Counting and Probability

537 Views

Download Presentation
## Counting and Probability

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Counting and Probability**Sets and Counting Permutations & Combinations Probability**Sets and Counting**A set is a well-defined collection of distinct objects. Well-defined means there is a rule that enables us to determine whether a given object is an element of the set. If a set has no elements, it is called the empty set, or null set, and is denoted by the symbol**If two sets A and B have precisely the same elements, then**A and B are said to be equal and write A = B.**A**B U**A**B U**A**B U**A**B U**A**A U**In survey of 50 people, 21 said they owned stocks, 32 said**they owned bonds and 12 said they owned both stocks and bonds. How many of the 50 people owned stocks or bonds? How many owned neither? A: person owns stock B: person owns bonds = 21 + 32 - 12 = 41 50 - 41 = 9 owned neither**Universe is 50 people.**In A = 21 owned stocks. In B = 32 owned bonds. In AB = 12 owned both stocks and bonds. In AB = 53-12 = 41 owned stocks or bonds. In (AB)= 50-41 = 9 owned neither. B 32 (AB) _ _ 9 A 21 AB_12**Permutations and Combinations**This is a tricky subject where even the text author makes mistakes. The next three slides are to distinguish some of the subtleties. After these are the slides from the text set and that help to elaborate some cases. Some situations can be very difficult to evaluate.**Counting Permutation Cases I**A permutation is an ordered arrangement of r objects from n objects. To find the total number of possible cases, the easy types of situations are: Multiplication of p,q,r,s, ... ways of selection => Total number of cases is N = p•q•r•s•... The number of permutations of r distinct objects with allowed repetition from n distinct objects (order is important) =>N = nr The number of permutations of r distinct objects with no repetition from n distinct objects (order is important) =>N = P(n, r) = n!/(n - r)!**Counting Permutation Cases II**A permutation is an ordered arrangement of r objects from n objects. An important harder situation of finding the total number of possible cases is when there are k distinct types each of non-distinct objects, with n1 of the 1st type, n2 of the 2nd type, ... nk of the kth type, and n = n1 + n2 + n3 + ... nk. Permutation of n, somenon-distinct, objects with allowed repetition from k distinct types of objects (order is important) => N = n!/[n1!•n2! ... •nk!].**Counting Combination Cases**A combination is an arrangement with no regard to orderof r distinct objects without repetition from n distinct objects (r<n). For finding the total number of possible cases, the easy type of situation is: The number of combinations of r distinct objects without repetition from n distinct objects (order is not important) => N = C(n, r) = n!/[r!(n - r)!].**Permutations and Combinations**Theorem: Multiplication Principle of Counting If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in different ways.**If a license plate consists of a letter, then 5 numbers, how**many different types of license plates are possible?**A permutation is an ordered arrangement of n distinct**objects without repetitions. The symbol P(n, r) represents the number of permutations of n distinct objects, taken r at a time,where r<n.**Theorem: Number of Permutations of n Distinct**Objects Taken r at a Time The number of different arrangements from selecting r objects from a set of n objects (r< n), in which 1. the n objects are distinct 2. once an object is used, it cannot be repeated 3. order is important is given by the formula**A combination is an arrangement, without regard to order, of**n distinct objects without repetitions. The symbol C(n, r) represents the number of combinations of n distinct objects taken r at a time, where r<n.**Theorem: Number of Combinations of n**Distinct Objects Taken r at a Time The number of different arrangements from selecting r objects from a set of n objects (r< n), in which 1. the n objects are distinct 2. once an object is used, it cannot be repeated 3. order is not important is given by the formula**Probability**An event is an outcome from an experiment. Its probability gives the likelihood it occurs. A probability model lists the different outcomes from an experiment and their corresponding probabilities. To construct probability models, we need to know the sample space of the experiment. This is the set S that lists all the possible outcomes of the experiment.**Determine the sample space resulting from the experiment of**rolling a die. S = {1, 2, 3, 4, 5, 6}**Determine which of the following are probability models from**rolling a single die. Not a probability model. The sum of all probabilities is not 1.**All probabilities between 0 and 1 inclusive and the sum of**all probabilities is 1.**Not a probability model. The event “roll a 6” has a**negative probability.**Theorem: Probability for Equally Likely**Outcomes If an experiment has n equally likely outcomes, and if the number of ways an event E can occur is m, then the probability of E is**A classroom contains 20 students: 7 Freshman, 5 Sophomores,**6 Juniors, and 2 Seniors. A student is selected at random. Construct a probability model for this experiment.**What is the probability of selecting an Ace or Diamond from**a standard deck of cards?**Let S denote the sample space of an experiment and let E**denote an event. The complement of E, denoted E, is the set of all outcomes in the sample space S that are not outcomes in the event E.**Theorem: Computing Probabilities of Complementary**Events If E represents any event and E represents the complement of E, then**The probability of having 4 boys in a four child family is**0.0625. What is the probability of having at least one girl? Sample Space: {4 boys; 3 boys, 1 girl, 2 boys, 2 girls; 1 boy, 3 girls; 4 girls} E = “at least one girl” E = “4 boys” P(E) = 1 - P(E) = 1 - 0.0625 = 0.9375**What is the probability of obtaining 3 of a kind when 5**cards are drawn from a standard 52-card deck? P(3 of a kind) This answer from the text slides is just wrong. For the correct value of 2.11% and similar examples see either of the poker sites: http://www.math.sfu.ca/~alspach/comp18/ http://www.pvv.ntnu.no/~nsaa/poker.html**What is the probability of obtaining 3 of a kind when 5**cards are drawn from a standard 52-card deck? Done correctly, one has P(3 of a kind) Why?