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Ch. 13 Notes MATH 2400 Mr. J. Dustin Tench Recap Ch. 12 - P(A or B) = - P(A) + P(B) – P(A∩B) - If A and B are disjoint, P(A∩B) = 0, so we get P(A) + P(B). - P(A and B) = - P(A|B) • P(B) - If A and B are independent, P(A|B) = P(A), so we get P(A) • P(B) When are A and B disjoint? When the sample spaces for A and B contain no common elements, they are considered disjoint. Ex: Rolling a 2 or a 4 on a die. Ex: Drawing a black ace or a red ace out of a standard deck of cards. Ex: Selecting a senior citizen or a teenager at random for a prize from a collection of surveys. Venn Diagram of Disjoint Events P(A∩B)=0 When are A and B Non-Disjoint? When A and B contain common elements in their sample space. Ex: Drawing a red card or a king from a standard deck of cards. Ex: Rolling a 4 or an even number. Ex: Buying a used car or buying a Toyota. Venn Diagram of Disjoint Events P(A∩B) ≠ 0 When are A and B Independent? Events A and B are independent when the outcome of A cannot affect the outcome of B. Ex: Rolling a 6 and flipping Heads. Ex: Having twins and the first being a boy and the second being a girl. Conditional Probability Conditional probability exists when we are only considering a subset of the original sample space. Generally, the word “given” is used, but not always. Ex: Of the freshmen, what’s the probability they are a science major? Ex: Given a student has a job, what’s the probability they have above a 3.5 GPA? Example The Clemson University Fact Book for 2007 shows that 123 of the university’s 338 assistant professors were women, along with 76 of the 263 associate professors and 73 of the 375 full professors. 1) What is the probability that a randomly chosen Clemson professor is a woman? 2) Given a selected individual is a full professor, what is the probability that the person is a woman? 3) Are the rank and gender of Clemson professors independent? Example A recent study has shown that 75% of teenagers own cell phones. The study also indicates that 65% of teenagers own a phone and text on their phones. a) What is the probability that a randomly selected teen is a “texter” if we know they own a phone. b) Of the teens that own phones, 15% send more than 6,000 texts a month. What percent of all teens own a phone, are texters, and send more than 6,000 texts a month? Ch. 13 Notes – FINALLY!!! Binomial Distributions are simply situations in which only two different outcomes are possible for each trial. Ex: Shooting free throws in a basketball game. Ex: Having a baby. Ex: Computerized telephone survey successfully calling a residence. Combinatorics Permutations & Combinations A permutation is an arrangement of objects in which the order they are arranged can be counted differently than a different arrangement of the same objects ABC is a different arrangement than ACB. A combination is a grouping of objects in which order does not matter and the group is counted as the same group if it contains that same elements ABC is the same group as ACB. Permutations Consider assigning three people 3 different roles in an organization. A – President B – Vice-President C – Secretary We can arrange these same 3 people 6 different ways. ABC, ACB, BAC, BCA, CAB, CBA Permutations Consider choosing the president, we have 3 choices. Then, consider choosing the vice-resident, we have 2 choices. Finally, consider choosing the secretary, we have 1 choice. So, the number of arrangements of 3 people is 3*2*1 = 3! = 6. Factorial: n! = n*(n-1)*(n-2)*(n-3)*…*3*2*1 Permutations What if we don’t use all the objects in a group. Consider the case where a President and VicePresident are to be chosen from a group of 4 people. How many different ways can we assign these two positions? For the president, we have 4 choices. For the vice-president, we have 3 choices. So, the number of ways we can assign these positions is 4*3 = 12. Permutations – A Formula There is a formula, but it is much easier just to think about the problem and count the number of ways each part can happen. Combinations Consider forming a team where each person has the same role. If 3 people are to be used from a pool of…3……people…..then there is only 1 group that can be formed. But, what if the pool was larger? Consider a terrible situation in which there are 4 people and a team of 3 is to be formed. How many ways can this happen? ABC ACD ABD BCD There are 4 ways. Combinations Combinations, unlike Permutations, have an accepted notation used. For the situation we just looked at, n=4 and r=3, so = 4. On The Ti-84 Permutations: MATH, PRB tab, 2: nPr Combinations: MATH, PRB tab, 3: nCr Type in what you want to calculate like… 4 3 and hit ENTER. Okay, Back to Binomial Distributions We use combinations to count the number of ways an event can occur. For example, if Jordan makes 70% of his free throws, what is the probability he makes 4 of the 6 free throws he attempts in a game? The number of ways he can make 4 of the 6 is = 15. Jordan There are 15 different ways Jordan can make 4 of his 6 free throws. The probability Jordan hits any one of the free throws is .7. The probability Jordan misses is .3. Lets consider the case where Jordan makes the first four free throws, but misses the next two. The probability of this happening is (.7)(.7)(.7)(.7)(.3)(.3) = .021609 Consider the case where Jordan makes the first three, misses two, but makes the last. (.7)(.7)(.7)(.3)(.3)(.7) = .021609 They are exactly the same! Jordan continued… So, there are 15 ways of making 4 out of 6 free throws, so the probability that Jordan makes exactly 4 out of the 6 free throws is. 15*(.7)(.7)(.7)(.7)(.3)(.3) = 15* (.7)4(.3)2 = .324135 More Jordan 1. Calculate the probability that Jordan makes at least 2 of the 6 free throws. P(X≥2) = 2. Calculate the probability that Jordan makes at most 4 of the 6 free throws. P(X≤4) = 3. Calculate the probability that Jordan makes 2, 3, or 4 of the 6 free throws. P(2≤X≤4) = On the Calculator 2nd, DISTR - 0:binompdf(n,p,r), for P(x = r) - A:binomcdf(n,p,r), for P(x <,≤,>, or ≥ r) So, for Jordan… P(X=4) = binompdf(6,0.7,4) P(X≥2) = 1 - binomcdf(6,0.7,4) P(X≤4) = binomcdf(6,0.7,4) P(2≤X≤4) = binomcdf(6,0.7,4) – binomcdf (6,0.7,2) μ and σ for Binomial Distributions 1) μ = np where μ is the mean, n is the number of trials, and p is the probability of success for each trial. 2) σ = 𝒏𝒑(𝟏 − 𝒑) where σ is the standard deviation. An Application Delta has been collecting data regarding its flights. Data indicates that 82% of the flights from Atlanta to Las Vegas depart on time. 1) If there are 227 flights from Atlanta to Las Vegas this year, what is the expected number (mean) of flights that should depart on time? 2) What is the standard deviation using this data? HW 13.1, 13.2, 13.3, 13.4, 13.5, 13.24, 13.26