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Probability Fundamentals Stat 201 Prof. Yanni Papadakis Key Terms § Probability Axioms § Rules for Addition § Rules for Multiplication § Venn Diagram § Tree Diagram § Bayes’ Rule Terminology Example: Throw of 2 6-face dice § Outcome (? ): e.g., (5,2) § Sample Space (O): all 36 outcomes {(1,1),(1,2),(1,3),…,(6,5),(6,6)} § Event (A): Any subset of O e.g. sum of 5 {(1,4),(2,3),(3,2),(4,1)} or at least one die is 5 {(1,5),(2,5),(3,5),(4,5),(5,5),(6,5), (5,1),(5,2),(5,3),(5,4),(5,6)} Venn Diagram Set Notation Logical Meaning ω∈A A ? ? realizes A belongs B A A∩ B = ∅ A and B Intersection are Incompatible Equals Null Set B A A⊂ B A implies B Subset Venn Diagram Set Notation Logical Meaning B A∪ B − A∩ B One And Only One A A or B − A and B Event of A and B is Realized Union minus Intersection B A∩ B A Both A and B A or B are realized Intersection O B A A+ B +C = Ω A,B,C are A∩ B = ∅ exhaustive and C A∩C = ∅ mutually exclusive B ∩C = ∅ Venn Diagram Set Notation Logical Meaning O A A’ A' = Ω − A All but A Complement is realized B A∩ B A| B = A ∩ B but Ω = B A given B is realized Given Work with partner § Common sources of caffeine in diet are coffee, tea, and cola drinks. Suppose that § 55% of adults drink coffee § 25% of adults drink tea § 45% of adults drink cola § Also § 15% drink both coffee and tea § 5% drink all three beverages § 25% drink both coffee and cola § 5% drink tea only § Draw Venn Diagram § What percent drinks cola only? § What percent drinks none of the above beverages? Probability Rules § P(A) = 0 P(A) = 1 § P(O) = 1 P(Æ) = 0 Addition Rule § P(A + B) = P(A) + P(B) - P(A and B) P(A + B) = P(A) + P(B) For Mutually Exclusive Events P(A’) = 1 - P(A) § P( A|B ) = P(A and B) / P(B) ,or Multiplication Rule § P(A and B) = P(A|B) P(B) Combining Outcomes § A: First die shows 4 One die sample space is {1,2,3,4,5,6} P(A)=1/6 § B: Second die shows 5 Other die sample space is {1,2,3,4,5,6} § Throw two dice together New space {(1,1),(1,2),…,(6,6)} § Extend first die space, then Event is {(4,1),(4,2),…,(4,6)} P(A) = 6/36 = 1/6 Event Independence § P(A and B) = P(A) P(B) INDEPENDENCE TEST 1 but § P(A and B) = P(A|B) P(B) thus § P(A|B) = P(A) INDEPENDENCE TEST 2 § P(A and B) = P((4,5)) = 1/36 § P(A) = 6/36 [first die shows 4] § P(B) = 6/36 [second die shows 5] Not Independent Events Example § [dice sum to 5] P(A) = 4/36 § [second die shows 5] P(B) = 6/36 § P(A|B) = 0 but § P(A) P(B) = 24/362 = 2 / 108 Probability of Becoming a Professional Athlete § The probability a male high school athlete will go to college is 5%. If they do not go to college, then they have a chance of 0.01% to become professional athletes. Of those who do go to college, 98.3% do not continue their career as professional athletes. § What are the odds for a male high school athlete to make it to professional sports? § Draw the relevant tree diagram § How do you explain the tree diagram to a high school kid contemplating a career in professional sports? Problem 5.15 Gas Well Completions 1986 D D’ Dry Not Dry N North America 14,131 31,575 45,706 N’ South America 404 2,563 2,967 14,535 34,138 48,673 Draw Venn Diagram Identify Event ( N and D ) What is its probability? Probability Tree: Problem 5.15 • Location first D 14,131/45,706 N 14,131/48,673 45,706/48,673 D’ 31,575/45,706 31,575/48,673 D 404/2,967 404/48,673 N’ 2,967/48,673 D’ 2,563/2,967 2,563/48,673 Are Events N and D Independent? What should the number of dry wells in So. America be, for N and D to be independent? Bayes’ Rule (form 1) P( A and B) P( A | B) = P( B) P( B | A) P( A) P( A | B) = P( B | A) P( A) + P( B | A' ) P( A' ) Bayes’ Rule (form 2) P( A and B) P( A | B) = P( B) P( B | Ai ) P( Ai ) P( Ai | B) = ∑ P( B | Ai ) P( Ai ) i where A i disjoint and exhaustive subsets of Ω HS Athlete College Professional B|A 0.05x0.017= 0.00085 0.017 A 0.05 0.983 B’|A 0.05x0.983= 0.04915 B|A’ 0.95x0.00001=0.0000095 0.0001 0.95 A’ 0.9999 B’|A’ 0.95x0.9999=0.949991 P(A|B)=? Work on this in groups § To detect naileria (an imaginary nail disease), doctors apply a test, which, if a patient really suffers from the disease, gives a positive result with probability 99%. However, it may happen that a healthy subject obtains a positive result. This probability of false positives is 2%. It is known that the prevalence (frequency of occurrence in the population) of naileria is 0.1%. What is the probability that a patient who gets a positive on the test really suffers from naileria? Assume the test is administered to the population at large and not only to patients with naileria symptoms. Albinism Example § People with Albinism have little pigment in their skin, hair, and eyes. The gene that governs albinism has two forms (called alleles), denoted as a and A. Each person has a pair of alleles inherited from parents with probability 0.5. Albinism gene is recessive, one has disease only if both alleles are aa. 1. Beth’s parents are not albino but she has an albino brother. This implies both her parents are type aA. Why? 2. Which of the type aa, Aa, AA could a child of Beth’s parents have? What is the probability of each type? 3. Beth is not an albino. What are the conditional probabilities for Beth’s possible genetic types, given this fact? Albinism continued § Beth knows the probabilities for her genetic types from previous analysis. She marries Bob, who is albino (type aa). 1. What is the conditional probability that Beth and Bob’s child is non-albino, if Beth is type Aa? What is the answer to the latter if Beth is type AA? 2. Beth and Bob’s first child is non-albino. What is the conditional probability that Beth is a carrier, type Aa? § Solve using tree diagram Work on this in groups § Factories A and B manufacture watches. Factory A produces defectives on average at rate 1/100. Factory B produces defectives at rate 1/200. A retailer receives a case of watches from one the above factories, without knowing from which one (assume equal probabilities). She checks the first watch and it works. § What is the probability that the second watch in the case works? § Is this independent from the first watch working? Card Magic § A magician shuffles three cards: one with two red faces, one with two white faces, and one with one white and one red face. She randomly selects a card, places it on the table and shows its open face being red. § What is the probability that the covered face is also red? § No, the answer is not 50%