Download Probability Theory: Rules, Examples, and Applications and more Exams Business Statistics in PDF only on Docsity! Probability Fundamentals Stat 201 Prof. Yanni Papadakis Key Terms § Probability Axioms § Rules for Addition § Rules for Multiplication § Venn Diagram § Tree Diagram § Bayes’ Rule Logical MeaningSet NotationVenn Diagram A,B,C are exhaustive and mutually exclusive Both A and B are realized Intersection One And Only One Event of A and B is Realized Union minus Intersection BABA BABA and or − ∩−∪ Bor A BA∩ A B AB ∅=∩ ∅=∩ ∅=∩ CB CA BA C O Ω=++ CBA A B Logical MeaningSet NotationVenn Diagram A given B is realized Given All but A is realizedComplement AA −Ω=' BBABA =Ω∩= but | A’A O B BA∩ 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? 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? Bayes’ Rule (form 1) )'()'|()()|( )()|( )|( )( ) and ( )|( APABPAPABP APABP BAP BP BAP BAP + = = Bayes’ Rule (form 2) Ω = = ∑ of subsets exhaustive anddisjoint A where )()|( )()|( )|( )( ) and ( )|( i i ii ii i APABP APABP BAP BP BAP BAP HS Athlete College Professional A A’ B|A 0.05x0.017= 0.00085 B’|A 0.05x0.983= 0.04915 B|A’ 0.95x0.00001=0.0000095 B’|A’ 0.95x0.9999=0.949991 0.05 0.95 0.017 0.983 0.0001 0.9999 P(A|B)=? 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%
