S be a sample space with equally likely outcomes, and let event E be a subset of S. Then the probability that event E occurs, written P(E), is where n(E) and n(S) represent the number of elements in sets E and S.

  4. Combinations tend to be used more often than permutations in probability because many times the final result does not depend on the order of events. • However, order does matter in some cases, and permutations would need to be used. • In some cases, a mixture of permutations and combinations are needed to solve the problem.

  5. Example: You and your friends order twelve burritos to go from a Mexican restaurant, five with hot peppers and seven without; however, the restaurant forgot to label them. If you pick three burritos at random, what is the probability that all three have hot peppers? Find the number of ways you can choose 3 burritos from 12. 12C3 = 220 This is your sample space. Find the number of ways you can choose 3 burritos with hot peppers. 5C3 • 7C0 = 10 P(3 w/hot peppers) = __10__ = 0.045 220

  6. Example: A basketball coach was asked to rank order his top three players from the starting five players. Suppose one of the players is named Jack. a.) What is the probability that Jack is one of the top three players? Find the number of ways the top 3 players can be chosen. 5C3 = 10 This is your sample space. Find the number of ways that Jack could be chosen as one of the three ranked players. 4C2 = 6 Since we want Jack to be one of the three chosen players, we only need to find out how many ways the coach can select the other two from the remaining 4 players. Determine the probability Jack is one of the top three. P(top three) = 6 = 0.6 10

  7. Example: A basketball coach was asked to rank order his top three players from the starting five players. Suppose one of the players is named Jack. b.) What is the probability that Jack is ranked # 1? Find the number of ways the top 3 players can be ranked. 5P3 = 60 This is your sample space. Find the number of ways that Jack could be chosen as the number one ranked player. _1_ • 4 • 3 = 12 Since we want Jack to be the number one ranked player, we only need to find out how many ways the coach can select the second and third ranked players from the remaining 4. Determine the probability Jack is the top-ranked player. P( rank #1 ) = 12 = 0.2 60

  8. Example: A standard poker hand consists of 5 cards. Find the probability that a person is dealt a straight flush (5 in a row in a single suit, but not a royal flush). Assume Aces are either high or low. A flush could start with an Ace, 2, 3, 4, 5, 6, 7, 8, or 9. This gives 9 choices in each of 4 suits, so there are 36 choices in all. ( 9 • 4 = 36 ) P(straight flush) = ___36____ 52 C 5 36______ 2, 598, 960 = -5 ≈ 0.00001385 = 1.385 x 10

  9. Example: Two hundred people apply for three jobs. Sixty of the applicants are women. If three persons are selected at random, what is the probability that: a.) all are women? b.) exactly two are men? c.) at least one is female?

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