Counting Principles and Probabilities in College Algebra
Learn about counting principles, permutations, combinations, subsets, and binomial coefficients in College Algebra to enhance your understanding of probability and counting techniques for solving mathematical problems efficiently.
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Probability and Counting Principles College Algebra
Counting Principles According to the Addition Principle, if one event can occur in ? ways and a second event with no common outcomes can occur in ? ways, then the first or second event can occur in ? + ? ways. According to the Multiplication Principle, if one event can occur in ? ways and a second event can occur in ? ways after the first event has occurred, then the two events can occur in ? ? ways. This is also known as the Fundamental Counting Principle. Example: On a restaurant menu, there are 3 appetizer options, 2 vegetarian entr e options and 5 meat entr e options, and 2 dessert options. There are 3 2 + 5 2 = 42 different choices for a three-course dinner.
Permutations of Distinct Objects Given ? distinct objects, the number of ways to select ? objects from the set in order is: ?! ? ?,? = ? ? ! Example: A club with six people need to elect a president, a vice president and a treasurer. How many ways can the officers be elected? 6 3 !=6 5 4 3 2 1 6! Solution: ? 6,3 = = 6 5 4 = 120 3 2 1
Permutations of Non-Distinct Objects If there are ? elements in a set and ?1are alike, ?2are alike, ?3are alike, and so on through ??, the number of permutations can be found by: ?! ?1!?2! ??! Example: Find the number of rearrangements of the letters in the word DISTINCT. Solution: There are 8 letters in the word, but I and T are repeated two times each. Therefore, the number of permutations is: 2! 2!=8 7 6 5 4 3 2 1 8! = 10080 2 2
Combinations When we are selecting objects and the order does not matter, we are dealing with combinations. Given ? distinct objects, the number of ways to select ? objects from the set is: ?! ? ?,? = ?! ? ? ! Example: An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split? 10! 10! 3!7!=10 9 8 7! =720 Solution: ? 10,3 = 3! 10 3 != 6= 120 3 2 7!
Number of Subsets of a Set A set containing ? distinct objects has 2?subsets. Example: A restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a baked potato? Solution: There are 4 options, so there are 24= 16 possible ways to order a baked potato. This result is the same as: ? 4,0 + ? 4,1 + ? 4,2 + ? 4,3 + ? 4,4 = 1 + 4 + 6 + 4 + 1 = 16
Binomial Coefficients In the shortcut to finding ? + ??we use combinations to find the coefficients that will appear in the expansion of the binomial. If ? and ? are positive integers with ? ?, then the binomial coefficient is: ? ? Note that ? ?= Example: 2= ?! = ? ?,? = ?! ? ? ! ? ? ?. 9! 2!7!=9 8 7! 9! 9 2! 9 2 != 2 7!= 36 9! 7!2!=9 8 7! 9! 9 7= 7! 9 7 != 7! 2= 36
Binomial Theorem The Binomial Theorem is a formula that can be used to expand any binomial. ? ??? ??? ? 1?? 1? + ? ? + ??= ?=0 ? 2?? 1?2+ + ? = ??+ ? 1??? 1+ ?? Example: Expand ? + ?5 5 0 5 1 5 2 5 3 5 4 5 5 ?5?0+ ?4?1+ ?3?2+ ?2?3+ ?1?4+ ?0?5 = = ?5+ 5?4? + 10?3?2+ 10?2?3+ 5??4+ ?5
Using the Binomial Theorem to Find a Single Term The (? + 1)th term of the binomial expansion of (? + ?)?is: ? ? ?? ??? Example: Find the sixth term of 3? ?9without fully expanding the binomial. Solution: Let ? = 5 for the sixth term, and use 3? and ? for the two variables. 9 5 4! 5! 3?9 5 ?5=9 8 7 6 5! 34?4( 1)5?5= 10206?4?5
Probabilities The likelihood of an event is known as probability. The probability of an event ? is a number that always satisfies 0 ? 1, where 0 indicates an impossible event and 1 indicates a certain event. A probability model is a mathematical description of an experiment listing all possible outcomes and their associated probabilities. The probability of an event ? in an experiment with sample space ? with equally likely outcomes is given by ? ? =number of elements in ? number of elements in ?=?(?) ?(?) ? is a subset of ?, so it is always true that 0 ?(?) 1.
Probability for Multiple Events The probability of the union of two events ? and ? (written ? ?) equals the sum of the probability of ? and the probability of ? minus the probability of ? and ? occurring together (which is called the intersection of ? and ? and is written as ? ?). ? ? ? = ? ? + ? ? ?(? ?) Example: A card is drawn from a standard deck. Find the probability of drawing a heart or a 7. ? =13 4 52, and ? 7 = 1 52=16 1 52 Solution: 52, ? 7 = ? 7 =13 4 52 4 13 52+ 52=
Computing the Probability of Mutually Exclusive Events The probability of the union of two mutually exclusive events ? and ? is given by ? ? ? = ? ? + ? ? Example: A card is drawn from a standard deck. Find the probability of drawing a heart or a spade. The events drawing a heart and drawing a spade are mutually exclusive because they cannot occur at the same time. The probability of drawing a heart is of drawing a heart or a spade is 1 1 4, and the probability of drawing a spade is also 1 4, so the probability 4+1 4=1 2
Probability That an Event Will Not Happen The complement of an event ?, denoted ? , is the set of outcomes in the sample space that are not in ?. ? ? = 1 ?(?) Example: Two six-sided dice are rolled. What is the probability that the sum of the numbers is greater than 3? Solution: The sample space is the set of all 36 possible outcomes from 1 + 1 to 6 + 6. It is easier to consider the 3 possible totals not greater than 3: 1 + 1, 1 + 2, and 2 + 1. Therefore, if ? ? = 36= 3 1 12, then ? ? =11 12.
Computing Probability Using Counting Theory Many probability problems use permutations and combinations to find the number of elements in events and sample spaces. Example: A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears. Find the probability that 2 bears and 3 dogs are chosen. Solution: There are 6 bears, so there are ?(6,2) ways to choose 2 bears. There are 5 dogs, so there are ?(5,3) ways to choose 3 dogs. There are ?(6,2) ?(5,3) ways to choose 2 bears and 3 dogs. There are 14 toys, so there are ? 14,5 ways to choose 5 toys. ? ? =?(6,2) ?(5,3) =15 10 2002 75 1001 = ?(14,5)
Quick Review What is the Fundamental Counting Principle? What is the difference between a permutation and a combination? What is the formula for the number of permutations of non-distinct items? How many subsets are there for a set of ? distinct items? What is the formula for a binomial coefficient? How is the Binomial Theorem used? What is the sum of probabilities for all possible events in a given probability model? How do you compute the probability of the union of two events?
